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Balbharti Maharashtra State Board 11th Maths Book Solutions Pdf Chapter 7 Limits Ex 7.1 Questions and Answers.

Maharashtra State Board 11th Maths Solutions Chapter 7 Limits Ex 7.1

I. Evaluate the following limits:

Question 1.
\(\lim _{z \rightarrow-3}\left[\frac{\sqrt{Z+6}}{Z}\right]\)
Solution:

Question 2.
\(\lim _{y \rightarrow-3}\left[\frac{y^{5}+243}{y^{3}+27}\right]\)
Solution:

Question 3.
\(\lim _{z \rightarrow-5}\left[\frac{\left(\frac{1}{z}+\frac{1}{5}\right)}{z+5}\right]\)
Solution:

II. Evaluate the following limits:

Question 1.
\(\lim _{x \rightarrow 3}\left[\frac{\sqrt{2 x+6}}{x}\right]\)
Solution:

Question 2.
\(\lim _{x \rightarrow 2}\left[\frac{x^{-3}-2^{-3}}{x-2}\right]\)
Solution:

Question 3.
\(\lim _{x \rightarrow 5}\left[\frac{x^{3}-125}{x^{5}-3125}\right]\)
Solution:

Question 4.
If \(\lim _{x \rightarrow 1}\left[\frac{x^{4}-1}{x-1}\right]=\lim _{x \rightarrow a}\left[\frac{x^{3}-a^{3}}{x-a}\right]\), find all possible values of a.
Solution:

III. Evaluate the following limits:

Question 1.
\(\lim _{x \rightarrow 1}\left[\frac{x+x^{2}+x^{3}+\ldots \ldots \ldots+x^{n}-n}{x-1}\right]\)
Solution:

Question 2.
\(\lim _{x \rightarrow 7}\left[\frac{(\sqrt[3]{x}-\sqrt[3]{7})(\sqrt[3]{x}+\sqrt[3]{7})}{x-7}\right]\)
Solution:

Question 3.
If \(\lim _{x \rightarrow 5}\left[\frac{x^{k}-5^{k}}{x-5}\right]\) = 500, find all possible values of k.
Solution:

Question 4.
\(\lim _{x \rightarrow 0}\left[\frac{(1-x)^{8}-1}{(1-x)^{2}-1}\right]\)
Solution:

Question 5.
\(\lim _{x \rightarrow 0}\left[\frac{\sqrt[3]{1+x}-\sqrt{1+x}}{x}\right]\)
Solution:

Question 6.
\(\lim _{y \rightarrow 1}\left[\frac{2 y-2}{\sqrt[3]{7+y}-2}\right]\)
Solution:

Question 7.
\(\lim _{z \rightarrow a}\left[\frac{(z+2)^{\frac{3}{2}}-(a+2)^{\frac{3}{2}}}{z-a}\right]\)
Solution:

Question 8.
\(\lim _{x \rightarrow 7}\left[\frac{x^{3}-343}{\sqrt{x}-\sqrt{7}}\right]\)
Solution:

Question 9.
\(\lim _{x \rightarrow 1}\left(\frac{x+x^{3}+x^{5}+\ldots+x^{2 n-1}-n}{x-1}\right)\)
Solution:

IV. In the following examples, given ∈ > 0, find a δ > 0 such that whenever, |x – a| < δ, we must have |f(x) – l| < ∈.

Question 1.
\(\lim _{x \rightarrow 2}(2 x+3)=7\)
Solution:
We have to find some δ so that
\(\lim _{x \rightarrow 2}(2 x+3)=7\)
Here a = 2, l = 1 and f(x) = 2x + 3
Consider ∈ > 0 and |f(x) – l| < ∈
∴ |(2x + 3) – 7| < ∈
∴ |2x + 4| < ∈
∴ 2(x – 2)|< ∈
∴ |x – 2| < \(\frac{\epsilon}{2}\)
∴ δ ≤ \(\frac{\epsilon}{2}\) such that
|2x + 4| < δ ⇒ |f(x) – 7| < ∈

Question 2.
\(\lim _{x \rightarrow-3}(3 x+2)=-7\)
Solution:
We have to find some δ so that
\(\lim _{x \rightarrow-3}(3 x+2)=-7\)
Here a = -3, l = -7 and f(x) = 3x + 2
Consider ∈ > 0 and |f(x) – l| < ∈
∴ |3x + 2 – (-7)| < ∈
∴ |3x + 9| < ∈
∴ |3(x + 3)| < ∈
∴ |x + 3| < \(\frac{\epsilon}{3}\)
∴ δ < \(\frac{\epsilon}{3}\) such that
|x + 3| ≤ δ ⇒ |f(x) + 7| < ∈

Question 3.
\(\lim _{x \rightarrow 2}\left(x^{2}-1\right)=3\)
Solution:
We have to find some δ > 0 such that
\(\lim _{x \rightarrow 2}\left(x^{2}-1\right)=3\)
Here, a = 2, l = 3 and f(x) = x² – 1
Consider ∈ > 0 and |f(x) – l| < ∈
∴ |(x² – 1) – 3| < ∈
∴ |x² – 4| < ∈
∴ |(x + 2)(x – 2)| < ∈ …..(i)
We have to get rid of the factor |x + 2|
As |x – 2| < δ
-δ < x – 2 < δ
∴ 2 – δ < x < 2 + δ
Since δ can be assumed as very small, let us choose δ < 1
∴ 1 < x < 3
∴ 3 < x + 2 < 5 …..(Adding 2 throughout)
∴ |x + 2| < 5
∴ |(x + 2)(x – 2)| < 5|x – 2| ……(ii)
From (i) and (ii), we get
5|x – 2|< ∈
∴ x – 2 < \(\frac{\epsilon}{5}\)
If δ = \(\frac{\epsilon}{5}\), |x – 2| < δ ⇒ |x² – 4| < ∈
∴ We choose δ = min{\(\frac{\epsilon}{5}\), 1} then
|x – 2| < δ ⇒ |f(x) – 3| < ∈

Question 4.
\(\lim _{x \rightarrow 1}\left(x^{2}+x+1\right)=3\)
Solution:
We have to find some δ > 0 such that
\(\lim _{x \rightarrow 1}\left(x^{2}+x+1\right)=3\)
Here a = 1, l = 3 and f(x) = x² + x + 1
Consider ∈ > 0 and |f(x) – l| < ∈
∴ |x² + x + 1 – 3| < ∈
∴ |x² + x – 2| < ∈
∴ |(x + 2)(x – 1)| < ∈ …..(i)
We have to get rid of the factor |x + 2|
As |x – 1| < δ
-δ < x – 1 < δ
∴ 1 – δ < x < 1 + δ
Since δ can be assumed as very small, let us choose δ < 1
∴ 0 < x < 2
∴ 2 < x + 2 < 4
∴ |x + 2| < 4
∴ |(x + 2)(x – 1)|< 4 |x – 1| …..(ii)
From (i) and (ii), we get
4|x – 1| < ∈
∴ |x – 1| < \(\frac{\epsilon}{4}\)
If δ = \(\frac{\epsilon}{4}\),
|x – 1| < δ ⇒ x² + x – 2 < ∈
∴ We choose δ = min{\(\frac{\epsilon}{4}\), 1} then
|x – 1| < δ ⇒ |f(x) – 3| < ∈

Balbharti Maharashtra State Board 11th Maths Book Solutions Pdf Chapter 6 Functions Miscellaneous Exercise 6 Questions and Answers.

Maharashtra State Board 11th Maths Solutions Chapter 6 Functions Miscellaneous Exercise 6

(I) Select the correct answer from the given alternatives.

Question 1.
If log (5x – 9) – log (x + 3) = log 2, then x = ________
(A) 3
(B) 5
(C) 2
(D) 7
Answer:
(B) 5
Hint:
log (5x – 9) – log (x + 3) = log 2
∴ \(\frac{5 x-9}{x+3}\) = 2
∴ 3x = 9 + 6
∴ x = 5

Question 2.
If log₁₀ (log₁₀ (log₁₀ x)) = 0, then x = ________
(A) 1000
(B) 10¹⁰
(C) 10
(D) 0
Answer:
(B) 10¹⁰
Hint:
log₁₀ log₁₀ log₁₀ x = 0
∴ log₁₀ (log₁₀ (x)) = 10⁰ = 1
∴ log₁₀ x = 10¹ = 10
∴ x = 10¹⁰

Question 3.
Find x, if 2 log₂ x = 4
(A) 4, -4
(B) 4
(C) -4
(D) not defined
Answer:
(B) 4
Hint:
2 log₂ x = 4, x > 0
∴ log₂ (x²) = 4
∴ x² = 16
∴ x = ±4
∴ x = 4

Question 4.
The equation \(\log _{x^{2}} 16+\log _{2 x} 64=3\) has,
(A) one irrational solution
(B) no prime solution
(C) two real solutions
(D) one integral solution
Answer:
(A), (B), (C), (D)
Hint:
\(\log _{x^{2}} 16+\log _{2 x} 64=3\)
∴ \(\frac{\log 16}{\log x^{2}}+\frac{\log 64}{\log 2 x}=3\)
∴ 4 log 2 [log x + log 2] + (6 log 2) (2 log x) = 3 (2 log x) (log 2 + log x)
Let log 2 = a, log x = t. Then
∴ 4at + 4a² + 12at = 6at + 6t²
∴ 6t² – 10at – 4a² = 0
∴ 3t² – 5at – 2a² = 0
∴ (3t + a) (t – 2a) = 0
∴ t = \(-\frac{1}{3}\)a, 2a
∴ log x = \(\log (2)^{-\frac{1}{3}}\), log (2²)
∴ x = \(2^{-\frac{1}{3}}\), 4
∴ x = \(\frac{1}{\sqrt[3]{2}}\), 4

Question 5.
If f(x) = \(\frac{1}{1-x}\), then f(f(f(x))) is
(A) x – 1
(B) 1 – x
(C) x
(D) -x
Answer:
(C) x
Hint:

Question 6.
If f: R → R is defined by f(x) = x³, then f^-1 (8) is equal to:
(A) {2}
(B) {-2.2}
(C) {-2}
(D) (-2.2)
Answer:
(A) {2}
Hint:

Question 7.
Let the function f be defined by f(x) = \(\frac{2 x+1}{1-3 x}\) then f^-1 (x) is:
(A) \(\frac{x-1}{3 x+2}\)
(B) \(\frac{x+1}{3 x-2}\)
(C) \(\frac{2 x+1}{1-3 x}\)
(D) \(\frac{3 x+2}{x-1}\)
Answer:
(A) \(\frac{x-1}{3 x+2}\)
Hint:
f(x) = \(\frac{2 x+1}{1-3 x}\) = y, say. then
2x + 1 = y(1 – 3x)
∴ y – 1 = x(2 + 3y)
∴ x = \(\frac{y-1}{2+3 y}\) = f^-1 (y)
∴ f^-1 (x) = \(\frac{x-1}{2+3 x}\)

Question 8.
If f(x) = 2x² + bx + c and f(0) = 3 and f(2) = 1, then f(1) is equal to
(A) -2
(B) 0
(C) 1
(D) 2
Answer:
(B) 0
Hint:
f(x) = 2x² + bx + c
f(0) = 3
∴ 2(0) + b(0) + c = 3
∴ c = 3 ……..(i)
∴ f(2) = 1
∴ 2(4) + 2b + c = 1
∴ 2b + c = -7
∴ 2b + 3 = -7 …..[From (i)]
∴ b = -5
∴ f(x) = 2x² – 5x + 3
∴ f(1) = 2(1)² – 5(1) + 3 = 0

Question 9.
The domain of \(\frac{1}{[x]-x}\), where [x] is greatest integer function is
(A) R
(B) Z
(C) R – Z
(D) Q – {0}
Answer:
(C) R – Z
Hint:
f(x) = \(\frac{1}{[x]-x}=\frac{1}{-\{x\}}\)
For f to be defined, {x} ≠ 0
∴ x cannot be integer.
∴ Domain = R – Z

Question 10.
The domain and range of f(x) = 2 – |x – 5| are
(A) R+, (-∞, 1]
(B) R, (-∞, 2]
(C) R, (-∞, 2)
(D) R+, (-∞, 2]
Answer:
(B) R, (-∞, 2]
Hint:

f(x) = 2 – |x – 5|
= 2 – (5 – x), x < 5
= 2 – (x – 5), x ≥ 5
∴ f(x) = x – 3, x < 5
= 7 – x, x ≥ 5
Domain = R,
Range (from graph) = (-∞, 2]

(II) Answer the following:

Question 1.
Which of the following relations are functions? If it is a function determine its domain and range.
(i) {(2, 1), (4, 2), (6, 3), (8, 4), (10, 5) (12, 6), (14, 7)}
(ii) {(0, 0), (1, 1), (1, -1), (4, 2), (4, -2) (9, 3), (9, -3), (16, 4), (16, -4)}
(iii) {(2, 1), (3, 1), (5, 2)}
Solution:
(i) {(2, 1), (4, 2), (6, 3), (8, 4), (10, 5) (12, 6), (14, 7)}

Every element of set A has been assigned a unique element in set B
∴ Given relation is a function
Domain = {2, 4, 6, 8, 10, 12, 14}, Range = {1, 2, 3, 4, 5, 6, 7}

(ii) {(0, 0), (1, 1), (1, -1), (4, 2), (4, -2) (9, 3), (9, -3) (16, 4), (16, -4)}
∵ (1, 1), (1, -1) ∈ the relation
∴ Given relation is not a function.
As element 1 of the domain has not been assigned a unique element of co-domain.

(iii) {(2, 1), (3, 1), (5, 2)}

Every element of set A has been assigned a unique element in set B.
∴ Given relation is a function.
Domain = {2, 3, 5}, Range = {1, 2}

Question 2.
Find whether the following functions are one-one.
(i) f: R → R defined by f(x) = x² + 5
(ii) f: R – {3} → R defined by f(x) = \(\frac{5 x+7}{x-3}\) for x ∈ R – {3}
Solution:
(i) f: R → R, defined by f(x) = x² + 5
Note that f(-x) = f(x) = x² + 5
∴ f is not one-one (i.e., many-one) function.

(ii) f: R – {3} → R, defined by f(x) = \(\frac{5 x+7}{x-3}\)
Let f(x₁) = f(x₂)
∴ \(\frac{5 x_{1}+7}{x_{1}-3}=\frac{5 x_{2}+7}{x_{2}-3}\)
∴ 5x₁ x₂ – 15x₁ + 7x₂ – 21 = 5x₁ x₂ – 15x₂ + 7x₁ – 21
∴ 22(x₁ – x₂) = 0
∴ x₁ = x₂
∴ f is a one-one function.

Question 3.
Find whether the following functions are onto or not.
(i) f: Z → Z defined by f(x) = 6x – 7 for all x ∈ Z
(ii) f: R → R defined by f(x) = x² + 3 for all x ∈ R
Solution:
(i) f(x) = 6x – 7 = y (say)
(x, y ∈ Z)
∴ x = \(\frac{7+y}{6}\)
Since every integer y does not give integer x, f is not onto.

(ii) f(x) = x² + 3 = y (say)
(x, y ∈ R)
Clearly y ≥ 3 …..[x² ≥ 0]
∴ All the real numbers less than 3 from codomain R, have not been pre-assigned any element from the domain R.
∴ f is not onto.

Question 4.
Let f: R → R be a function defined by f(x) = 5x³ – 8 for all x ∈ R. Show that f is one-one and onto. Hence find f^-1.
Solution:

Question 5.
A function f: R → R defined by f(x) = \(\frac{3 x}{5}\) + 2, x ∈ R. Show that f is one-one and onto. Hence find f^-1.
Solution:

Question 6.
A function f is defined as f(x) = 4x + 5, for -4 ≤ x < 0. Find the values of f(-1), f(-2), f(0), if they exist.
Solution:
f(x) = 4x + 5, -4 ≤ x < 0
f(-1) = 4(-1) + 5 = -4 + 5 = 1
f(-2) = 4(-2) + 5 = -8 + 5 = -3
x = 0 ∉ domain of f
∴ f(0) does not exist.

Question 7.
A function f is defined as f(x) = 5 – x for 0 ≤ x ≤ 4. Find the values of x such that
(i) f(x) = 3
(ii) f(x) = 5
Solution:
(i) f(x) = 3
∴ 5 – x = 3
∴ x = 5 – 3 = 2

(ii) f(x) = 5
∴ 5 – x = 5
∴ x = 0

Question 8.
If f(x) = 3x⁴ – 5x² + 7, find f(x – 1).
Solution:
f(x) = 3x⁴ – 5x² + 7
∴ f(x – 1) = 3(x – 1)⁴ – 5(x – 1)² + 7
= 3(x⁴ – ⁴C₁ x³ + ⁴C₂ x² – ⁴C₃ x + ⁴C₄) – 5(x² – 2x + 1) + 7
= 3(x⁴ – 4x³ + 6x² – 4x + 1) – 5(x² – 2x + 1) + 7
= 3x⁴ – 12x³ + 18x² – 12x + 3 – 5x² + 10x – 5 + 7
= 3x⁴ – 12x³ + 13x² – 2x + 5

Question 9.
If f(x) = 3x + a and f(1) = 7, find a and f(4).
Solution:
f(x) = 3x + a, f(1) = 7
∴ 3(1) + a = 7
∴ a = 7 – 3 = 4
∴ f(x) = 3x + 4
∴ f(4) = 3(4) + 4 = 12 + 4 = 16

Question 10.
If f(x) = ax² + bx + 2 and f(1) = 3, f(4) = 42, find a and b.
Solution:
f(x) = ax² + bx + 2
f(1) = 3
∴ a(1)² + b(1) + 2 = 3
∴ a + b = 1 ….(i)
f(4) = 42
∴ a(4)² + b(4) + 2 = 42
∴ 16a + 4b = 40
Dividing by 4, we get
4a + b = 10 …..(ii)
Solving (i) and (ii), we get
a = 3, b = -2

Question 11.
Find composite of f and g:
(i) f = {(1, 3), (2, 4), (3, 5), (4, 6)}
g = {(3, 6), (4, 8), (5, 10), (6, 12)}
(ii) f = {(1, 1), (2, 4), (3, 4), (4, 3)}
g = {(1, 1), (3, 27), (4, 64)}
Solution:
(i) f = {(1, 3), (2, 4), (3, 5), (4, 6)}
g = {(3, 6), (4, 8), (5, 10), (6, 12)}
∴ f(1) = 3, g(3) = 6
f(2) = 4, g(4) = 8
f(3) = 5, g(5)=10
f(4) = 6, g(6) = 12
(gof) (x) = g (f(x))
(gof)(1) = g(f(1)) = g(3) = 6
(gof)(2) – g(f(2)) = g(4) = 8
(gof)(3) = g(f(3)) = g(5) = 10
(gof)(4) = g(f(4)) = g(6) = 12
∴ gof = {(1, 6), (2, 8), (3, 10), (4, 12)}

(ii) f = {(1, 1), (2, 4), (3, 4), (4, 3)}
g = {(1, 1), (3, 27), (4, 64)}
f(1) = 1, g(1) = 1
f(2) = 4, g(3) = 27
f(3) = 4, g(4) = 64
f(4) = 3
(gof) (x) = g(f(x))
(gof) (1) = g(f(1)) = g(1) = 1
(gof) (2) = g(f(2)) = g(4) = 64
(gof) (3) = g(f(3)) = g(4) = 64
(gof) (4) = g(f(4)) = g(3) = 27
∴ gof = {(1, 1), (2, 64), (3, 64), (4, 27)}

Question 12.
Find fog and gof:
(i) f(x) = x² + 5, g(x) = x – 8
(ii) f(x) = 3x – 2, g(x) = x²
(iii) f(x) = 256x⁴, g(x) = √x
Solution:
(i) f(x) = x² + 5, g(x) = x – 8
(fog) (x) = f(g(x))
= f(x – 8)
= (x – 8)² + 5
= x² – 16x + 64 + 5
= x² – 16x + 69
(gof) (x) = g(f(x))
= g(x² + 5)
= x² + 5 – 8
= x – 3

(ii) f(x) = 3x – 2, g(x) = x²
(fog) (x) = f(g(x)) = f(x²) = 3x² – 2
(gof) (x) = g(f(x))
= g(3x – 2)
= (3x – 2)²
= 9x² – 12x + 4

(iii) f(x) = 256x⁴, g(x) = √x
(fog) (x) = f(g(x)) = f (√x) = 256 (√x)⁴ = 256x²
(gof) (x) = g(f(x)) = g(256x⁴) = \(\sqrt{256 x^{4}}\) = 16x²

Question 13.
If f(x) = \(\frac{2 x-1}{5 x-2}, x \neq \frac{2}{5}\), show that (fof) (x) = x.
Solution:

Question 14.
If f(x) = \(\frac{x+3}{4 x-5}\), g(x) = \(\frac{3+5 x}{4 x-1}\), then show that (fog) (x) = x.
Solution:
f(x) = \(\frac{x+3}{4 x-5}\), g(x) = \(\frac{3+5 x}{4 x-1}\)
(fog)(x) = f(g(x))
= f(\(\frac{3+5 x}{4 x-1}\))

Question 15.
Let f: R – {2} → R be defined by f(x) = \(\frac{x^{2}-4}{x-2}\) and g: R → R be defined by g(x) = x + 2. Examine whether f = g or not.
Solution:
f(x) = \(\frac{x^{2}-4}{x-2}\), x ≠ 2
∴ f(x) = x + 2, x ≠ 2 and g(x) = x + 2,
The domain of f = R – {2}
The domain of g = R
Here, f and g have different domains.
∴ f ≠ g

Question 16.
Let f: R → R be given by f(x) = x + 5 for all x ∈ R. Draw its graph.
Solution:
f(x) = x + 5

Question 17.
Let f: R → R be given by f(x) = x³ + 1 for all x ∈ R. Draw its graph.
Solution:
Let y = f(x) = x³ + 1

Question 18.
For any base show that log(1 + 2 + 3) = log 1 + log 2 + log 3
Solution:
L.H.S. = log(1 + 2 + 3) = log 6
R.H.S. = log 1 + log 2 + log 3
= 0 + log (2 × 3)
= log 6
∴ L.H.S. = R.H.S.

Question 19.
Find x, if x = \(3^{3 \log _{3} 2}\).
Solution:
x = \(3^{3 \log _{3} 2}\)
= \(3^{\log 3\left(2^{3}\right)}\)
= 2³ ….[\(a^{\log _{a} b}\) = b]
= 8

Question 20.
Show that, log|\(\sqrt{x^{2}+1}\) + x| + log|\(\sqrt{x^{2}+1}\) – x| = 0.
Solution:
L.H.S. = log|\(\sqrt{x^{2}+1}\) + x| + log|\(\sqrt{x^{2}+1}\) – x|
= \(\log \left|\left(\sqrt{x^{2}+1}+x\right)\left(\sqrt{x^{2}+1}-x\right)\right|\)
= log|x² + 1 – x²|
= log 1
= 0
= R.H.S.

Question 21.
Show that \(\log \frac{\mathrm{a}^{2}}{\mathrm{bc}}+\log \frac{\mathrm{b}^{2}}{\mathrm{ca}}+\log \frac{\mathrm{c}^{2}}{\mathrm{ab}}=0\).
Solution:

Question 22.
Simplify log (log x⁴) – log(log x).
Solution:
log (log x⁴) – log (log x)
= log (4 log x) – log (log x) …..[log mⁿ = n log m]
= log 4 + log (log x) – log (log x) …..[log (mn) = log m + log n]
= log 4

Question 23.
Simplify \(\log _{10} \frac{28}{45}-\log _{10} \frac{35}{324}+\log _{10} \frac{325}{432}-\log _{10} \frac{13}{15}\)
Solution:

Question 24.
If log (\(\frac{a+b}{2}\)) = \(\frac{1}{2}\) (log a + log b), then show that a = b.
Solution:
log (\(\frac{a+b}{2}\)) = \(\frac{1}{2}\) (log a + log b)
∴ 2 log (\(\frac{a+b}{2}\)) = log a + log b
∴ log \(\left(\frac{a+b}{2}\right)^{2}\) = log ab
∴ \(\frac{(a+b)^{2}}{4}\) = ab
∴ a² + 2ab + b² = 4ab
∴ a² + 2ab – 4ab + b² = 0
∴ a² – 2ab + b² = 0
∴ (a – b)² = 0
∴ a – b = 0
∴ a = b

Question 25.
If b² = ac. Prove that, log a + log c = 2 log b.
Solution:
b² = ac
Taking log on both sides, we get
log b² = log ac
∴ 2 log b = log a + log c
∴ log a + log c = 2 log b

Question 26.
Solve for x, log_x (8x – 3) – log_x 4 = 2.
Solution:
log_x (8x – 3) – log_x 4 = 2
∴ \(\log _{x}\left(\frac{8 x-3}{4}\right)\) = 2
∴ x² = \(\frac{8 x-3}{4}\)
∴ 4x² = 8x – 3
∴ 4x² – 8x + 3 = 0
∴ 4x² – 2x – 6x + 3 = 0
∴ 2x(2x – 1) – 3(2x – 1) = 0
∴ (2x – 1)(2x – 3) = 0
∴ 2x – 1 = 0 or 2x – 3 = 0
∴ x = \(\frac{1}{2}\) or x = \(\frac{3}{2}\)

Question 27.
If a² + b² = 7ab, show that \(\log \left(\frac{a+b}{3}\right)=\frac{1}{2} \log a+\frac{1}{2} \log b\)
Solution:
a² + b² = 7ab
a² + 2ab + b² = 7ab + 2ab
(a + b)² = 9ab
\(\frac{(a+b)^{2}}{9}\) = ab
\(\left(\frac{a+b}{3}\right)^{2}\) = ab
Taking log on both sides, we get
log \(\left(\frac{a+b}{3}\right)^{2}\) = log (ab)
2 log \(\left(\frac{a+b}{3}\right)\) = log a + log b
Dividing throughout by 2, we get
\(\log \left(\frac{a+b}{3}\right)=\frac{1}{2} \log a+\frac{1}{2} \log b\)

Question 28.
If \(\log \left(\frac{x-y}{5}\right)=\frac{1}{2} \log x+\frac{1}{2} \log y\), show that x² + y² = 27xy.
Solution:

Question 29.
If log₃ [log₂ (log₃ x)] = 1, show that x = 6561.
Solution:
log₃ [log₂ (log₃ x)] = 1
∴ log₂ (log₃ x) = 3¹
∴ log₃ x = 2³
∴ log₃ x = 8
∴ x = 3⁸
∴ x = 6561

Question 30.
If f(x) = log(1 – x), 0 ≤ x < 1, show that f(\(\frac{1}{1+x}\)) = f(1 – x) – f(-x).
Solution:

Question 31.
Without using log tables, prove that \(\frac{2}{5}\) < log₁₀ 3 < \(\frac{1}{2}\).
Solution:
We have to prove that, \(\frac{2}{5}\) < log₁₀ 3 < \(\frac{1}{2}\)
i.e., to prove that \(\frac{2}{5}\) < log₁₀ 3 and log₁₀ 3 < \(\frac{1}{2}\)
i.e., to prove that 2 < 5 log₁₀ 3 and 2 log₁₀ 3 < 1
i.e., to prove that 2 log₁₀ 10 < 5 log₁₀ 3 and 2 log₁₀ 3 < log₁₀ 10 ……[∵ log_a a = 1]
i.e., to prove that log₁₀ 10² < log₁₀ 3⁵ and log₁₀ 3² < log₁₀ 10
i.e., to prove that 10² < 3⁵ and 3² < 10
i.e., to prove that 100 < 243 and 9 < 10 which is true
∴ \(\frac{2}{5}\) < log₁₀ 3 < \(\frac{1}{2}\)

Question 32.
Show that \(7 \log \left(\frac{15}{16}\right)+6 \log \left(\frac{8}{3}\right)+5 \log \left(\frac{2}{5}\right)+\log \left(\frac{32}{25}\right)\) = log 3
Solution:

Question 33.
Solve : \(\sqrt{\log _{2} x^{4}}+4 \log _{4} \sqrt{\frac{2}{x}}=2\)
Solution:

Question 34.
Find the value of \(\frac{3+\log _{10} 343}{2+\frac{1}{2} \log _{10}\left(\frac{49}{4}\right)+\frac{1}{2} \log _{10}\left(\frac{1}{25}\right)}\)
Solution:

Question 35.
If \(\frac{\log a}{x+y-2 z}=\frac{\log b}{y+z-2 x}=\frac{\log c}{z+x-2 y}\), show that abc = 1.
Solution:
Let \(\frac{\log a}{x+y-2 z}=\frac{\log b}{y+z-2 x}=\frac{\log c}{z+x-2 y}\) = k
∴ log a = k(x + y – 2z), log b = k(y + z – 2x), log c = k(z + x – 2y)
log a + log b + log c = k(x + y – 2z) + k(y + z – 2x) + k(z + x – 2y)
= k(x + y – 2z + y + z – 2x + z + x – 2y)
= k(0)
= 0
∴ log (abc) = log 1 …….[∵ log 1 = 0]
∴ abc = 1

Question 36.
Show that, log_y x³ . log_z y⁴ . log_x z⁵ = 60.
Solution:

Question 37.
If \(\frac{\log _{2} \mathrm{a}}{4}=\frac{\log _{2} \mathrm{~b}}{6}=\frac{\log _{2} \mathrm{c}}{3 \mathrm{k}}\) and a³b²c = 1, find the value of k.
Solution:

Question 38.
If a² = b³ = c⁴ = d⁵, show that loga bcd = \(\frac{47}{30}\).
Solution:

Question 39.
Solve the following for x, where |x| is modulus function, [x] is the greatest integer function, {x} is a fractional part function.
(i) 1 < |x – 1| < 4
(ii) |x² – x – 6| = x + 2
(iii) |x² – 9| + |x² – 4| = 5
(iv) -2 < [x] ≤ 7
(v) 2[2x – 5] – 1 = 7
(vi) [x]2 – 5 [x] + 6 = 0
(vii) [x – 2] + [x + 2] + {x} = 0
(viii) \(\left[\frac{x}{2}\right]+\left[\frac{x}{3}\right]=\frac{5 x}{6}\)
Solution:
(i) 1 < |x – 1| < 4
∴ -4 < x – 1 < -1 or 1 < x – 1 < 4
∴ -3 < x < 0 or 2 < x < 5
∴ Solution set = (-3, 0) ∪ (2, 5)

(ii) |x² – x – 6| = x + 2 …..(i)
R.H.S. must be non-negative
∴ x ≥ -2 …..(ii)
|(x – 3) (x + 2)| = x + 2
∴ (x + 2) |x – 3| = x + 2 as x + 2 ≥ 0
∴ |x – 3| = 1 if x ≠ -2
∴ x – 3 = ±1
∴ x = 4 or 2
∴ x = -2 also satisfies the equation
∴ Solution set = {-2, 2, 4}

(iii) |x² – 9| + |x² – 4| = 5
∴ |(x – 3) (x + 3)| + |(x – 2) ( x + 2)| = 5 ………(i)
Case I: x < -3
Also, x < -2, x < 2, x < 3
∴ (x – 3) (x + 3) > 0 and (x – 2) (x + 2) > 0
Equation (i) reduces to
x² – 9 + x² – 4 = 5
∴ 2x² = 18
∴ x = -3 or 3 (both rejected as x < -3)

Case II: -3 ≤ x < -2
As x < -2, x < 3
∴ (x – 3) (x + 3) < 0, (x – 2) (x + 2) > 0
Equation (i) reduces to
-(x² – 9) + x² – 4 = 5
∴ 5 = 5 (true)
-3 ≤ x < -2 is a solution ….(ii)

Case III: -2 ≤ x < 2 As x > -3, x < 3
∴ (x – 3) (x + 3) < 0,
(x – 2) (x + 2) < 0
Equation (i) reduces to
9 – x² + 4 – x2 = 5
∴ 2x² = 13 – 5
∴ x² = 4
∴ x = -2 is a solution …..(iii)

Case IV: 2 ≤ x < 3 As x > -3, x > -2
∴ (x – 3) (x + 3) < 0, (x – 2) (x + 2) > 0
Equation (i) reduces to
9 – x² + x² – 4 = 5
∴ 5 = 5 (true)
∴ 2 ≤ x < 3 is a solution ……(iv)

Case V: 3 ≤ x As x > -3, x > -2, x > 2
∴ (x + 3) (x – 3) > 0,
(x – 2) (x + 2) > 0
Equation (i) reduces to
x² – 9 + x² – 4 = 5
∴ 2x² = 18
∴ x² = 9
∴ x = 3 …..(v)
(x = -3 rejected as x ≥ 3)
From (ii), (iii), (iv), (v), we get
∴ Solution set = [-3, -2] ∪ [2, 3]

(iv) -2 < [x] ≤ 7
∴ -2 < x < 8
∴ Solution set = (-2, 8)

(v) 2[2x – 5] – 1 = 7
∴ [2x – 5] = \(\frac{7+1}{2}\) = 4
∴ [2x] – 5 = 4
∴ [2x] = 9
∴ 9 ≤ 2x < 10
∴ \(\frac{9}{2}\) ≤ x < 5
∴ Solution set = [\(\frac{9}{2}\), 5)

(vi) [x]² – 5[x] + 6 = 0
∴ ([x] – 3)([x] – 2) = 0
∴ [x] = 3 or 2
If [x] = 2, then 2 ≤ x < 3
If [x] = 3, then 3 ≤ x < 4
∴ Solution set = [2, 4)

(vii) [x – 2] + [x + 2] + {x} = 0
∴ [x] – 2 + [x] + 2 + {x} = 0
∴ [x] + x = 0 …..[{x} + [x] = x]
∴ x = 0

(viii) \(\left[\frac{x}{2}\right]+\left[\frac{x}{3}\right]=\frac{5 x}{6}\)
L.H.S. = an integer
R.H.S. = an integer
∴ x = 6k, where k is an integer

Question 40.
Find the domain of the following functions.
(i) f(x) = \(\frac{x^{2}+4 x+4}{x^{2}+x-6}\)
(ii) f(x) = \(\sqrt{x-3}+\frac{1}{\log (5-x)}\)
(iii) f(x) = \(\sqrt{1-\sqrt{1-\sqrt{1-x^{2}}}}\)
(iv) f(x) = x!
(v) f(x) = \({ }^{5-x} P_{x-1}\)
(vi) f(x) = \(\sqrt{x-x^{2}}+\sqrt{5-x}\)
(vii) f(x) = \(\sqrt{\log \left(x^{2}-6 x+6\right)}\)
Solution:
(i) f(x) = \(\frac{x^{2}+4 x+4}{x^{2}+x-6}=\frac{x^{2}+4 x+4}{(x+3)(x-2)}\)
For f to be defined, x ≠ -3, 2
∴ Domain of f = (-∞, -3) ∪ (-3, 2) ∪ (2, ∞)

(ii) f(x) = \(\sqrt{x-3}+\frac{1}{\log (5-x)}\)
For f to be defined,
x – 3 ≥ 0, 5 – x > 0 and 5 – x ≠ 1
x ≥ 3, x < 5 and x ≠ 4
∴ Domain of f = [3, 4) ∪ (4, 5)

(iii) f(x) = \(\sqrt{1-\sqrt{1-\sqrt{1-x^{2}}}}\)

Equation (i) gives solution set = [-1, 1]
∴ Domain of f = [-1, 1]

(iv) f(x) = x!
∴ Domain of f = set of whole numbers (W)

(v) f(x) = \({ }^{5-x} P_{x-1}\)
5 – x > 0, x – 1 ≥ 0, x – 1 ≤ 5 – x
∴ x < 5, x ≥ 1 and 2x ≤ 6
∴ x ≤ 3
∴ Domain of f = {1, 2, 3}

(vi) f(x) = \(\sqrt{x-x^{2}}+\sqrt{5-x}\)
x – x² ≥ 0
∴ x² – x ≤ 0
∴ x(x – 1) ≤ 0
∴ 0 ≤ x ≤ 1 …..(i)
5 – x ≥ 0
∴ x ≤ 5 …..(ii)
Intersection of intervals given in (i) and (ii) gives
Solution set = [0, 1]
∴ Domain of f = [0, 1]

(vii) f(x) = \(\sqrt{\log \left(x^{2}-6 x+6\right)}\)
For f to be defined,
log (x² – 6x + 6) ≥ 0
∴ x² – 6x + 6 ≥ 1
∴ x² – 6x + 5 ≥ 0
∴ (x – 5)(x – 1) ≥ 0
∴ x ≤ 1 or x ≥ 5 …..(i)
[∵ The solution of (x – a) (x – b) ≥ 0 is x ≤ a or x ≥ b, for a < b]
and x² – 6x + 6 > 0
∴ (x – 3)² > -6 + 9
∴ (x – 3)² > 3
∴ x < 3 – √3 0r x > 3 + √3 ……..(ii)
From (i) and (ii), we get
x ≤ 1 or x ≥ 5
Solution set = (-∞, 1] ∪ [5, ∞)
∴ Domain of f = (-∞, 1] ∪ [5, ∞)

Question 41.
(i) f(x) = |x – 5|
(ii) f(x) = \(\frac{x}{9+x^{2}}\)
(iii) f(x) = \(\frac{1}{1+\sqrt{x}}\)
(iv) f(x) = [x] – x
(v) f(x) = 1 + 2x + 4x
Solution:
(i) f(x) = |x – 5|

∴ Range of f = [0, ∞)

(ii) f(x) = \(\frac{x}{9+x^{2}}\) = y (say)
∴ x²y – x + 9y = 0
For real x, Discriminant > 0
∴ 1 – 4(y)(9y) ≥ 0
∴ y² ≤ \(\frac{1}{36}\)
∴ \(\frac{-1}{6}\) ≤ y ≤ \(\frac{1}{6}\)
∴ Range of f = [\(\frac{-1}{6}\), \(\frac{1}{6}\)]

(iii) f(x) = \(\frac{1}{1+\sqrt{x}}\) = y, (say)
∴ √x y + y = 1
∴ √x = \(\frac{1-y}{y}\) ≥ 0
∴ \(\frac{y-1}{y}\) ≤ 0
∴ o < y ≤ 1
∴ Range of f = (0, 1]

(iv) f(x) = [x] – x = -{x}
∴ Range of f = (-1, 0] …..[0 ≤ {x} < 1]

(v) f(x) = 1 + 2x + 4x
Since, 2x > 0, 4x > 0
∴ f(x) > 1
∴ Range of f = (1, ∞)

Question 42.
Find (fog) (x) and (gof) (x)
(i) f(x) = e^x, g(x) = log x
(ii) f(x) = \(\frac{x}{x+1}\), g(x) = \(\frac{x}{1-x}\)
Solution:
(i) f(x) = e^x, g(x) = log x
(fog) (x) = f(g(x))
= f(log x)
= e^{log x}
= x
(gof) (x) = g(f(x))
= g(e^x)
= log (e^x)
= x log e
= x …..[∵ log e = 1]

(ii) f(x) = \(\frac{x}{x+1}\), g(x) = \(\frac{x}{1-x}\)

Question 43.
Find f(x), if
(i) g(x) = x² + x – 2 and (gof) (x) = 4x² – 10x + 4
(ii) g(x) = 1 + √x and f [g(x)] = 3 + 2√x + x.
Solution:
(i) g(x) = x² + x – 2
(gof) (x) = 4x² – 10x + 4
= (2x – 3)² + (2x – 3) – 2
= g(2x – 3)
= g(f(x))
∴ f(x) = 2x – 3
(gof) (x) = 4x² – 10x + 4
= (-2x + 2)² + (-2x + 2) – 2
= g(-2x + 2)
= g(f(x))
∴ f(x) = -2x + 2

(ii) g(x) = 1 + √x
f(g(x)) = 3 + 2√x + x
= x + 2√x + 1 + 2
= (√x + 1)² + 2
f(√x + 1) = (√x + 1)² + 2
∴ f(x) = x² + 2

Question 44.
Find (fof) (x) if
(i) f(x) = \(\frac{x}{\sqrt{1+x^{2}}}\)
(ii) f(x) = \(\frac{2 x+1}{3 x-2}\)
Solution:

Balbharti Maharashtra State Board 11th Maths Book Solutions Pdf Chapter 6 Functions Ex 6.2 Questions and Answers.

Maharashtra State Board 11th Maths Solutions Chapter 6 Functions Ex 6.2

Question 1.
If f(x) = 3x + 5, g(x) = 6x – 1, then find
(i) (f + g) (x)
(ii) (f – g) (2)
(iii) (fg) (3)
(iv) (f/g) (x) and its domain
Solution:
f(x) = 3x + 5, g (x) = 6x – 1
(i) (f + g) (x) = f (x) + g (x)
= 3x + 5 + 6x – 1
= 9x + 4

(ii) (f – g) (2) = f(2) – g(2)
= [3(2) + 5] – [6(2) – 1]
= 6 + 5 – 12 + 1
= 0

(iii) (fg) (3) = f (3) g(3)
= [3(3) + 5] [6(3) – 1]
= (14) (17)
= 238

(iv) \(\left(\frac{\mathrm{f}}{\mathrm{g}}\right)(x)=\frac{\mathrm{f}(x)}{\mathrm{g}(x)}=\frac{3 x+5}{6 x-1}, x \neq \frac{1}{6}\)
Domain = R – {\(\frac{1}{6}\)}

Question 2.
Let f: (2, 4, 5} → {2, 3, 6} and g: {2, 3, 6} → {2, 4} be given by f = {(2, 3), (4, 6), (5, 2)} and g = {(2, 4), (3, 4), (6, 2)}. Write down gof.
Solution:
f = {(2, 3), (4, 6), (5, 2)}
∴ f(2) = 3, f(4) = 6, f(5) = 2
g ={(2, 4), (3, 4), (6, 2)}
∴ g(2) = 4, g(3) = 4, g(6) = 2
gof: {2, 4, 5} → {2, 4}
(gof) (2) = g(f(2)) = g(3) = 4
(gof) (4) = g(f(4)) = g(6) = 2
(gof) (5) = g(f(5)) = g(2) = 4
∴ gof = {(2, 4), (4, 2), (5, 4)}

Question 3.
If f(x) = 2x² + 3, g(x) = 5x – 2, then find
(i) fog
(ii) gof
(iii) fof
(iv) gog
Solution:
f(x) = 2x² + 3, g(x) = 5x – 2
(i) (fog) (x) = f(g(x))
= f(5x – 2)
= 2(5x – 2)² + 3
= 2(25x² – 20x + 4) + 3
= 50x² – 40x + 11

(ii) (gof) (x) = g(f(x))
= g(2x² + 3)
= 5(2x + 3) – 2
= 10x² + 13

(iii) (fof) (x) = f(f(x))
= f(2x² + 3)
= 2(2x² + 3)² + 3
= 2(4x⁴ + 12x² + 9) + 3
= 8x⁴ + 24x² + 21

(iv) (gog) (x) = g(g(x))
= g(5x – 2)
= 5(5x – 2) – 2
= 25x – 12

Question 4.
Verify that f and g are inverse functions of each other, where
(i) f(x) = \(\frac{x-7}{4}\), g(x) = 4x + 7
(ii) f(x) = x³ + 4, g(x) = \(\sqrt[3]{x-4}\)
(iii) f(x) = \(\frac{x+3}{x-2}\), g(x) = \(\frac{2 x+3}{x-1}\)
Solution:
(i) f(x) = \(\frac{x-7}{4}\)
Replacing x by g(x), we get
f[g(x)] = \(\frac{g(x)-7}{4}=\frac{4 x+7-7}{4}\) = x
g(x) = 4x + 7
Replacing x by f(x), we get
g[f(x)] = 4f(x) + 7 = 4(\(\frac{x-7}{4}\)) + 7 = x
Here, f[g(x)] = x and g[f(x)] = x.
∴ f and g are inverse functions of each other.

(ii) f(x) = x³ + 4
Replacing x by g(x), we get
f[g(x)] = [g(x)]³ + 4
= \((\sqrt[3]{x-4})^{3}+4\)
= x – 4 + 4
= x
g(x) = \(\sqrt[3]{x-4}\)
Replacing x by f(x), we get
g[f(x)] = \(\sqrt[3]{f(x)-4}=\sqrt[3]{x^{3}+4-4}=\sqrt[3]{x^{3}}\) = x
Here, f[g(x)] = x and g[f(x)] = x
∴ f and g are inverse functions of each other.

(iii) f(x) = \(\frac{x+3}{x-2}\)
Replacing x by g(x), we get

Here, f[g(x)] = x and g[f(x)] = x.
∴ f and g are inverse functions of each other.

Question 5.
Check if the following functions have an inverse function. If yes, find the inverse function.
(i) f(x) = 5x²
(ii) f(x) = 8
(iii) f(x) = \(\frac{6 x-7}{3}\)
(iv) f(x) = \(\sqrt{4 x+5}\)
(v) f(x) = 9x³ + 8
(vi) f(x) =
Solution:
(i) f(x) = 5x² = y (say)

For two values (x₁ and x₂) of x, values of the function are equal.
∴ f is not one-one.
∴ f does not have an inverse.

(ii) f(x) = 8 = y (say)
For every value of x, the value of the function f is the same.
∴ f is not one-one i.e. (many-one) function.
∴ f does not have the inverse.

(iii) f(x) = \(\frac{6 x-7}{3}\)
Let f(x₁) = f(x₂)
∴ \(\frac{6 x_{1}-7}{3}=\frac{6 x_{2}-7}{3}\)
∴ x₁ = x₂
∴ f is a one-one function.
f(x) = \(\frac{6 x-7}{3}\) = y (say)
∴ x = \(\frac{3y+7}{6}\)
∴ For every y, we can get x
∴ f is an onto function.
∴ x = \(\frac{3y+7}{6}\) = f^-1 (y)
Replacing y by x, we get
f-1 (x) = \(\frac{3x+7}{6}\)

(iv) f(x) = \(\sqrt{4 x+5}, x \geq \frac{-5}{4}\)
Let f(x₁) = f(x₂)
∴ \(\sqrt{4 x_{1}+5}=\sqrt{4 x_{2}+5}\)
∴ x₁ = x₂
∴ f is a one-one function.
f(x) = \(\sqrt{4 x+5}\) = y, (say) y ≥ 0
Squaring on both sides, we get
y² = 4x + 5
∴ x = \(\frac{y^{2}-5}{4}\)
∴ For every y we can get x.
∴ f is an onto function.
∴ x = \(\frac{y^{2}-5}{4}\) = f^-1 (y)
Replacing y by x, we get
f^-1 (x) = \(\frac{x^{2}-5}{4}\)

(v) f(x) 9x³ + 8
Let f(x₁) = f(x₂)
∴ \(9 x_{1}^{3}+8=9 x_{2}^{3}+8\)
∴ x₁ = x₂
∴ f is a one-one function.
∴ f(x) = 9x³ + 8 = y, (say)
∴ x = \(\sqrt[3]{\frac{y-8}{9}}\)
∴ For every y we can get x.
∴ f is an onto function.
∴ x = \(\sqrt[3]{\frac{y-8}{9}}\) = f^-1 (y)
Replacing y by x, we get
f^-1 (x) = \(\sqrt[3]{\frac{x-8}{9}}\)

(vi) f(x) = x + 7, x < 0
= 8 – x, x ≥ 0

We observe from the graph that for two values of x, say x₁, x₂ the values of the function are equal.
i.e. f(x₁) = f(x₂)
∴ f is not one-one (i.e. many-one) function.
∴ f does not have inverse.

Question 6.
If f(x) = , then find
(i) f(3)
(ii) f(2)
(iii) f(0)
Solution:
f(x) = x² + 3, x ≤ 2
= 5x + 7, x > 2
(i) f(3) = 5(3) + 7
= 15 + 7
= 22

(ii) f(2) = 2² + 3
= 4 + 3
= 7

(iii) f(0) = 0² + 3 = 3

Question 7.
If f(x) = , then find
(i) f(-4)
(ii) f(-3)
(iii) f(1)
(iv) f(5)
Solution:
f(x) = 4x – 2, x ≤ -3
= 5, -3 < x < 3
= x², x ≥ 3
(i) f(-4) = 4(-4) – 2
= -16 – 2
= -18

(ii) f(-3) = 4(-3) – 2
= -12 – 2
= -14

(iii) f(1) = 5

(iv) f(5) = 5² = 25

Question 8.
If f(x) = 2 |x| + 3x, then find
(i) f(2)
(ii) f(-5)
Solution:
f(x) = 2 |x| + 3x
(i) f(2) = 2|2| + 3(2)
= 2 (2) + 6 ….. [∵ |x| = x, x > 0]
= 10

(ii) f(-5) = 2 |-5| + 3(-5)
= 2(5) – 15 …..[∵ |x| = -x, x < 0]
= 10 – 15
= -5

Question 9.
If f(x) = 4[x] – 3, where [x] is greatest integer function of x, then find
(i) f(7.2)
(ii) f(0.5)
(iii) \(f\left(-\frac{5}{2}\right)\)
(iv) f(2π), where π = 3.14
Solution:
f(x) = 4[x] – 3
(i) f(7.2) = 4 [7.2] – 3
= 4(7) – 3 ………[∵ 7 ≤ 7.2 < 8, [7.2] = 7]
= 25

(ii) f(0.5) = 4[0.5] – 3
= 4(0) – 3 ………[∵ 0 ≤ 0.5 < 1, [0.5] = 0]
= -3

(iii) \(f\left(-\frac{5}{2}\right)\) = f(-2.5)
= 4[-2.5] – 3
= 4(-3) – 3 …….[∵-3 ≤ -2.5 ≤ -2, [-2.5] = -3]
= -15

(iv) f(2π) = 4[2π] – 3
= 4[6.28] – 3 …..[∵ π = 3.14]
= 4(6) – 3 …….[∵ 6 ≤ 6.28 < 7, [6.28] = 6]
= 21

Question 10.
If f(x) = 2{x} + 5x, where {x} is fractional part function of x, then find
(i) f(-1)
(ii) f(\(\frac{1}{4}\))
(iii) f(-1.2)
(iv) f(-6)
Solution:
f(x) = 2{x} + 5x
(i) {-1} = -1 – [-1] = -1 + 1 = 0
∴ f(-1) = 2 {-1} + 5(-1)
= 2(0) – 5
= -5

(ii) {\(\frac{1}{4}\)} = \(\frac{1}{4}\) – [latex]\frac{1}{4}[/latex] = \(\frac{1}{4}\) – 0 = \(\frac{1}{4}\)
f(\(\frac{1}{4}\)) = 2{\(\frac{1}{4}\)} + 5(\(\frac{1}{4}\))
= 2(\(\frac{1}{4}\)) + \(\frac{5}{4}\)
= \(\frac{7}{4}\)
= 1.75

(iii) {-1.2} = -1.2 – [-1.2] = -1.2 + 2 = 0.8
f(-1.2) = 2{-1.2} + 5(-1.2)
= 2(0.8) + (-6)
= -4.4

(iv) {-6} = -6 – [-6] = -6 + 6 = 0
f(-6) = 2{-6} + 5(-6)
= 2(0) – 30
= -30

Question 11.
Solve the following for x, where |x| is modulus function, [x] is the greatest integer function, {x} is a fractional part function.
(i) |x + 4| ≥ 5
(ii) |x – 4| + |x – 2| = 3
(iii) x² + 7|x| + 12 = 0
(iv) |x| ≤ 3
(v) 2|x| = 5
(vi) [x + [x + [x]]] = 9
(vii) {x} > 4
(viii) {x} = o
(ix) {x} = 0.5
(x) 2{x} = x + [x]
Solution:
(i) |x + 4| ≥ 5
The solution of |x| ≥ a is x ≤ -a or x ≥ a
∴ |x + 4| ≥ 5 gives
∴ x + 4 ≤ -5 or x + 4 ≥ 5
∴ x ≤ -5 – 4 or x ≥ 5 – 4
∴ x ≤ -9 or x ≥ 1
∴ The solution set = (-∞, – 9] ∪ [1, ∞)

(ii) |x – 4| + |x – 2| = 3 …..(i)
Case I: x < 2
Equation (i) reduces to
4 – x + 2 – x = 3 …….[x < 2 < 4, x – 4 < 0, x – 2 < 0]
∴ 6 – 3 = 2x
∴ x = \(\frac{3}{2}\)

Case II: 2 ≤ x < 4
Equation (i) reduces to
4 – x + x – 2 = 3
∴ 2 = 3 (absurd)
There is no solution in [2, 4)

Case III: x ≥ 4
Equation (i) reduces to
x – 4 + x – 2 = 3
∴ 2x = 6 + 3 = 9
∴ x = \(\frac{9}{2}\)
∴ x = \(\frac{3}{2}\), \(\frac{9}{2}\) are solutions.
The solution set = {\(\frac{3}{2}\), \(\frac{9}{2}\)}

(iii) x² + 7|x| + 12 = 0
∴ (|x|)² + 7|x| + 12 = 0
∴ (|x| + 3) (|x| + 4) = 0
∴ There is no x that satisfies the equation.
The solution set = { } or Φ

(iv) |x| ≤ 3 The solution set of |x| ≤ a is -a ≤ x ≤ a
∴ The required solution is -3 ≤ x ≤ 3
∴ The solution set is [-3, 3]

(v) 2|x| = 5
∴ |x| = \(\frac{5}{2}\)
∴ x = ±\(\frac{5}{2}\)

(vi) [x + [x + [x]]] = 9
∴ [x + [x] + [x] ] = 9 …….[[x + n] = [x] + n, if n is an integer]
∴ [x + 2[x]] = 9
∴ [x] + 2[x] = 9 …..[[2[x] is an integer]]
∴ [x] = 3
∴ x ∈ [3, 4)

(vii) {x} > 4
This is a meaningless statement as 0 ≤ {x} < 1
∴ The solution set = { } or Φ

(viii) {x} = 0
∴ x is an integer
∴ The solution set is Z.

(ix) {x} = 0.5
∴ x = ….., -2.5, -1.5, -0.5, 0.5, 1.5, …..
∴ The solution set = {x : x = n + 0.5, n ∈ Z}

(x) 2{x} = x + [x]
= [x] + {x} + [x] ……[x = [x] + {x}]
∴ {x} = 2[x]
R.H.S. is an integer
∴ L.H.S. is an integer
∴ {x} = 0
∴ [x] = 0
∴ x = 0

Balbharti Maharashtra State Board 11th Maths Book Solutions Pdf Chapter 6 Functions Ex 6.1 Questions and Answers.

Maharashtra State Board 11th Maths Solutions Chapter 6 Functions Ex 6.1

Question 1.
Check if the following relations are functions.
(a)

Solution:
Yes.
Reason: Every element of set A has been assigned a unique element in set B.

(b)

Solution:
No.
Reason: An element of set A has been assigned more than one element from set B.

(c)

Solution:
No.
Reason:
Not every element of set A has been assigned an image from set B.

Question 2.
Which sets of ordered pairs represent functions from A = {1, 2, 3, 4} to B = {-1, 0, 1, 2, 3}? Justify.
(i) {(1, 0), (3, 3), (2, -1), (4, 1), (2, 2)}
(ii) {(1, 2), (2, -1), (3, 1), (4, 3)}
(iii) {(1, 3), (4, 1), (2, 2)}
(iv) {(1, 1), (2, 1), (3, 1), (4, 1)}
Solution:
(i) {(1, 0), (3, 3), (2, -1), (4, 1), (2, 2)} does not represent a function.
Reason: (2, -1), (2, 2), show that element 2 ∈ A has been assigned two images -1 and 2 from set B.

(ii) {(1, 2), (2, -1), (3, 1), (4, 3)} represents a function.
Reason: Every element of set A has been assigned a unique image in set B.

(iii) {(1, 3), (4, 1), (2, 2)} does not represent a function.
Reason:
3 ∈ A does not have an image in set B.

(iv) {(1, 1), (2, 1), (3, 1), (4, 1)} represents a function
Reason: Every element of set A has been assigned a unique image in set B.

Question 3.
Check if the relation given by the equation represents y as function of x.
(i) 2x + 3y = 12
(ii) x + y² = 9
(iii) x² – y = 25
(iv) 2y + 10 = 0
(v) 3x – 6 = 21
Solution:
(i) 2x + 3y = 12
∴ y = \(\frac{12-2 x}{3}\)
∴ For every value of x, there is a unique value of y.
∴ y is a function of x.

(ii) x + y² = 9
∴ y² = 9 – x
∴ y = ±\(\sqrt{9-x}\)
∴ For one value of x, there are two values of y.
∴ y is not a function of x.

(iii) x² – y = 25
∴ y = x² – 25
∴ For every value of x, there is a unique value of y.
∴ y is a function of x.

(iv) 2y + 10 = 0
∴ y = -5
∴ For every value of x, there is a unique value of y.
∴ y is a function of x.

(v) 3x – 6 = 21
∴ x = 9
∴ x = 9 represents a point on the X-axis.
There is no y involved in the equation.
So the given equation does not represent a function.

Question 4.
If f(m) = m² – 3m + 1, find
(i) f(0)
(ii) f(-3)
(iii) f(\(\frac{1}{2}\))
(iv) f(x + 1)
(v) f(-x)
(vi) \(\left(\frac{\mathbf{f}(2+h)-f(2)}{h}\right)\), h ≠ 0.
Solution:
f(m) = m² – 3m + 1
(i) f(0) = 0² – 3(0) + 1 = 1

(ii) f (-3) = (-3)² – 3(-3) + 1
= 9 + 9 + 1
= 19

(iii) f(\(\frac{1}{2}\)) = \(\left(\frac{1}{2}\right)^{2}-3\left(\frac{1}{2}\right)+1\)
= \(\frac{1}{4}-\frac{3}{2}+1\)
= \(\frac{1-6+4}{4}\)
= \(-\frac{1}{4}\)

(iv) f(x + 1) = (x + 1)² – 3(x + 1) + 1
= x² + 2x + 1 – 3x – 3 + 1
= x² – x – 1

(v) f(-x) = (-x)2 – 3(-x) + 1 = x2 + 3x + 1

(vi) \(\left(\frac{\mathbf{f}(2+h)-f(2)}{h}\right)\)
= \(\frac{(2+h)^{2}-3(2+h)+1-\left(2^{2}-3(2)+1\right)}{h}\)
= \(\frac{\mathrm{h}^{2}+\mathrm{h}}{\mathrm{h}}\)
= h + 1

Question 5.
Find x, if g(x) = 0 where
(i) g(x) = \(\frac{5 x-6}{7}\)
(ii) g(x) = \(\frac{18-2 x^{2}}{7}\)
(iii) g(x) = 6x² + x – 2
(iv) g(x) = x³ – 2x² – 5x + 6
Solution:
(i) g(x) = \(\frac{5 x-6}{7}\)
g(x) = 0
∴ \(\frac{5 x-6}{7}\) = 0
∴ x = \(\frac{6}{5}\)

(ii) g(x) = \(\frac{18-2 x^{2}}{7}\)
g(x) = 0
\(\frac{18-2 x^{2}}{7}\) = 0
∴ 18 – 2x² = 0
∴ x² = 9
∴ x = ±3

(iii) g(x) = 6x² + x – 2
g(x) = 0
∴ 6x² + x – 2 = 0
∴ (2x – 1) (3x + 2) = 0
∴ 2x – 1 = 0 or 3x + 2 = 0
∴ x = \(\frac{1}{2}\) or x = \(\frac{-2}{3}\)

(iv) g(x) = x³ – 2x² – 5x + 6
= ( x- 1) (x² – x – 6)
= (x – 1) (x + 2) (x – 3)
g(x) = 0
∴ (x – 1) (x + 2) (x – 3) = 0
∴ x – 1 = 0 or x + 2 = 0 or x – 3 = 0
∴ x = 1, -2, 3

Question 6.
Find x, if f(x) = g(x) where
(i) f(x) = x⁴ + 2x², g(x) = 11x²
(ii) f(x) = √x – 3, g(x) = 5 – x
Solution:
(i) f(x) = x⁴ + 2x², g(x) = 11x²
f(x) = g(x)
∴ x⁴ + 2x² = 11x²
∴ x⁴ – 9x² = 0
∴ x² (x² – 9) = 0
∴ x² = 0 or x² – 9 = 0
∴ x = 0 or x² = 9
∴ x = 0, ±3

(ii) f(x) = √x – 3, g(x) = 5 – x
f(x) = g(x)
∴ √x – 3 = 5 – x
∴ √x = 5 – x + 3
∴ √x = 8 – x
on squaring, we get
x = 64 + x² – 16x
∴ x² – 17x + 64 = 0
∴ x = \(\frac{17 \pm \sqrt{(-17)^{2}-4(64)}}{2}\)
∴ x = \(\frac{17 \pm \sqrt{289-256}}{2}\)
∴ x = \(\frac{17 \pm \sqrt{33}}{2}\)

Question 7.
If f(x) = \(\frac{a-x}{b-x}\), f(2) is undefined, and f(3) = 5, find a and b.
Solution:
f(x) = \(\frac{a-x}{b-x}\)
Given that,
f(2) is undefined
b – 2 = 0
∴ b = 2 …..(i)
f(3) = 5
∴ \(\frac{a-3}{b-3}\) = 5
∴ \(\frac{a-3}{2-3}\) = 5 ….. [From (i)]
∴ a – 3 = -5
∴ a = -2
∴ a = -2, b = 2

Question 8.
Find the domain and range of the following functions.
(i) f(x) = 7x² + 4x – 1
Solution:
f(x) = 7x² + 4x – 1
f is defined for all x.
∴ Domain of f = R (i.e., the set of real numbers)

∴ Range of f = [\(-\frac{11}{7}\), ∞)

(ii) g(x) = \(\frac{x+4}{x-2}\)
Solution:
g(x) = \(\frac{x+4}{x-2}\)
Function g is defined everywhere except at x = 2.
∴ Domain of g = R – {2}
Let y = g(x) = \(\frac{x+4}{x-2}\)
∴ (x – 2) y = x + 4
∴ x(y – 1) = 4 + 2y
∴ For every y, we can find x, except for y = 1.
∴ y = 1 ∉ range of function g
∴ Range of g = R – {1}

(iii) h(x) = \(\frac{\sqrt{x+5}}{5+x}\)
Solution:
h(x) = \(\frac{\sqrt{x+5}}{5+x}=\frac{1}{\sqrt{x+5}}\), x ≠ -5
For x = -5, function h is not defined.
∴ x + 5 > 0 for function h to be well defined.
∴ x > -5
∴ Domain of h = (-5, ∞)
Let y = \(\frac{1}{\sqrt{x+5}}\)
∴ y > 0
Range of h = (0, ∞) or R+

(iv) f(x) = \(\sqrt[3]{x+1}\)
Solution:
f(x) = \(\sqrt[3]{x+1}\)
f is defined for all real x and the values of f(x) ∈ R
∴ Domain of f = R, Range of f = R

(v) f(x) = \(\sqrt{(x-2)(5-x)}\)
Solution:
f(x) = \(\sqrt{(x-2)(5-x)}\)
For f to be defined,
(x – 2)(5 – x) ≥ 0
∴ (x – 2)(x – 5) ≤ 0
∴ 2 ≤ x ≤ 5 ……[∵ The solution of (x – a) (x – b) ≤ 0 is a ≤ x ≤ b, for a < b]
∴ Domain of f = [2, 5]
(x – 2) (5 – x) = -x² + 7x – 10
= \(-\left(x-\frac{7}{2}\right)^{2}+\frac{49}{4}-10\)
= \(\frac{9}{4}-\left(x-\frac{7}{2}\right)^{2} \leq \frac{9}{4}\)
∴ \(\sqrt{(x-2)(5-x)} \leq \sqrt{\frac{9}{4}} \leq \frac{3}{2}\)
Range of f = [0, \(\frac{3}{2}\)]

(vi) f(x) = \(\sqrt{\frac{x-3}{7-x}}\)
Solution:
f(x) = \(\sqrt{\frac{x-3}{7-x}}\)
For f to be defined,
\(\sqrt{\frac{x-3}{7-x}}\) ≥ 0, 7 – x ≠ 0
∴ \(\sqrt{\frac{x-3}{7-x}}\) ≤ 0 and x ≠ 7
∴ 3 ≤ x < 7
Let a < b, \(\frac{x-a}{x-b}\) ≤ 0 ⇒ a ≤ x < b
∴ Domain of f = [3, 7)
f(x) ≥ 0 … [∵ The value of square root function is non-negative]
∴ Range of f = [0, ∞)

(vii) f(x) = \(\sqrt{16-x^{2}}\)
Solution:
f(x) = \(\sqrt{16-x^{2}}\)
For f to be defined,
16 – x² ≥ 0
∴ x² ≤ 16
∴ -4 ≤ x ≤ 4
∴ Domain of f = [-4, 4]
Clearly, f(x) ≥ 0 and the value of f(x) would be maximum when the quantity subtracted from 16 is minimum i.e. x = 0
∴ Maximum value of f(x) = √16 = 4
∴ Range of f = [0, 4]

Question 9.
Express the area A of a square as a function of its
(a) side s
(b) perimeter P
Solution:
(a) area (A) = s²
(b) perimeter (P) = 4s
∴ s = \(\frac{\mathrm{P}}{4}\)
Area (A) = s² = \(\left(\frac{\mathrm{P}}{4}\right)^{2}\)
∴ A = \(\frac{\mathrm{P}^{2}}{16}\)

Question 10.
Express the area A of a circle as a function of its
(i) radius r
(ii) diameter d
(iii) circumference C
Solution:
(i) Area (A) = πr²

(ii) Diameter (d) = 2r
∴ r = \(\frac{\mathrm{d}}{2}\)
∴ Area (A) = πr² = \(\frac{\pi \mathrm{d}^{2}}{4}\)

(iii) Circumference (C) = 2πr
∴ r = \(\frac{C}{2 \pi}\)
Area (A) = πr² = \(\pi\left(\frac{\mathrm{C}}{2 \pi}\right)^{2}\)
∴ A = \(\frac{C^{2}}{4 \pi}\)

Question 11.
An open box is made from a square of cardboard of 30 cms side, by cutting squares of length x centimeters from each corner and folding the sides up. Express the volume of the box as a function of x. Also, find its domain.
Solution:

Length of the box = 30 – 2x
Breadth of the box = 30 – 2x
Height of the box = x
Volume = (30 – 2x)² x, x < 15, x ≠ 15, x > 0
= 4x(15 – x)², x ≠ 15, x > 0
Domain (0, 15)

Question 12.
Let f be a subset of Z × Z defined by f = {(ab, a + b): a, b ∈ Z}. Is f a function from Z to Z? Justify?
Solution:
f = {(ab, a + b): a, b ∈ Z}
Let a = 1, b = 1. Then, ab = 1, a + b = 2
∴ (1, 2) ∈ f
Let a = -1, b = -1. Then, ab = 1, a + b = -2
∴ (1, -2) ∈ f
Since (1, 2) ∈ f and (1, -2) ∈ f,
f is not a function as element 1 does not have a unique image.

Question 13.
Check the injectivity and surjectivity of the following functions.
(i) f : N → N given by f(x) = x²
Solution:
f: N → N given by f(x) = x²

∴ f is injective.
For every y = x² ∈ N, there does not exist x ∈ N.
Example: 7 ∈ N (codomain) for which there is no x in domain N such that x² = 7
∴ f is not surjective.

(ii) f : Z → Z given by f(x) = x²
Solution:
f: Z → Z given by f(x) = x2

∴ f is not injective.
(Example: f(-2) = 4 = f(2). So, -2, 2 have the same image. So, f is not injective.)
Since x² ≥ 0,
f(x) ≥ 0
Therefore all negative integers of codomain are not images under f.
∴ f is not surjective.

(iii) f : R → R given by f(x) = x²
Solution:
f : R → R given by f(x) = x²

∴ f is not injective.
f(x) = x² ≥ 0
Therefore all negative integers of codomain are not images under f.
∴ f is not surjective.

(iv) f : N → N given by f(x) = x³
Solution:
f: N → N given by f(x) = x³

∴ f is injective.
Numbers from codomain which are not cubes of natural numbers are not images under f.
∴ f is not surjective.

(v) f : R → R given by f(x) = x³
Solution:
f: R → R given by f(x) = x³

∴ For every y ∈ R, there is some x ∈ R.
∴ f is surjective.

Question 14.
Show that if f : A → B and g : B → C are one-one, then gof is also one-one.
Solution:
f is a one-one function.
Let f(x₁) = f(x₂)
Then, x₁ = x₂ for all x₁, x₂ …..(i)
g is a one-one function.
Let g(y₁) = g(y₂)
Then, y₁ = y₂ for all y₁, y₂ …..(ii)
Let (gof) (x₁) = (gof) (x₂)
∴ g(f(x₁)) = g(f(x₂))
∴ g(y₁) = g(y₂),
where y₁ = f(x₁), y₂ = f(x₂) ∈ B
∴ y₁ = y₂ …..[From (ii)]
i.e., f(x₁) = f(x₂)
∴ x₁ = x₂ ….[From (i)]
∴ gof is one-one.

Question 15.
Show that if f : A → B and g : B → C are onto, then gof is also onto.
Solution:
Since g is surjective (onto),
there exists y ∈ B for every z ∈ C such that
g(y) = z …….(i)
Since f is surjective,
there exists x ∈ A for every y ∈ B such that
f(x) = y …….(ii)
(gof) x = g(f(x))
= g(y) ……[From (ii)]
= z …..[From(i)]
i.e., for every z ∈ C, there is x in A such that (gof) x = z
∴ gof is surjective (onto).

Question 16.
If f(x) = 3(4^x+1), find f(-3).
Solution:
f(x) = 3(4^x+1)
∴ f(-3) = 3(4^-3+1)
= 3(4^-2)
= \(\frac{3}{16}\)

Question 17.
Express the following exponential equations in logarithmic form:
(i) 2⁵ = 32
(ii) 54⁰ = 1
(iii) 23¹ = 23
(iv) \(9^{\frac{3}{2}}\) = 27
(v) 3^-4 = \(\frac{1}{81}\)
(vi) 10^-2 = 0.01
(vii) e² = 7.3890
(viii) \(e^{\frac{1}{2}}\) = 1.6487
(ix) e^-x = 6
Solution:

Question 18.
Express the following logarithmic equations in exponential form:
(i) log₂ 64 = 6
(ii) \(\log _{5} \frac{1}{25}\) = -2
(iii) log₁₀ 0.001 = -3
(iv) \(\log _{\frac{1}{2}}\)(8) = -3
(v) ln 1 = 0
(vi) ln e = 1
(vii) ln \(\frac{1}{2}\) = -0.693
Solution:
(i) log₂ 64 = 6
∴ 64 = 2⁶, i.e., 2⁶ = 64

Question 19.
Find the domain of
(i) f(x) = ln (x – 5)
(ii) f(x) = log10 (x² – 5x + 6)
Solution:
(i) f(x) = ln (x – 5)
f is defined, when x – 5 > 0
∴ x > 5
∴ Domain of f = (5, ∞)

(ii) f(x) = log₁₀ (x² – 5x + 6)
x² – 5x + 6 = (x – 2) (x – 3)
f is defined, when (x – 2) (x – 3) > 0
∴ x < 2 or x > 3
Solution of (x – a) (x – b) > 0 is x < a or x > b where a < b
∴ Domain of f = (-∞, 2) ∪ (3, ∞)

Question 20.
Write the following expressions as sum or difference of logarithms:
(a) \(\log \left(\frac{p q}{r s}\right)\)
Solution:

(b) \(\log (\sqrt{x} \sqrt[3]{y})\)
Solution:

(c) \(\ln \left(\frac{a^{3}(a-2)^{2}}{\sqrt{b^{2}+5}}\right)\)
Solution:

(d) \(\ln \left[\frac{\sqrt[3]{x-2}(2 x+1)^{4}}{(x+4) \sqrt{2 x+4}}\right]^{2}\)
Solution:

Question 21.
Write the following expressions as a single logarithm.
(i) 5 log x + 7 log y – log z
Solution:

(ii) \(\frac{1}{3}\) log(x – 1) + \(\frac{1}{2}\) log(x)
Solution:

(iii) ln (x + 2) + ln (x – 2) – 3 ln (x + 5)
Solution:

Question 22.
Given that log 2 = a and log 3 = b, write log √96 terms of a and b.
Solution:
log 2 = a and log 3 = b
log √96 = \(\frac{1}{2}\) log (96)
= \(\frac{1}{2}\) log (2⁵ x 3)
= \(\frac{1}{2}\) (log 2⁵ + log 3) …..[∵ log mn = log m + log n]
= \(\frac{1}{2}\) (5 log 2 + log 3) ……[∵ log mⁿ = n log m]
= \(\frac{5 a+b}{2}\)

Question 23.
Prove that:
(a) \(b^{\log _{b} a}=a\)
Solution:
We have to prove that \(b^{\log _{b} a}=a\)
i.e., to prove that (logb a) (log_b b) = log_b a
(Taking log on both sides with base b)
L.H.S. = (log_b a) (log_b b)
= log_b a …..[∵ log_b b = 1]
= R.H.S.

(b) \(\log _{b^{m}} a=\frac{1}{m} \log _{b} a\)
Solution:

(c) \(a^{\log _{c} b}=b^{\log _{c} a}\)
Solution:

Question 24.
If f(x) = ax² – bx + 6 and f(2) = 3 and f(4) = 30, find a and b.
Solulion:
f(x) = ax² – bx + 6
f(2) = 3
∴ a(2)² – b(2) + 6 = 3
∴ 4a – 2b + 6 = 3
∴ 4a – 2b + 3 = 0 …..(i)
f(4) = 30
∴ a(4)² – b(4) + 6 = 30
∴ 16a – 4b + 6 = 30
∴ 16a – 4b – 24 = 0 …..(ii)
By (ii) – 2 × (i), we get
8a – 30 = 0
∴ a = \(\frac{30}{8}=\frac{15}{4}\)
Substiting a = \(\frac{15}{4}\) in (i), we get
4(\(\frac{15}{4}\)) – 2b + 3 = 0
∴ 2b = 18
∴ b = 9
∴ a = \(\frac{15}{4}\), b = 9

Question 25.
Solve for x:
(i) log 2 + log (x + 3) – log (3x – 5) = log 3
Solution:
log 2 + log (x + 3) – log (3x – 5) = log 3
∴ log 2(x + 3) – log(3x – 5) = log 3 …..[∵ log m + log n = log mn]
∴ log \(\frac{2(x+3)}{3 x-5}\) = log 3 …..[∵ log m – log n = log \(\frac{m}{n}\)]
∴ \(\frac{2(x+3)}{3 x-5}\) = 3
∴ 2x + 6 = 9x – 15
∴ 7x = 21
∴ x = 3

Check:
If x = 3 satisfies the given condition, then our answer is correct.
L.H.S. = log 2 + log (x + 3) – log (3x – 5)
= log 2 + log (3 + 3) – log (9 – 5)
= log 2 + log 6 – log 4
= log (2 × 6) – log 4
= log \(\frac{12}{4}\)
= log 3
= R.H.S.
Thus, our answer is correct.

(ii) 2log₁₀ x = 1 + \(\log _{10}\left(x+\frac{11}{10}\right)\)
Solution:


∴ x² = 10x + 11
∴ x² – 10x – 11 = 0
∴ (x – 11)(x + 1) = 0
∴ x = 11 or x = -1
But log of a negative numbers does not exist
∴ x ≠ -1
∴ x = 11

(iii) log₂ x + log₄ x + log₁₆ x = \(\frac{21}{4}\)
Solution:

(iv) x + log₁₀ (1 + 2^x) = x log₁₀ 5 + log₁₀ 6
Solution:

∴ a + a² = 6
∴ a² + a – 6 = 0
∴ (a + 3)(a – 2) = 0
∴ a + 3 = 0 or a – 2 = 0
∴ a = -3 or a = 2
Since 2x = -3 is not possible,
2x = 2 = 21
∴ x = 1

Question 26.
If log \(\left(\frac{x+y}{3}\right)\) = \(\frac{1}{2}\) log x + \(\frac{1}{2}\) log y, show that \(\frac{x}{y}+\frac{y}{x}\) = 7.
Solution:

Question 27.
If log\(\left(\frac{x-y}{4}\right)\) = log√x + log√y, show that (x + y)² = 20xy.
Solution:

Question 28.
If x = log_abc, y = log_b ca, z = log_c ab, then prove that \(\frac{1}{1+x}+\frac{1}{1+y}+\frac{1}{1+z}\) = 1.
Solution:

Balbharti Maharashtra State Board 11th Maths Book Solutions Pdf Chapter 5 Sets and Relations Miscellaneous Exercise 5 Questions and Answers.

Maharashtra State Board 11th Maths Solutions Chapter 5 Sets and Relations Miscellaneous Exercise 5

(I) Select the correct answer from the given alternative.

Question 1.
For the set A = {a, b, c, d, e} the correct statement is
(A) {a, b} ∈ A
(B) {a} ∈ A
(C) a ∈ A
(D) a ∉ A
Answer:
(C) a ∈ A

Question 2.
If aN = {ax : x ∈ N}, then set 6N ∩ 8N =
(A) 8N
(B) 48N
(C) 12N
(D) 24N
Answer:
(D) 24N
Hint:
6N = {6x : x ∈ N} = {6, 12, 18, 24, 30, ……}
8N = {8x : x ∈ N} = {8, 16, 24, 32, ……}
∴ 6N ∩ 8N = {24, 48, 72, …..}
= {24x : x ∈ N}
= 24N

Question 3.
If set A is empty set then n[P[P[P(A)]]] is
(A) 6
(B) 16
(C) 2
(D) 4
Answer:
(D) 4
Hint:
A = Φ
∴ n(A) = 0
∴ n[P(A)] = 2^n(A) = 2⁰ = 1
∴ n[P[P(A)]] = 2^n[P(A)] = 2¹ = 2
∴ n[P[P[P(A)]]] = 2^n[P[P(A)]] = 2² = 4

Question 4.
In a city 20% of the population travels by car, 50% travels by bus and 10% travels by both car and bus. Then, persons travelling by car or bus are
(A) 80%
(B) 40%
(C) 60%
(D) 70%
Answer:
(C) 60%
Hint:
Let C = Population travels by car
B = Population travels by bus
n(C) = 20%, n(B) = 50%, n(C ∩ B) = 10%
n(C ∪ B) = n(C) + n(B) – n(C ∩ B)
= 20% + 50% – 10%
= 60%

Question 5.
If the two sets A and B are having 43 elements in common, then the number of elements common to each of the sets A × B and B × A is
(A) 43²
(B) 2⁴³
(C) 43⁴³
(D) 2⁸⁶
Answer:
(A) 43²

Question 6.
Let R be a relation on the set N be defied by {(x, y) / x, y ∈ N, 2x + y = 41} Then R is
(A) Reflexive
(B) Symmetric
(C) Transitive
(D) None of these
Answer:
(D) None of these

Question 7.
The relation “>” in the set of N (Natural number) is
(A) Symmetric
(B) Reflexive
(C) Transitive
(D) Equivalent relation
Answer:
(C) Transitive
Hint:
For any a ∈ N, a ≯ a
∴ (a, a) ∉ R
∴ > is not reflexive.
For any a, b ∈ N, if a > b, then b ≯ a.
∴ > is not symmetric.
For any a, b, c ∈ N,
if a > b and b > c, then a > c
∴ > is transitive.

Question 8.
A relation between A and B is
(A) only A × B
(B) An Universal set of A × B
(C) An equivalent set of A × B
(D) A subset of A × B
Answer:
(D) A subset of A × B

Question 9.
If (x, y) ∈ N × N, then xy = x² is a relation that is
(A) Symmetric
(B) Reflexive
(C) Transitive
(D) Equivalence
Answer:
(D) Equivalence
Hint:
Let x ∈ R, then xx = x²
∴ x is related to x.
∴ Given relation is reflexive.
Letx = 0 and y = 2,
then xy = 0 × 2 = 0 = x²
∴ x is related to y.
Consider, yx = 2 × 0 = 0 ≠ y²
∴ y is not related to x.
∴ Given relation is not symmetric.
Let x be related to y and y be related to z.
∴ xy = x² and yz = y²
∴ x = \(\frac{x^{2}}{y}\) and z = \(\frac{y^{2}}{y}\) = y …..[if y ≠ 0]
Consider, xz = \(\frac{x^{2}}{y}\) × y = x²
∴ x is related to z.
∴ Given relation is transitive.

Question 10.
If A = {a, b, c}, The total no. of distinct relations in A × A is
(A) 3
(B) 9
(C) 8
(D) 29
Answer:
(D) 29

(II) Answer the following.

Question 1.
Write down the following sets in set builder form:
(i) {10, 20, 30, 40, 50}
(ii) {a, e, i, o, u}
(iii) {Sunday, Monday, Tuesday, Wednesday, Thursday, Friday, Saturday}
Solution:
(i) Let A = {10, 20, 30, 40, 50}
∴ A = {x/x = 10n, n ∈ N and n ≤ 5}

(ii) Let B = {a, e, i, o, u}
∴ B = {x/x is a vowel of English alphabets}

(iii) Let C = {Sunday, Monday, Tuesday, Wednesday, Thursday, Friday, Saturday}
∴ C = {x/x is a day of a week}

Question 2.
If U = {x/x ∈ N, 1 ≤ x ≤ 12}, A = {1,4, 7,10}, B = {2, 4, 6, 7, 11}, C = {3, 5, 8, 9, 12}. Write down the sets.
(i) A ∪ B
(ii) B ∩ C
(iii) A – B
(iv) B ∩ C’
(v) A ∪ B ∪ C
(vi) A ∩ (B ∪ C)
Solution:
U = {x/x ∈ N, 1 ≤ x ≤ 12} = {1, 2, 3, …., 12}
A = {1, 4, 7, 10}, B = {2, 4, 6, 7, 11}, C = {3, 5, 8, 9, 12}
(i) A ∪ B = {1, 2, 4, 6, 7, 10, 11}

(ii) B ∩ C = {}

(iii) A – B = {1, 10}

(iv) C’ = {1, 2, 4, 6, 7, 10, 11}
∴ B ∩ C’ = {2, 4, 6, 7, 11}

(v) A ∪ B ∪ C = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}

(vi) B ∪ C = {2, 3, 4, 5, 6, 7, 8, 9, 11, 12}
∴ A ∩ (B ∪ C) = {4, 7}

Question 3.
In a survey of 425 students in a school, it was found that 115 drink apple juice, 160 drink orange juice, and 80 drink both apple as well as orange juice. How many drinks neither apple juice nor orange juice?
Solution:
Let A = set of students who drink apple juice
B = set of students who drink orange juice
X = set of all students
∴ n(X) = 425, n(A) = 115, n(B) = 160, n(A ∩ B) = 80

No. of students who neither drink apple juice nor orange juice = n(A’ ∩ B’) = n(A ∪ B)’
= n(X) – n(A ∪ B)
= 425 – [n(A) + n(B) – n(A ∩ B)]
= 425 – (115 + 160 – 80)
= 230

Question 4.
In a school, there are 20 teachers who teach Mathematics or Physics. Of these, 12 teach Mathematics and 4 teach both Physics and Mathematics. How many teachers teach Physics?
Solution:
Let A = set of teachers who teach Mathematics
B = set of teachers who teach Physics
∴ n(A ∪ B) = 20, n(A) = 12, n(A ∩ B) = 4

Since n(A ∪ B) = n(A) + n(B) – n(A ∩ B),
20 = 12 + n(B) – 4
∴ n(B) = 12
∴ Number of teachers who teach physics = 12

Question 5.
(i) If A = {1, 2, 3} and B = {2, 4}, state the elements of A × A, A × B, B × A, B × B, (A × B) ∩ (B × A).
(ii) If A = {-1, 1}, find A × A × A.
Solution:
(i) A = {1, 2, 3} and B = {2, 4}
A × A = {(1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (3, 3)}
A × B = {(1, 2), (1, 4), (2, 2), (2, 4), (3, 2), (3, 4)}
B × A = {(2, 1), (2, 2), (2, 3), (4, 1), (4, 2), (4, 3)}
B × B = {(2, 2), (2, 4), (4, 2), (4, 4)}
∴ (A × B) ∩ (B × A) = {(2, 2)}

(ii) A = {-1, 1}
∴ A × A × A = {(-1, -1, -1), (-1, -1, 1), (-1, 1, -1), (-1, 1, 1), (1, -1, -1), (1, -1, 1), (1, 1, -1), (1, 1, 1)}

Question 6.
If A = {1, 2, 3}, B = {4, 5, 6}, check if the following are relations from A to B. Also, write its domain and range.
(i) R₁ = {(1, 4), (1, 5), (1, 6)}
(ii) R₂ = {(1, 5), (2, 4), (3, 6)}
(iii) R₃ = {(1, 4), (1, 5), (3, 6), (2, 6), (3, 4)}
(iv) R₄ = {(4, 2), (2, 6), (5, 1), (2, 4)}
Solution:
A = {1, 2, 3}, B = {4, 5, 6}
∴ A × B = {(1, 4), (1, 5), (1, 6), (2,4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6)}
(i) R₁ = {(1, 4), (1, 5), (1, 6)}
Since R₁ ⊆ A × B,
R₁ is a relation from A to B.
Domain (R₁) = Set of first components of R₁ = {1}
Range (R₁) = Set of second components of R₁ = {4, 5, 6}

(ii) R₂ = {(1, 5),(2, 4),(3, 6)}
Since R₂ ⊆ A × B,
R₂ is a relation from A to B.
Domain (R₂) = Set of first components of R₂ = {1, 2, 3}
Range (R₂) = Set of second components of R₂ = {4, 5, 6}

(iii) R₃ = {(1, 4), (1, 5), (3, 6), (2, 6), (3, 4)}
Since R₃ ⊆ A × B,
R₃ is a relation from A to B.
Domain (R₃) = Set of first components of R₃ = {1, 2, 3}
Range (R₃) = Set of second components of R₃ = {4, 5, 6}

(iv) R₄ = {(4, 2), (2, 6), (5, 1), (2, 4)}
Since (4, 2) ∈ R₄, but (4, 2) ∉ A × B,
R₄ ⊄ A × B
∴ R₄ is not a relation from A to B.

Question 7.
Determine the domain and range of the following relations.
(i) R = {(a, b) / a ∈ N, a < 5, b = 4}
(ii) R = {(a, b) / b = |a – 1|, a ∈ Z, |a| < 3}
Solution:
(i) R = {(a, b) / a ∈ N, a < 5, b = 4}
∴ Domain (R) = {a / a ∈ N, a < 5} = {1, 2, 3, 4}
Range (R) = {b / b = 4} = {4}

(ii) R = {(a, b) / b = |a – 1|, a ∈ Z, |a| < 3}
Since a ∈ Z and |a| < 3,
a < 3 and a > -3
∴ -3 < a < 3
∴ a = -2, -1, 0, 1, 2
b = |a – 1|
When a = -2, b = 3
When a = -1, b = 2
When a = 0, b = 1
When a = 1, b = 0
When a = 2, b = 1
Domain (R) = {-2, -1, 0, 1, 2}
Range (R) = {0, 1, 2, 3}

Question 8.
Find R : A → A when A = {1, 2, 3, 4} such that
(i) R = {(a, b) / a – b = 10}
(ii) R = {(a, b) / |a – b| ≥ 0}
Solution:
R : A → A, A = {1, 2, 3,4}
(i) R = {(a, b)/a – b = 10} = { }

(ii) R = {(a, b) / |a – b| ≥ 0}
= {(1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (2, 4), (3, 1), (3, 2), (3, 3), (3, 4), (4, 1), (4, 2), (4, 3), (4, 4)}
A × A = {(1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (2, 4), (3, 1), (3, 2), (3, 3), (3, 4), (4, 1), (4, 2), (4, 3), (4, 4)}
∴ R = A × A

Question 9.
R : {1, 2, 3} → {1, 2, 3} given by R = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 3)}. Check if R is
(i) reflexive
(ii) symmetric
(iii) transitive
Solution:
R = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 3)}
(i) Here, (x, x) ∈ R, for x ∈ {1, 2, 3}
∴ R is reflexive.

(ii) Here, (1, 2) ∈ R, but (2, 1) ∉ R.
∴ R is not symmetric.

(iii) Here, (1, 2), (2, 3) ∈ R,
But (1, 3) ∉ R.
∴ R is not transitive.

Question 10.
Check if R : Z → Z, R = {(a, b) | 2 divides a – b} is an equivalence relation.
Solution:
(i) Since 2 divides a – a,
(a, a) ∈ R
∴ R is reflexive. .

(ii) Let (a, b) ∈ R
Then 2 divides a – b
∴ 2 divides b – a
∴ (b, a) ∈ R
∴ R is symmetric.

(iii) Let (a, b) ∈ R, (b, c) ∈ R
Then a – b = 2m, b – c = 2n,
∴ a – c = 2(m + n), where m, n are integers.
∴ 2 divides a – c
∴ (a, c) ∈ R
∴ R is transitive.
Thus, R is an equivalence relation.

Question 11.
Show that the relation R in the set A = {1, 2, 3, 4, 5} Given by R = {(a, b) / |a – b| is even} is an equivalence relation.
Solution:
(i) Since |a – a| is even,
∴ (a, a) ∈ R
∴ R is reflexive.

(ii) Let (a, b) ∈ R
Then |a – b| is even
∴ |b – a| is even
∴ (b, a) ∈ R
∴ R is symmetric.

(iii) Let (a, b), (b, c) ∈ R
Then a – b = ±2m, b – c = ±2n
∴ a – c = ±2(m + n), where m, n are integers.
∴ (a, c) ∈ R
∴ R is transitive
Thus, R is an equivalence relation.

Question 12.
Show that the following are equivalence relations:
(i) R in A is set of all books given by R = {(x, y) / x and y have same number of pages}
(ii) R in A = {x ∈ Z | 0 ≤ x ≤ 12} given by R = {(a, b) / |a – b| is a multiple of 4}
(iii) R in A = (x ∈ N/x ≤ 10} given by R = {(a, b) | a = b}
Solution:
(i) a. Clearly (x, x) ∈ R
∴ R is reflexive.

b. If (x, y) ∈ R then (y, x) ∈ R.
∴ R is symmetric.

c. Let (x, y) ∈ R, (y, x) ∈ R.
Then x, y, and z are 3 books having the same number of pages.
∴ (x, z) ∈ R as x, z has the same number of pages.
∴ R is transitive.
Thus, R is an equivalence relation.

(ii) a. Since |a – a| is a multiple of 4,
(a, a) ∈ R
∴ R is reflexive.

b. Let (a, b) ∈ R
Then a – b = ±4m,
∴ b – a = ±4m, where m is an integer
∴ (b, a) ∈ R
∴ R is symmetric.

c. Let (a, b), (b, c) ∈ R
a – b = ± 4m, b – c = ± 4n,
∴ a – c = ±4(m + n), where m, n are integers
∴ (a, c) ∈ R
∴ R is transitive
Thus, R is an equivalence relation.

(iii) a. Since a = a
∴ (a, a) ∈ R
∴ R is reflexive.

b. Let (a, b) ∈ R Then a = b
∴ b = a
∴ (b, a) ∈ R
∴ R is symmetric.

c. Let (a, b), (b, c) ∈ R
Then, a = b, b = c
∴ a = c
∴ (a, c) ∈ R
∴ R is transitive.
Thus, R is an equivalence relation.

Balbharti Maharashtra State Board 11th Maths Book Solutions Pdf Chapter 5 Sets and Relations Ex 5.2 Questions and Answers.

Maharashtra State Board 11th Maths Solutions Chapter 5 Sets and Relations Ex 5.2

Question 1.
If (x – 1, y + 4) = (1, 2), find the values of x and y.
Solution:
(x – 1, y + 4) = (1, 2)
By the definition of equality of ordered pairs, we have
x – 1 = 1 and y + 4 = 2
∴ x = 2 and y = -2

Question 2.
If \(\left(x+\frac{1}{3}, \frac{y}{3}-1\right)=\left(\frac{1}{2}, \frac{3}{2}\right)\), find x and y.
Solution:
\(\left(x+\frac{1}{3}, \frac{y}{3}-1\right)=\left(\frac{1}{2}, \frac{3}{2}\right)\)
By the definition of equality of ordered pairs, we have
x + \(\frac{1}{3}\) = \(\frac{1}{2}\) and \(\frac{y}{3}\) – 1 = \(\frac{3}{2}\)
∴ x = \(\frac{1}{2}\) – \(\frac{1}{3}\) and \(\frac{y}{3}\) = \(\frac{3}{2}\) + 1
∴ x = \(\frac{1}{6}\) and y = \(\frac{15}{2}\)

Question 3.
If A = {a, b, c}, B = {x, y}, find A × B, B × A, A × A, B × B.
Solution:
A = (a, b, c}, B = {x, y}
A × B = {(a, x), (a, y), (b, x), (b, y), (c, x), (c, y)}
B × A = {(x, a), (x, b), (x, c), (y, a), (y, b), (y, c)}
A × A = {(a, a), (a, b), (a, c), (b, a), (b, b), (b, c), (c, a), (c, b), (c, c)}
B × B = {(x, x), (x, y), (y, x), (y, y)}

Question 4.
If P = {1, 2, 3} and Q = {1, 4}, find sets P × Q and Q × P.
Solution:
P = {1, 2, 3}, Q = {1, 4}
∴ P × Q = {(1, 1), (1, 4), (2, 1), (2, 4), (3, 1), (3, 4)}
and Q × P = {(1, 1), (1, 2), (1, 3), (4, 1), (4, 2), (4, 3)}

Question 5.
Let A = {1, 2, 3, 4}, B = {4, 5, 6}, C = {5, 6}. Verify,
(i) A × (B ∩ C) = (A × B) ∩ (A × C)
(ii) A × (B ∪ C) = (A × B) ∪ (A × C)
Solution:
A = {1, 2, 3, 4}, B = {4, 5, 6}, C = {5, 6}
(i) B ∩ C = {5, 6}
A × (B ∩ C) = = {(1, 5), (1, 6), (2, 5), (2, 6), (3, 5), (3, 6), (4, 5), (4, 6)}
A × B = {(1, 4), (1, 5), (1, 6), (2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6), (4, 4), (4, 5), (4, 6)}
A × C = {(1, 5), (1, 6), (2, 5), (2, 6), (3, 5), (3, 6), (4, 5), (4, 6)}
∴ (A × B) ∩ (A × C) = {(1, 5), (1, 6), (2, 5), (2, 6), (3, 5), (3, 6), (4, 5), (4, 6)}
∴ A × (B ∩ C) = (A × B) ∩ (A × C)

(ii) B ∪ C = {4, 5, 6}
A × (B ∪ C) = {(1, 4), (1, 5), (1, 6), (2, 4), (2, 5), (2, 6), (3,4), (3, 5), (3, 6), (4, 4), (4, 5), (4, 6)}
A × B = {(1, 4), (1, 5), (1, 6), (2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6), (4, 4), (4, 5), (4, 6)}
A × C = {(1, 5), (1, 6), (2, 5), (2, 6), (3, 5), (3, 6), (4, 5), (4, 6)}
∴ (A × B) ∪ (A × C) = {(1, 4), (1, 5), (1, 6), (2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6), (4, 4), (4, 5), (4, 6)}
∴ A × (B ∪ C) = (A × B) ∪ (A × C)

Question 6.
Express {(x, y) / x² + y² = 100, where x, y ∈ W} as a set of ordered pairs.
Solution:
{(x, y) / x² + y² = 100, where x, y ∈ W}
We have, x² + y² = 100
When x = 0 and y = 10,
x² + y² = 0² + 10² = 100
When x = 6 and y = 8,
x² + y² = 6² + 8² = 100
When x = 8 and y = 6,
x² + y² = 8² + 6² = 100
When x = 10 and y = 0,
x² + y² = 10² + 0² = 100
∴ Set of ordered pairs = {(0, 10), (6, 8), (8, 6), (10, 0)}

Question 7.
Let A = {6, 8} and B = {1, 3, 5}. Show that R₁ = {(a, b) / a ∈ A, b ∈ B, a – b is an even number} is a null relation, R₂ = {(a, b) / a ∈ A, b ∈ B, a + b is an odd number} is a universal relation.
Solution:
A = {6, 8}, B = {1, 3, 5}
R₁ = {(a, b)/ a ∈ A, b ∈ B, a – b is an even number}
a ∈ A
∴ a = 6, 8
b ∈ B
∴ b = 1, 3, 5
When a = 6 and b = 1, a – b = 5, which is odd
When a = 6 and b = 3, a – b = 3, which is odd
When a = 6 and b = 5, a – b = 1, which is odd
When a = 8 and b = 1, a – b = 7, which is odd
When a = 8 and b = 3, a – b = 5, which is odd
When a = 8 and b = 5, a – b = 3, which is odd
Thus, no set of values of a and b gives a – b as even.
∴ R₁ has a null relation from A to B.
A × B = {(6, 1), (6, 3), (6, 5), (8, 1), (8, 3), (8, 5)}
When a = 6 and b = 1, a + b = 7, which is odd
When a = 6 and b = 3, a + b = 9, which is odd
When a = 6 and b = 5, a + b = 11, which is odd
When a = 8 and b = 1, a + b = 9, which is odd
When a = 8 and b = 3, a + b = 11, which is odd
When a = 8 and b = 5, a + b = 13, which is odd
∴ R₂ = {(6, 1), (6, 3), (6, 5), (8, 1), (8, 3), (8, 5)}
Here, R₂ = A × B
∴ R₂ has a universal relation from A to B.

Question 8.
Write the relation in the Roster form. State its domain and range.
(i) R₁ = {(a, a²) / a is a prime number less than 15}
(ii) R₂ = {(a, \(\frac{1}{a}\)) / 0 < a ≤ 5, a ∈ N}
(iii) R₃ = {(x, y / y = 3x, y ∈ {3, 6, 9, 12}, x ∈ {1, 2, 3}}
(iv) R₄ = {(x, y) / y > x + 1, x = 1, 2 and y = 2, 4, 6}
(v) R₅ = {(x, y) / x + y = 3, x, y ∈ {0, 1, 2, 3}}
(vi) R₆ = {(a, b) / a ∈ N, a < 6 and b = 4}
(vii) R₇ = {(a, b) / a, b ∈ N, a + b = 6}
(viii) R₈ = {(a, b)/ b = a + 2, a ∈ Z, 0 < a < 5}
Solution:
(i) R₁ = {(a, a²) / a is a prime number less than 15}
∴ a = 2, 3, 5, 7, 11, 13
∴ a² = 4, 9, 25, 49, 121, 169
∴ R₁ = {(2, 4), (3, 9), (5, 25), (7, 49), (11, 121), (13, 169)}
∴ Domain (R₁) = {a/a is a prime number less than 15}
= {2, 3, 5, 7, 11, 13}
Range (R₁) = {a²/a is a prime number less than 15}
= {4, 9, 25, 49, 121, 169}

(iii) R₃ = {(x, y) / y = 3x, x ∈ {1, 2, 3}, y ∈ {3, 6, 9, 12}}
Here y = 3x
When x = 1, y = 3(1) = 3
When x = 2, y = 3(2) = 6
When x = 3, y = 3(3) = 9
∴ R₃ = {(1, 3), (2, 6), (3, 9)}
∴ Domain (R₃) ={1, 2, 3}
∴ Range (R₃) = {3, 6, 9}

(iv) R₄ = {(x, y) / y > x + 1, x = 1, 2 and y = 2, 4, 6}
Here, y > x + 1
When x = 1 and y = 2, 2 ≯  1 + 1
When x = 1 and y = 4, 4 > 1 + 1
When x = 1 and y = 6, 6 > 1 + 1
When x = 2 and y = 2, 2 ≯  2 + 1
When x = 2 and y = 4, 4 > 2 + 1
When x = 2 and y = 6, 6 > 2 + 1
∴ R₄ = {(1, 4), (1, 6), (2, 4), (2, 6)}
Domain (R₄) = {1, 2}
Range (R₄) = {4, 6}

(v) R₅ = {{x, y) / x + y = 3, x, y ∈ (0, 1, 2, 3)}
Here, x + y = 3
When x = 0, y = 3
When x = 1, y = 2
When x = 2, y = 1
When x = 3, y = 0
∴ R₅ = {(0, 3), (1, 2), (2, 1), (3, 0)}
Domain (R₅) = {0, 1, 2, 3}
Range (R₅) = {3, 2, 1, 0}

(vi) R₆ = {(a, b)/ a ∈ N, a < 6 and b = 4}
a ∈ N and a < 6
∴ a = 1, 2, 3, 4, 5 and b = 4
R₆ = {(1, 4), (2, 4), (3, 4), (4, 4), (5, 4)}
Domain (R₆) = {1, 2, 3, 4, 5}
Range (R₆) = {4}

(vii) R₇ = {(a, b) / a, b ∈ N, a + b = 6}
Here, a + b = 6
When a = 1, b = 5
When a = 2, b = 4
When a = 3, b = 3
When a = 4, b = 2
When a = 5, b = 1
∴ R₇ = {(1, 5), (2, 4), (3, 3), (4, 2), (5, 1)}
Domain (R₇) = {1, 2, 3, 4, 5}
Range (R₇) = {5, 4, 3, 2, 1}

(viii) R₈ = {(a, b) / b = a + 2, a ∈ Z, 0 < a < 5}
Here, b = a + 2
When a = 1, b = 3
When a = 2, b = 4
When a = 3, b = 5
When a = 4, b = 6
∴ R₈ = {(1, 3), (2, 4), (3, 5), (4, 6)}
Domain (R₈) = {1, 2, 3, 4}
Range (R₈) = {3, 4, 5, 6}

Question 9.
Identify which of the following relations are reflexive, symmetric, and transitive.

Solution:

(i) Given, R = {(a, b): a, b ∈ Z, a – b is an integer}
Let a ∈ Z, then a – a ∈ Z
∴ (a, a) ∈ R
∴ R is reflexive.
Let (a, b) ∈ R
∴ a – b ∈ Z
∴ -(a – b) ∈ Z, i.e., b – a ∈ Z
∴ (b, a) ∈ R
∴ R is symmetric.
Let (a, b) and (b, c) ∈ R
∴ a – b ∈ Z and b – c ∈ Z
∴ (a – b) + (b – c) ∈ Z
∴ a – c ∈ Z
∴ (a, c) ∈ R
∴ R is transitive.

(ii) Given, R = {(a, b) : a, b ∈ N, a + b is even}
Let a ∈ N, then a + a = 2a, which is even.
∴ (a, a) ∈ R
∴ R is reflexive.
Let (a, b) ∈ R
∴ a + b is even
∴ b + a is even
∴ (b, a) ∈ R
∴ R is symmetric.
Let (a, b) and (b, c) ∈ R
∴ a + b and b + c is even
Let a + b = 2x and b + c = 2y for x, y ∈ N
∴ (a + b) + (b + c) = 2x + 2y
∴ a + 2b + c = 2(x + y)
∴ a + c = 2(x + y) – 2b = 2(x + y – b)
∴ a + c is even ……..[∵ x, y, b ∈ N, x + y – b ∈ N]
∴ (a, c) ∈ R
∴ R is transitive.

(iii) Given, R = {(a, b) : a, b ∈ N, a divides b}
Let a ∈ N, then a divides a.
∴ (a, a) ∈ R
∴ R is reflexive.
Let a = 2 and b = 8, then 2 divides 8
∴ (a, b) ∈ R
But 8 does not divide 2.
∴ (b, a) ∉ R
∴ R is not symmetric.
Let (a, b) and (b, c) ∈ R
∴ a divides b and b divides c.
Let b = ax and c = by for x, y ∈ N.
∴ c = (ax) y = a(xy)
i.e., a divides c.
∴ (a, c) ∈ R
∴ R is transitive.

(iv) Given, R = {(a, b) : a, b ∈ N, a² – 4ab + 3b² = 0}
Let a ∈ N, then a² – 4aa + 3a² = a² – 4a² + 3a² = 0
∴ (a, a) ∈ R
∴ R is reflexive.
Let a = 3 and b = 1,
then a² – 4ab + 3b² = 9 – 12 + 3 = 0
∴ (a, b) ∈ R
Consider, b² – 4ba + 3a² = 1 – 12 + 9 = -2 ≠ 0
∴ (b, a) ∉ R
∴ R is not symmetric.
Let a = 3, b = 1 and c = \(\frac{1}{3}\),
then a² – 4ab + 3b² = 9 – 12 + 3 = 0 and
b² – 4bc + 3c² = 1 – \(\frac{4}{3}\) + \(\frac{1}{3}\) = 1 – 1 = 0
∴ we get (a, b) and (b, c) ∈ R.
Consider, a² – 4ac + 3c² = 9 – 4 + \(\frac{1}{3}\) = \(\frac{16}{3}\) ≠ 0
∴ (a, c) ∉ R
∴ R is not transitive.

(v) Given, R = {(a, b) : a is sister of b and a, b ∈ G = Set of girls}
Let a ∈ G, then ‘a’ cannot be a sister of herself.
∴ (a, a) ∉ R
∴ R is not reflexive.
Let (a, b) ∈ R
∴ ‘a’ is a sister of ‘b’.
∴ ‘b’ is a sister of ‘a’.
∴ (b, c) ∈ R
∴ R is symmetric.
Let (a, b) and (b, c) ∈ R
∴ ‘a’ is a sister of ‘b’ and ‘b’ is a sister of ‘c’
∴ ‘a’ is a sister of ‘c’.
∴ (a, c) ∈ R
∴ R is transitive.

(vi) Given, R = {(a, b) : Line a is perpendicular to line b in a plane}
Let a be any line in the plane, then a cannot be perpendicular to itself.
∴ (a, a) ∉ R
∴ R is not reflexive.
Let (a, b) ∈ R
∴ a is perpendicular to b.
∴ b is perpendicular to a.
∴ (b, a) ∈ R.
∴ R is symmetric.
Let (a, b) and (b, c) ∈ R.
∴ a is perpendicular to b and b is perpendicular to c.
∴ a is parallel to c.
∴ (a, c) ∉ R
∴ R is not transitive.

(vii) Given, R = {(a, b) : a, b ∈ R, a < b}
Let a ∈ R, then a ≮ a.
∴ (a, a) ∉ R
∴ R is not reflexive.
Let a = 1 and b = 2, then 1 < 2
∴ (a, b) ∈ R
But 2 ≮ 1
∴ (b, a) ∉ R
∴ R is not symmetric.
Let (a, b) and (b, c) ∈ R
∴ a < b and b < c
∴ a < c
∴ (a, c) ∈ R
∴ R is transitive.

(viii) Given, R = {(a, b) : a, b ∈ R, a ≤ b³}
Let a = -3, then a³ = -27.
Here, a ≮ a
∴ (a, a) ∉ R
∴ R is not reflexive.
Let a = 2 and b = 9, then b³ = 729
Here, a < b³
∴ (a, b) ∈ R
Consider, a3 = 8
Here, b ≮ a³
∴ (b, a) ∉ R
∴ R is not symmetric.
Let a = 10, b = 3, c = 2,
then b³ = 27 and c³ = 8
Here, a < b³ and b < c³.
∴ (a, b) and (b, c) ∈ R
But a ≮ c³
∴ (a, c) ∉ R.
∴ R is not transitive.

Balbharti Maharashtra State Board 11th Maths Book Solutions Pdf Chapter 3 Permutations and Combination Ex 3.4 Questions and Answers.

Maharashtra State Board 11th Maths Solutions Chapter 3 Permutations and Combination Ex 3.4

Question 1.
Find the number of permutations of letters in each of the following words.
(i) DIVYA
(ii) SHANTARAM
(iii) REPRESENT
(iv) COMBINE
(v) BAL BHARATI
Solution:
(i) There are 5 distinct letters in the word DIVYA.
∴ Number of permutations of the letters of the word DIVYA = 5! = 120

(ii) There are 9 letters in the word SHANTARAM in which ‘A’ is repeated 3 times.
∴ Number of permutations of the letters of the word SHANTARAM = \(\frac{9 !}{3 !}\)
= 9 × 8 × 7 × 6 × 5 × 4
= 60480

(iii) There are 9 letters in the word REPRESENT in which ‘E’ is repeated 3 times and ‘R’ is repeated 2 times.
∴ Number of permutations of the letters of the word REPRESENT = \(\frac{9 !}{3 ! 2 !}\)
= \(\frac{9 \times 8 \times 7 \times 6 \times 5 \times 4}{2}\)
= 30240

(iv) There are 7 distinct letters in the word COMBINE.
∴ Number of permutations of the letters of the word COMBINE = 7! = 5040

(v) There are 10 letters in the word BALBHARATI in which ‘B’ is repeated 2 times and ‘A’ is repeated 3 times.
∴ Number of permutations of the letters of the word BALBHARATI = \(\frac{10 !}{2 ! 3 !}\)
= \(\frac{10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}{2 \times 3 \times 2}\)
= 302400

Question 2.
You have 2 identical books on English, 3 identical books on Hindi, and 4 identical books on Mathematics. Find the number of distinct ways of arranging them on a shelf.
Solution:
There are total 9 books to be arranged on a shelf.
Out of these 9 books, 2 books on English, 3 books on Hindi and 4 books on mathematics are identical.
∴ Total number of arrangements possible = \(\frac{9 !}{2 ! 3 ! 4 !}\)
= \(\frac{9 \times 8 \times 7 \times 6 \times 5 \times 4 !}{2 \times 3 \times 2 \times 4 !}\)
= 9 × 4 × 7 × 5
= 1260

Question 3.
A coin is tossed 8 times. In how many ways can we obtain (a) 4 heads and 4 tails? (b) at least 6 heads?
Solution:
A coin is tossed 8 times. All heads are identical and all tails are identical.
(a) 4 heads and 4 tails are to be obtained.
∴ Number of ways it can be obtained = \(\frac{8 !}{4 ! 4 !}\)
= \(\frac{8 \times 7 \times 6 \times 5}{4 \times 3 \times 2}\)
= 70

(b) At least 6 heads are to be obtained.
∴ Outcome can be (6 heads and 2 tails) or (7 heads and 1 tail) or (8 heads)
∴ Number of ways it can be obtained = \(\frac{8 !}{6 ! 2 !}+\frac{8 !}{7 ! 1 !}+\frac{8 !}{8 !}\)
= \(\frac{8 \times 7}{2}\) + 8 + 1
= 28 + 8 + 1
= 37

Question 4.
A bag has 5 red, 4 blue, and 4 green marbles. If all are drawn one by one and their colours are recorded, how many different arrangements can be found?
Solution:
There is a total of 13 marbles in a bag.
Out of these 5 are Red, 4 Blue, and 4 are Green marbles.
All balls of the same colour are taken to be identical.
∴ Required number of arrangements = \(\frac{13 !}{5 ! 4 ! 4 !}\)

Question 5.
Find the number of ways of arranging letters of the word MATHEMATICAL. How many of these arrangements have all vowels together?
Solution:
There are 12 letters in the word MATHEMATICAL in which ‘M’ is repeated 2 times, ‘A’ repeated 3 times and ‘T’ repeated 2 times.
∴ Required number of arrangements = \(\frac{12 !}{2 ! 3 ! 2 !}\)
When all the vowels,
i.e., ‘A’, ‘A’, ‘A’, ‘E’, ‘I’ are to be kept together.
Let us consider them as one unit.
Number of arrangements of these vowels among themselves = \(\frac{5 !}{3 !}\) ways.
This unit is to be arranged with 7 other letters in which ‘M’ and ‘T’ repeated 2 times each.
∴ Number of such arrangements = \(\frac{8 !}{2 ! 2 !}\)
∴ Required number of arrangements = \(\frac{8 ! \times 5 !}{2 ! 2 ! 3 !}\)

Question 6.
Find the number of different arrangements of letters in the word MAHARASHTRA. How many of these arrangements have (a) letters R and H never together? (b) all vowels together?
Solution:
There are 11 letters in the word MAHARASHTRA in which ‘A’ is repeated 4 times, ‘H’ repeated 2 times, and ‘R’ repeated 2 times.
∴ Total number of words can be formed = \(\frac{11 !}{4 ! 2 ! 2 !}\)

(a) When letters R and H are never together.
Other than 2R, 2H there are 4A, 1S, 1T, 1M.
These letters can be arranged in \(\frac{7 !}{4 !}\) ways = 210.
These seven letters create 8 gaps in which 2R, 2H are to be arranged.
Number of ways to do = \(\frac{{ }^{8} \mathrm{P}_{4}}{2 ! 2 !}\) = 420
Required number of arrangements = 210 × 420 = 88200.

(b) When all vowels are together.
There are 4 vowels in the word MAHARASHTRA, i.e., A, A, A, A.
Let us consider these 4 vowels as one unit, which can be arranged among themselves in \(\frac{4 !}{4 !}\) = 1 way.
This unit is to be arranged with 7 other letters in which ‘H’ is repeated 2 times, ‘R’ is repeated 2 times.
∴ Total number of arrangements = \(\frac{8 !}{2 ! 2 !}\)

Question 7.
How many different words are formed if the letter R is used thrice and letters S and T are used twice each?
Solution:
To find the number of different words when ‘R’ is taken thrice, ‘S’ is taken twice and ‘T’ is taken twice.
∴ Total number of letters available = 7, of which ‘S’ and ‘T’ repeat 2 times each, ‘R’ repeats 3 times.
∴ Required number of words = \(\frac{7 !}{2 ! 2 ! 3 !}\)
= \(\frac{7 \times 6 \times 5 \times 4 \times 3 !}{2 \times 1 \times 2 \times 1 \times 3 !}\)
= 7 × 6 × 5
= 210

Question 8.
Find the number of arrangements of letters in the word MUMBAI so that the letter B is always next to A.
Solution:
There are 6 letters in the word MUMBAI.
These letters are to be arranged in such a way that ‘B’ is always next to ‘A’.
Let us consider AB as one unit.
This unit with the other 4 letters in which ‘M’ repeats twice is to be arranged.
∴ Required number of arrangements = \(\frac{5 !}{2 !}\)
= \(\frac{5 \times 4 \times 3 \times 2 !}{2 !}\)
= 60

Question 9.
Find the number of arrangements of letters in the word CONSTITUTION that begin and end with N.
Solution:
There are 12 letters in the word CONSTITUTION, in which ‘O’, ‘N’, ‘I’ repeat two times each, ‘T’ repeats 3 times.
When the arrangement starts and ends with ‘N’,
other 10 letters can be arranged between two N,
in which ‘O’ and ‘I’ repeat twice each and ‘T’ repeats 3 times.
∴ Required number of arrangements = \(\frac{10 !}{2 ! 2 ! 3 !}\)

Question 10.
Find the number of different ways of arranging letters in the word ARRANGE. How many of these arrangements do not have the two R’s and two A’s together?
Solution:
There are 7 letters in the word ARRANGE in which ‘A’ and ‘R’ repeat 2 times each.
∴ Number of ways to arrange the letters of word ARRANGE = \(\frac{7 !}{2 ! 2 !}\) = 1260
Consider the words in which 2A are together and 2R are together.
Let us consider 2A as one unit and 2R as one unit.
These two units with remaining 3 letters can be arranged in = \(\frac{5 !}{2 ! 2 !}\) = 30 ways.
Number of arrangements in which neither 2A together nor 2R are together = 1260 – 30 = 1230

Question 11.
How many distinct 5 digit numbers can be formed using the digits 3, 2, 3, 2, 4, 5.
Solution:
5 digit numbers are to be formed from 2, 3, 2, 3, 4, 5.
Case I: Numbers formed from 2, 2, 3, 4, 5 OR 2, 3, 3, 4, 5
Number of such numbers = \(\frac{5 !}{2 !}+\frac{5 !}{2 !}\) = 5! = 120
Case II: Numbers are formed from 2, 2, 3, 3 and any one of 4 or 5
Number of such numbers = \(\frac{5 !}{2 ! 2 !}+\frac{5 !}{2 ! 2 !}\) = 60
Required number of numbers = 120 + 60 = 180

Question 12.
Find the number of distinct numbers formed using the digits 3, 4, 5, 6, 7, 8, 9, so that odd positions are occupied by odd digits.
Solution:
A number is to be formed with digits 3, 4, 5, 6, 7, 8, 9 such that odd digits always occupy the odd places.
There are 4 odd digits, i.e. 3, 5, 7, 9.
∴ They can be arranged at 4 odd places among themselves in 4! = 24 ways.
There are 3 even digits, i.e. 4, 6, 8.
∴ They can be arranged at 3 even places among themselves in 3! = 6 ways.
∴ Required number of numbers formed = 24 × 6 = 144

Question 13.
How many different 6-digit numbers can be formed using digits in the number 659942? How many of them are divisible by 4?
Solution:
A 6-digit number is to be formed using digits of 659942, in which 9 repeats twice.
∴ Required number of numbers formed = \(\frac{6 !}{2 !}\)
= \(\frac{6 \times 5 \times 4 \times 3 \times 2 !}{2 !}\)
= 360
A 6-digit number is to be formed using the same digits that are divisible by 4.
For a number to be divisible by 4, the last two digits should be divisible by 4,
i.e. 24, 52, 56, 64, 92 or 96.
Case I: When the last two digits are 24, 52, 56 or 64.
As the digit 9 repeats twice in the remaining four numbers, the number of arrangements = \(\frac{4 !}{2 !}\) = 12
∴ 6-digit numbers that are divisible by 4 so formed are 12 + 12 + 12 + 12 = 48.
Case II: When the last two digits are 92 or 96.
As each of the remaining four numbers are distinct, the number of arrangements = 4! = 24
∴ 6-digit numbers that are divisible by 4 so formed are 24 + 24 = 48.
∴ Required number of numbers framed = 48 + 48 = 96

Question 14.
Find the number of distinct words formed from letters in the word INDIAN. How many of them have the two N’s together?
Solution:
There are 6 letters in the word INDIAN in which I and N are repeated twice.
Number of different words that can be formed using the letters of the word INDIAN = \(\frac{6 !}{2 ! 2 !}\)
= \(\frac{6 \times 5 \times 4 \times 3 \times 2 !}{2 \times 2 !}\)
= 180
When two N’s are together.
Let us consider the two N’s as one unit.
They can be arranged with 4 other letters in \(\frac{5 !}{2 !}\)
= \(\frac{5 \times 4 \times 3 \times 2 !}{2 !}\)
= 60 ways.
∴ 2 N can be arranged in \(\frac{2 !}{2 !}\) = 1 way.
∴ Required number of words = 60 × 1 = 60

Question 15.
Find the number of different ways of arranging letters in the word PLATOON if (a) the two O’s are never together. (b) consonants and vowels occupy alternate positions.
Solution:
There are 7 letters in the words PLATOON in which ‘O’ repeat 2 times.
(a) When the two O’s are never together.
Let us arrange the other 5 letters first, which can be done in 5! = 120 ways.
The letters P, L, A, T, N create 6 gaps, in which O’s are arranged.
Two O’s can take their places in ⁶P₂ ways.
But ‘O’ repeats 2 times.
∴ Two O’s can be arranged in \(\frac{{ }^{6} \mathrm{P}_{2}}{2 !}\)
= \(\frac{\frac{6 !}{(6-2) !}}{2 !}\)
= \(\frac{6 \times 5 \times 4 !}{4 ! \times 2 \times 1}\)
= 3 × 5
= 15 ways
∴ Required number of arrangements = 120 × 15 = 1800

(b) When consonants and vowels occupy alternate positions.
There are 4 consonants and 3 vowels in the word PLATOON.
∴ At odd places, consonants occur and at even places, vowels occur.
4 consonants can be arranged among themselves in 4! ways.
3 vowels in which O occurs twice and A occurs once.
∴ They can be arranged in \(\frac{3 !}{2 !}\) ways.
Now, vowels and consonants should occupy alternate positions.
∴ Required number of arrangements = 4! × \(\frac{3 !}{2 !}\)
= 4 × 3 × 2 × \(\frac{3 \times 2 !}{2 !}\)
= 72

Balbharti Maharashtra State Board 11th Maths Book Solutions Pdf Chapter 3 Permutations and Combination Ex 3.3 Questions and Answers.

Maharashtra State Board 11th Maths Solutions Chapter 3 Permutations and Combination Ex 3.3

Question 1.
Find n, if ⁿP₆ : ⁿP₃ = 120 : 1.
Solution:
ⁿP₆ : ⁿP₃ = 120 : 1

∴ (n – 3) (n – 4) (n – 5) = 120
∴ (n – 3) (n – 4) (n – 5) = 6 × 5 × 4
Comparing on both sides, we get
n – 3 = 6
∴ n = 9

Question 2.
Find m and n, if ^(m+n)P₂ = 56 and ^(m-n)P₂ = 12.
Solution:
^(m+n)P₂ = 56

(m + n) (m + n – 1) = 56
Let m + n = t
t(t – 1) = 56
t² – t – 56 = 0
(t – 8) (t + 7) = 0
t = 8 or t = -7
m + n = 8 or m + n = -7
But m + n ≠ -7
∴ m + n = 8 ……(i)
Also, ^(m-n)P₂ = 12

(m – n) (m – n – 1) = 12
Let m – n = a
a(a – 1) = 12
a² – a – 12 = 0
(a – 4)(a + 3) = 0
a = 4 or a = -3
m – n = 4 or m – n = -3
But m – n ≠ -3
∴ m – n = 4 ……(ii)
Adding (i) and (ii), we get
2m = 12
∴ m = 6
Substituting m = 6 in (ii), we get
6 – n = 4
∴ n = 2

Question 3.
Find r, if ¹²P_r-2 : ¹¹P_r-1 = 3 : 14.
Solution:

(14 – r)(13 – r) = 8 × 7
Comparing on both sides, we get
14 – r = 8
∴ r = 6

Question 4.
Show that (n + 1) (ⁿP_r) = (n – r + 1) [⁽ⁿ⁺¹⁾P_r]
Solution:

Question 5.
How many 4 letter words can be formed using letters in the word MADHURI, if (a) letters can be repeated (b) letters cannot be repeated.
Solution:
There are 7 letters in the word MADHURI.
(a) A 4 letter word is to be formed from the letters of the word MADHURI and repetition of letters is allowed.
∴ 1st letter can be filled in 7 ways.
2nd letter can be filled in 7 ways.
3rd letter can be filled in 7 ways.
4th letter can be filled in 7 ways.
∴ Total no. of ways a 4-letter word can be formed = 7 × 7 × 7 × 7 = 2401

(b) When repetition of letters is not allowed, the number of 4-letter words formed from the letters of the word MADHURI is ⁷P₄ = \(\frac{7 !}{(7-4) !}=\frac{7 \times 6 \times 5 \times 4 \times 3 !}{3 !}\) = 840

Question 6.
Determine the number of arrangements of letters of the word ALGORITHM if
(a) vowels are always together.
(b) no two vowels are together.
(c) consonants are at even positions.
(d) O is the first and T is the last letter.
Solution:
There are 9 letters in the word ALGORITHM.
(a) When vowels are always together.
There are 3 vowels in the word ALGORITHM (i.e., A, I, O).
Let us consider these 3 vowels as one unit.
This unit with 6 other letters is to be arranged.
∴ The number of arrangement = ⁷P₇ = 7! = 5040
3 vowels can be arranged among themselves in ³P₃ = 3! = 6 ways.
∴ Required number of arrangements = 7! × 3!
= 5040 × 6
= 30240

(b) When no two vowels are together.
There are 6 consonants in the word ALGORITHM,
they can be arranged among themselves in ⁶P₆ = 6! = 720 ways.
Let consonants be denoted by C.
_C _C_ C _C_C_C
There are 7 places marked by ‘_’ in which 3 vowels can be arranged.
∴ Vowels can be arranged in ⁷P₃ = \(\frac{7 !}{(7-3) !}=\frac{7 \times 6 \times 5 \times 4 !}{4 !}\) = 210 ways.
∴ Required number of arrangements = 720 × 210 = 151200

(c) When consonants are at even positions.
There are 4 even places and 6 consonants in the word ALGORITHM.
∴ 6 consonants can be arranged at 4 even positions in 6P4 = \(\frac{6 !}{(6-4) !}=\frac{6 \times 5 \times 4 \times 3 \times 2 !}{2 !}\) = 360 ways.
Remaining 5 letters (3 vowels and 2 consonants) can be arranged in odd position in ⁵P₅ = 5! = 120 ways.
∴ Required number of arrangements = 360 × 120 = 43200

(d) When O is the first and T is the last letter.
All the letters of the word ALGORITHM are to be arranged among themselves such that arrangement begins with O and ends with T.
∴ Position of O and T are fixed.
∴ Other 7 letters can be arranged between O and T among themselves in ⁷P₇ = 7! = 5040 ways.
∴ Required number of arrangements = 5040

Question 7.
In a group photograph, 6 teachers and principal are in the first row and 18 students are in the second row. There are 12 boys and 6 girls among the students. If the middle position is reserved for the principal and if no two girls are together, find the number of arrangements.
Solution:
In 1st row, 6 teachers can be arranged among themselves in ⁶P₆ = 6! ways.
In the 2nd row, 12 boys can be arranged among themselves in ¹²P₁₂ = 12! ways.
No two girls are together.
So, there are 13 places formed by 12 boys in which 6 girls occupy any 6 places in ¹³P₆ ways.
∴ Required number of arrangements

Question 8.
Find the number of ways so that letters of the word HISTORY can be arranged as
(a) Y and T are together
(b) Y is next to T
(c) there is no restriction
(d) begin and end with a vowel
(e) end in ST
(f) begin with S and end with T
Solution:
There are 7 letters in the word HISTORY
(a) When ‘Y’ and ‘T’ are together.
Let us consider ‘Y’ and ‘T’ as one unit.
This unit with other 5 letters are to be arranged.
∴ The number of arrangement of one unit and 5 letters = ⁶P₆ = 6! = 720
Also, ‘Y’ and ‘T’ can be arranged among themselves in ²P₂ = 2! = 2 ways.
∴ A total number of arrangements when Y and T are always together = 6! × 2!
= 120 × 2
= 1440

(b) When ‘Y’ is next to ‘T’.
Let us take this (‘Y’ next to ‘T’) as one unit.
This unit with 5 other letters is to be arranged.
∴ The number of arrangements of 5 letters and one unit = ⁶P₆ = 6! = 720
Also, ‘Y’ has to be always next to ‘T’.
∴ They can be arranged among themselves in 1 way only.
∴ Total number of arrangements possible when Y is next to T = 720 × 1 = 720

(c) When there is no restriction.
7 letters can be arranged among themselves in ⁷P₇ = 7! ways.
∴ The total number of arrangements possible if there is no restriction = 7!

(d) When begin and end with a vowel.
There are 2 vowels in the word HISTORY.
All other letters of the word HISTORY are to be arranged between 2 vowels such that the arrangement begins and ends with a vowel.
The other 5 letters can be filled between the two vowels in ⁵P₅ = 5! = 120 ways.
Also, 2 vowels can be arranged among themselves at first and last places in ²P₂ = 2! = 2 ways.
∴ Total number of arrangements when the word begins and ends with vowel = 120 × 2 = 240

(e) When a word ends in ST.
As the arrangement ends with ST,
the remaining 5 letters can be arranged among themselves in ⁵P₅ = 5! = 120 ways.
∴ Total number of arrangements when the word ends with ST = 120

(f) When a word begins with S and ends with T.
As arrangement begins with S and ends with T,
the remaining 5 letters can be arranged between S and T among themselves in ⁵P₅ = 5! = 120 ways.
Total number of arrangements when the word begins with S and ends with T = 120

Question 9.
Find the number of arrangements of the letters in the word SOLAPUR so that consonants and vowels are placed alternately.
Solution:
There are 4 consonants S, L, P, R, and 3 vowels A, O, U in the word SOLAPUR.
Consonants and vowels are to be alternated.
∴ Vowels must occur in even places and consonants in odd places.
∴ 3 vowels can be arranged at 3 even places in ³P₃ = 3! = 6 ways.
Also, 4 consonants can be arranged at 4 odd places in ⁴P₄ = 4! = 24 ways.
Required number of arrangements = 6 × 24 = 144

Question 10.
Find the number of 4-digit numbers that can be formed using the digits 1, 2, 4, 5, 6, 8 if
(a) digits can be repeated.
(b) digits cannot be repeated.
Solution:
(a) A 4 digit number is to be made from the digits 1, 2, 4, 5, 6, 8 such that digits can be repeated.
∴ Unit’s place digit can be filled in 6 ways.
10’s place digit can be filled in 6 ways.
100’s place digit can be filled in 6 ways.
1000’s place digit can be filled in 6 ways.
∴ Total number of numbers that can be formed = 6 × 6 × 6 × 6 = 1296

(b) A 4 different digit number is to be made from the digits 1, 2, 4, 5, 6, 8 without repetition of digits.
∴ 4 different digits are to be arranged from 6 given digits which can be done in ⁶P₄ ways.
∴ Total number of numbers that can be formed
= \(\frac{6 !}{(6-4) !}\)
= \(\frac{6 \times 5 \times 4 \times 3 \times 2 !}{2 !}\)
= 360

Question 11.
How many numbers can be formed using the digits 0, 1, 2, 3, 4, 5 without repetition so that resulting numbers are between 100 and 1000?
Solution:
A number between 100 and 1000 that can be formed from the digits 0, 1, 2, 3, 4, 5 is of 3 digits, and repetition of digits is not allowed.
∴ 100’s place can be filled in 5 ways as it is a non-zero number.
10’s place digits can be filled in 5 ways.
Unit’s place digit can be filled in 4 ways.
∴ Total number of ways the number can be formed = 5 × 5 × 4 = 100

Question 12.
Find the number of 6-digit numbers using the digits 3, 4, 5, 6, 7, 8 without repetition. How many of these numbers are (a) divisible by 5 (b) not divisible by 5
Solution:
A number of 6 different digits is to be formed from the digits 3, 4, 5, 6, 7, 8 which can be done in ⁶P₆ = 6! = 720 ways.
(a) If the number is to be divisible by 5,
the unit’s place digit can be 5 only.
∴ it can be arranged in 1 way only.
The other 5 digits can be arranged among themselves in ⁵P₅ = 5! = 120 ways.
∴ Required number of numbers divisible by 5 = 1 × 120 = 120

(b) If the number is not divisible by 5,
unit’s place can be any digit from 3, 4, 6, 7, 8.
∴ it can be arranged in 5 ways.
Other 5 digits can be arranged in ⁵P₅ = 5! = 120 ways.
∴ Required number of numbers not divisible by 5 = 5 × 120 = 600

Question 13.
A code word is formed by two different English letters followed by two non-zero distinct digits. Find the number of such code words. Also, find the number of such code words that end with an even digit.
Solution:
There is a total of 26 alphabets.
A code word contains 2 English alphabets.
∴ 2 alphabets can be filled in ²⁶P₂
= \(\frac{26 !}{(26-2) !}\)
= \(\frac{26 \times 25 \times 24 !}{24 !}\)
= 650 ways.
Also, alphabets to be followed by two distinct non-zero digits from 1 to 9 which can be filled in
⁹P₂ = \(\frac{9 !}{(9-2) !}=\frac{9 \times 8 \times 7 !}{7 !}\) = 72 ways.
∴ Total number of a code words = 650 × 72 = 46800.
To find the number of codewords end with an even integer.
2 alphabets can be filled in 650 ways.
The digit in the unit’s place should be an even number between 1 to 9, which can be filled in 4 ways.
Also, 10’s place can be filled in 8 ways.
∴ Total number of codewords = 650 × 4 × 8 = 20800

Question 14.
Find the number of ways in which 5 letters can be posted in 3 post boxes if any number of letters can be posted in a post box.
Solution:
There are 5 letters and 3 post boxes and any number of letters can be posted in all three post boxes.
∴ Each letter can be posted in 3 ways.
∴ Total number of ways 5 letters can be posted = 3 × 3 × 3 × 3 × 3 = 243

Question 15.
Find the number of arranging 11 distinct objects taken 4 at a time so that a specified object (a) always occurs (b) never occurs
Solution:
There are 11 distinct objects and 4 are to be taken at a time.
(a) The number of permutations of n distinct objects, taken r at a time, when one particular object will always occur is \(\mathbf{r} \times{ }^{(\mathbf{n}-1)} \mathbf{P}_{(\mathbf{r}-1)}\)

∴ In 2880 permutations of 11 distinct objects, taken 4 at a time, one particular object will always occur.

(b) When one particular object will not occur, then 4 objects are to be arranged from 10 objects which can be done in ¹⁰P₄ = 10 × 9 × 8 × 7 = 5040 ways.
∴ In 5040 permutations of 11 distinct objects, taken 4 at a time, one particular object will never occur.

Question 16.
In how many ways can 5 different books be arranged on a shelf if
(i) there are no restrictions
(ii) 2 books are always together
(iii) 2 books are never together
Solution:
(i) 5 books arranged in ⁵P₅ = 5! = 120 ways.

(ii) 2 books are together.
Let us consider two books as one unit. This unit with the other 3 books can be arranged in ⁴P₄ = 4! = 24 ways.
Also, two books can be arranged among themselves in ²P₂ = 2 ways.
∴ Required number of arrangements = 24 × 2 = 48

(iii) Say books are B₁, B₂, B₃, B₄, B₅ are to be arranged with B₁, B₂ never together.
B₃, B₄, B₅ can be arranged among themselves in ³P₃ = 3! = 6 ways.
B₃, B₄, B₅ create 4 gaps in which B₁, B₂ are arranged in ⁴P₂ = 4 × 3 = 12 ways.
∴ Required number of arrangements = 6 × 12 = 72

Question 17.
3 boys and 3 girls are to sit in a row. How many ways can this be done if
(i) there are no restrictions.
(ii) there is a girl at each end.
(iii) boys and girls are at alternate places.
(iv) all-boys sit together.
Solution:
3 boys and 3 girls are to be arranged in a row.
(i) When there are no restrictions.
∴ Required number of arrangements = 6! = 720

(ii) When there is a girl at each end.
3 girls can be arranged at two ends in
³P₂ = \(\frac{3 !}{1 !}\) = 3 × 2 = 6 ways.
And remaining 1 girl and 3 boys can be arranged between the two girls in ⁴P₄ = 4! = 24 ways.
∴ Required number of arrangements = 6 × 24 = 144

(iii) Boys and girls are at alternate places.
We can first arrange 3 girls among themselves in ³P₃ = 3! = 6 ways.
Let girls be denoted by G.
G – G – G –
There are 3 places marked by ‘-’ where 3 boys can be arranged in 3! = 6 ways.
∴ Total number of such arrangements = 6 × 6 = 36
OR
Similarly, we can first arrange 3 boys in 3! = 6 ways
and then arrange 3 girls alternately in 3! = 6 ways.
∴ Total number of such arrangements = 6 × 6 = 36
∴ Required number of arrangements = 36 + 36 = 72

(iv) All boys sit together.
Let us consider all boys as one group.
This one group with the other 3 girls can be arranged ⁴P₄ = 4! = 24 ways.
Also, 3 boys can be arranged among themselves in ³P₃ = 3! = 6 ways.
∴ Required number of arrangements = 24 × 6 = 144

Balbharti Maharashtra State Board 11th Maths Book Solutions Pdf Chapter 3 Permutations and Combination Ex 3.2 Questions and Answers.

Maharashtra State Board 11th Maths Solutions Chapter 3 Permutations and Combination Ex 3.2

Question 1.
Evaluate:
(i) 8!
Solution:
8!
= 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1
= 40320

(ii) 10!
Solution:
10!
= 10 × 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1
= 3628800

(iii) 10! – 6!
Solution:
10! – 6!
= 10 × 9 × 8 × 7 × 6! – 6!
= 6! (10 × 9 × 8 × 7 – 1)
= 6! (5040 – 1)
= 6 × 5 × 4 × 3 × 2 × 1 × 5039
= 3628080

(iv) (10 – 6)!
Solution:
(10 – 6)!
= 4!
= 4 × 3 × 2 × 1
= 24

Question 2.
Compute:
(i) \(\frac{12 !}{6 !}\)
Solution:
\(\frac{12 !}{6 !}=\frac{12 \times 11 \times 10 \times 9 \times 8 \times 7 \times 6 !}{6 !}\)
= 12 × 11 × 10 × 9 × 8 × 7
= 665280

(ii) \(\left(\frac{12}{6}\right) !\)
Solution:
\(\left(\frac{12}{6}\right) !\)
= 2!
= 2 × 1
= 2

(iii) (3 × 2)!
Solution:
(3 × 2)!
= 6!
= 6 × 5 × 4 × 3 × 2 × 1
= 720

(iv) 3! × 2!
Solution:
3! × 2!
= 3 × 2 × 1 × 2 × 1
= 12

(v) \(\frac{9 !}{3 ! 6 !}\)
Solution:
\(\frac{9 !}{3 ! 6 !}=\frac{9 \times 8 \times 7 \times 6 !}{(3 \times 2 \times 1) \times 6 !}=84\)

(vi) \(\frac{6 !-4 !}{4 !}\)
Solution:
\(\frac{6 !-4 !}{4 !}=\frac{6 \times 5 \times 4 !-4 !}{4 !}=\frac{4 !(6 \times 5-1)}{4 !}=29\)

(vii) \(\frac{8 !}{6 !-4 !}\)
Solution:
\(\frac{8 !}{6 !-4 !}=\frac{8 \times 7 \times 6 \times 5 \times 4 !}{6 \times 5 \times 4 !-4 !}\)
= \(\frac{8 \times 7 \times 6 \times 5 \times 4 !}{4 !(6 \times 5-1)}\)
= \(\frac{1680}{29}\)
= 57.93

(viii) \(\frac{8 !}{(6-4) !}\)
Solution:
\(\frac{8 !}{(6-4) !}=\frac{8 !}{2 !}\)
= \(\frac{8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 !}{2 !}\)
= 20160

Question 3.
Write in terms of factorials
(i) 5 × 6 × 7 × 8 × 9 × 10
Solution:
5 × 6 × 7 × 8 × 9 × 10 = 10 × 9 × 8 × 7 × 6 × 5
Multiplying and dividing by 4!, we get
= \(\frac{10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 !}{4 !}\)
= \(\frac{10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}{4 !}\)
= \(\frac{10 !}{4 !}\)

(ii) 3 × 6 × 9 × 12 × 15
Solution:
3 × 6 × 9 × 12 × 15
= 3 × (3 × 2) × (3 × 3) × (3 × 4) × (3 × 5)
= (3⁵) (5 × 4 × 3 × 2 × 1)
= 3⁵ (5!)

(iii) 6 × 7 × 8 × 9
Solution:
6 × 7 × 8 × 9 = 9 × 8 × 7 × 6
Multiplying and dividing by 5!, we get
= \(\frac{9 \times 8 \times 7 \times 6 \times 5 !}{5 !}\)
= \(\frac{9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}{5 !}\)
= \(\frac{9 !}{5 !}\)

(iv) 5 × 10 × 15 × 20
Solution:
5 × 10 × 15 × 20
= (5 × 1) × (5 × 2) × (5 × 3) × (5 × 4)
= (5⁴) (4 × 3 × 2 × 1)
= (5⁴) (4!)

Question 4.
Evaluate: \(\frac{n !}{r !(n-r) !}\) for
(i) n = 8, r = 6
(ii) n = 12, r = 12
(iii) n = 15, r = 10
(iv) n = 15, r = 8
Solution:

Question 5.
Find n, if
(i) \(\frac{n}{8 !}=\frac{3}{6 !}+\frac{1 !}{4 !}\)
Solution:

(ii) \(\frac{n}{6 !}=\frac{4}{8 !}+\frac{3}{6 !}\)
Solution:

(iii) \(\frac{1 !}{n !}=\frac{1 !}{4 !}-\frac{4}{5 !}\)
Solution:

(iv) (n + 1)! = 42 × (n -1)!
Solution:
(n + 1)! = 42(n – 1)!
∴ (n + 1) n (n – 1)! = 42(n – 1)!
∴ n² + n = 42
∴ n² + n – 42 = 0
∴ (n + 7)(n – 6) = 0
∴ n = -7 or n = 6
But n ≠ -7 as n ∈ N
∴ n = 6

(v) (n + 3)! = 110 × (n + 1)!
Solution:
(n + 3)! = (110) (n + 1)!
∴ (n + 3)(n + 2)(n + 1)! = 110(n + 1)!
∴ (n + 3) (n + 2) = (11) (10)
Comparing on both sides, we get
n + 3 = 11
∴ n = 8

Question 6.
Find n, if:
(i) \(\frac{(17-n) !}{(14-n) !}=5 !\)
Solution:

∴ (17 – n) (16 – n) (15 – n) = 6 × 5 × 4
Comparing on both sides, we get
17 – n = 6
∴ n = 11

(ii) \(\frac{(15-n) !}{(13-n) !}=12\)
Solution:
\(\frac{(15-n) !}{(13-n) !}=12\)
∴ \(\frac{(15-n)(14-n)(13-n) !}{(13-n) !}=12\)
∴ (15 – n) (14 – n) = 4 × 3
Comparing on both sides, we get
∴ 15 – n = 4
∴ n = 11

(iii) \(\frac{n !}{3 !(n-3) !}: \frac{n !}{5 !(n-5) !}=5: 3\)
Solution:

∴ 12 = (n – 3)(n – 4)
(n – 3)(n – 4) = 4 × 3
Comparing on both sides, we get
n – 3 = 4
∴ n = 7

(iv) \(\frac{n !}{3 !(n-3) !}: \frac{n !}{5 !(n-7) !}=1: 6\)
Solution:

∴ 120 = (n – 3)(n – 4) (n – 5)(n – 6)
∴ (n – 3)(n – 4) (n – 5)(n – 6) = 5 × 4 × 3 × 2
Comparing on both sides, we get
n – 3 = 5
∴ n = 8

(v) \(\frac{(2 n) !}{7 !(2 n-7) !}: \frac{n !}{4 !(n-4) !}=24: 1\)
Solution:


(2n – 1)(2n – 3)(2n – 5) = \(\frac{24 \times 7 \times 6 \times 5}{16}\)
∴ (2n – 1)(2n – 3)(2n – 5) = 9 × 7 × 5
Comparing on both sides. We get
∴ 2n – 1 = 9
∴ n = 5

Question 7.
Show that \(\frac{n !}{r !(n-r) !}+\frac{n !}{(r-1) !(n-r+1) !}=\frac{(n+1) !}{r !(n-r+1) !}\)
Solution:

Question 8.
Show that \(\frac{9 !}{3 ! 6 !}+\frac{9 !}{4 ! 5 !}=\frac{10 !}{4 ! 6 !}\)
Solution:

Question 9.
Show that \(\frac{(2 n) !}{n !}\) = 2n (2n – 1)(2n – 3)…5.3.1
Solution:

Question 10.
Simplify
(i) \(\frac{(2 n+2) !}{(2 n) !}\)
Solution:

(ii) \(\frac{(n+3) !}{\left(n^{2}-4\right)(n+1) !}\)
Solution:

(iii) \(\frac{1}{n !}-\frac{1}{(n-1) !}-\frac{1}{(n-2) !}\)
Solution:

(iv) n[n! + (n – 1)!] + n²(n – 1)! + (n + 1)!
Solution:

(v) \(\frac{n+2}{n !}-\frac{3 n+1}{(n+1) !}\)
Solution:

(vi) \(\frac{1}{(n-1) !}+\frac{1-n}{(n+1) !}\)
Solution:

(vii) \(\frac{1}{n !}-\frac{3}{(n+1) !}-\frac{n^{2}-4}{(n+2) !}\)
Solution:

(viii) \(\frac{n^{2}-9}{(n+3) !}+\frac{6}{(n+2) !}-\frac{1}{(n+1) !}\)
Solution:

Balbharti Maharashtra State Board 11th Maths Book Solutions Pdf Chapter 3 Permutations and Combination Ex 3.1 Questions and Answers.

Maharashtra State Board 11th Maths Solutions Chapter 3 Permutations and Combination Ex 3.1

Question 1.
A teacher wants to select the class monitor in a class of 30 boys and 20 girls. In how many ways can the monitor be selected if the monitor must be a girl?
Solution:
There are 30 boys and 20 girls.
A teacher can select any boy as a class monitor from 30 boys in 30 different ways and he can select any girl as a class monitor from 20 girls in 20 different ways.
∴ by the fundamental principle of addition, the total number of ways a teacher can select a class monitor = 30 + 20 = 50
Hence, there are 50 different ways to select a class monitor.

Question 2.
A Signal is generated from 2 flags by putting one flag above the other. If 4 flags of different colours are available, how many different signals can be generated?
Solution:
A signal is generated from 2 flags and there are 4 flags of different colours available.
∴ 1st flag can be any one of the available 4 flags.
∴ It can be selected in 4 ways.
Now, 2nd flag is to be selected for which 3 flags are available for a different signal.
∴ 2nd flag can be anyone from these 3 flags.
∴ It can be selected in 3 ways.
∴ By using the fundamental principle of multiplication, total no. of ways a signal can be generated = 4 × 3 = 12
∴ 12 different signals can be generated.

Question 3.
How many two-letter words can be formed using letters from the word SPACE, when repetition of letters (i) is allowed (ii) is not allowed
Solution:
A two-letter word is to be formed out of the letters of the word SPACE.
(i) When repetition of the letters is allowed 1st letter can be selected in 5 ways.
2nd letter can be selected in 5 ways.
∴ By using the fundamental principle of multiplication, the total number of 2-letter words = 5 × 5 = 25

(ii) When repetition of the letters is not allowed 1st letter can be selected in 5 ways.
2nd letter can be selected in 4 ways.
∴ By using the fundamental principle of multiplication, the total number of 2-letter words = 5 × 4 = 20

Question 4.
How many three-digit numbers can be formed from the digits 0, 1, 3, 5, 6 if repetitions of digits (i) are allowed (ii) are not allowed?
Solution:
A three-digit number is to be formed from the digits 0, 1, 3, 5, 6.
(i) When repetition of digits is allowed,
100’s place digit should be a non-zero number.
Hence, it can be anyone from digits 1, 3, 5,6.
∴ 100’s place digit can be selected in 4 ways.
10’s and unit’s place digit can be zero and digits can be repeated.
∴ 10’s place digit can be selected in 5 ways and the unit’s place digit can be selected in 5 ways.
∴ By using the fundamental principle of multiplication, the total number of three-digit numbers = 4 × 5 × 5 = 100

(ii) When repetition of digits is not allowed,
100’s place digit should be a non-zero number.
Hence, it can be anyone from digits 1, 3, 5, 6.
∴ 100’s place digit can be selected in 4 ways.
10’s and unit’s place digit can be zero and digits can’t be repeated.
∴ 10’s place digit can be selected in 4 ways and the unit’s place digit can be selected in 3 ways.
∴ By using the fundamental principle of multiplication, the total number of two-digit numbers = 4 × 4 × 3 = 48

Question 5.
How many three-digit numbers can be formed using the digits 2, 3, 4, 5, 6 if digits can be repeated?
Solution:
A 3-digit number is to be formed from the digits 2, 3, 4, 5, 6 where digits can be repeated.
∴ Unit’s place digit can be selected in 5 ways.
10’s place digit can be selected in 5 ways.
100’s place digit can be selected in 5 ways.
∴ By using fundamental principle of multiplication, total number of 3-digit numbers = 5 × 5 × 5 = 125

Question 6.
A letter lock contains 3 rings and each ring contains 5 letters. Determine the maximum number of false trails that can be made before the lock is opened.
Solution:
A letter lock has 3 rings, each ring containing 5 different letters.
∴ A letter from each ring can be selected in 5 ways.
∴ By using the fundamental principle of multiplication, a total number of trials that can be made = 5 × 5 × 5 = 125.
Out of these 124 wrong attempts are made and in the 125th attempt, the lock gets opened.
∴ A maximum number of false trials = 124.

Question 7.
In a test, 5 questions are of the form ‘state, true or false. No student has got all answers correct. Also, the answer of every student is different. Find the number of students who appeared for the test.
Solution:
Every question can be answered in 2 ways. (True or False)
∴ By using the fundamental principle of multiplication, the total number of set of answers possible = 2 × 2 × 2 × 2 × 2 = 32.
Since One of them is the case where all questions are answered correctly,
The number of wrong answers = 32 – 1 = 31.
Since no student has answered all the questions correctly, the number of students who appeared for the test are 31.

Question 8.
How many numbers between 100 and 1000 have 4 in the unit’s place?
Solution:
Numbers between 100 and 1000 are 3-digit numbers.
A 3-digit number is to be formed from the digits 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, where the unit place digit is 4.
Unit’s place digit is 4.
∴ it can be selected in 1 way only.
10’s place digit can be selected in 10 ways.
For a 3-digit number, 100’s place digit should be a non-zero number.
∴ 100’s place digit can be selected in 9 ways.
∴ By using the fundamental principle of multiplication,
total numbers between 100 and 1000 which have 4 in the units place = 1 × 10 × 9 = 90.

Question 9.
How many numbers between 100 and 1000 have the digit 7 exactly once?
Solution:
Numbers between 100 and 1000 are 3-digit numbers.
A 3-digit number is to be formed from the digits 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, where exactly one of the digits is 7.
When 7 is in unit’s place:
Unit’s place digit is 7.
∴ it can be selected in 1 way only.
10’s place digit can be selected in 9 ways.
100’s place digit can be selected in 8 ways.
Total numbers which have 7 in unit’s place = 1 × 9 × 8 = 72.

When 7 is in 10’s place:
Unit’s place digit can be selected in 9 ways.
10’s place digit is 7.
∴ it can be selected in 1 way only.
100’s place digit can be selected in 8 ways.
∴ A total number of numbers which have 7 in 10’s place = 9 × 1 × 8 = 72.

When 7 is in 100’s place:
Unit’s place digit can be selected in 9 ways.
10’s place digit can be selected in 9 ways.
100’s place digit is 7.
∴ it can be selected in 1 way only.
∴ Total numbers which have 7 in 100’s place = 9 × 9 × 1 = 81.
∴ Total numbers between 100 and 1000 having digit 7 exactly once = 72 + 72 + 81 = 225

Question 10.
How many four-digit numbers will not exceed 7432 if they are formed using the digits 2, 3, 4, 7 without repetition?
Solution:
Between any set of digits, the greatest number is possible when digits are arranged in descending order.
∴ 7432 is the greatest number, formed from the digits 2, 3, 4, 7.
Since a 4-digit number is to be formed from the digits 2, 3, 4, 7, where repetition of the digit is not allowed,
1000’s place digit can be selected in 4 ways,
100’s place digit can be selected in 3 ways,
10’s place digit can be selected in 2 ways,
Unit’s place digit can be selected in 1 way.
∴ Total number of numbers not exceeding 7432 that can be formed with the digits 2, 3, 4, 7 = Total number of four-digit numbers possible from the digits 2, 3, 4, 7
= 4 × 3 × 2 × 1
= 24

Question 11.
If numbers are formed using digits 2, 3, 4, 5, 6 without repetition, how many of them will exceed 400?
Solution:
Case I: Number of three-digit numbers formed from 2, 3, 4, 5, 6, greater than 400.
100’s place can be filled by any one of the numbers 4, 5, 6.
100’s place digit can be selected in 3 ways.
Since repetition is not allowed, 10’s place can be filled by any one of the remaining four numbers.
∴ 10’s place digit can be selected in 4 ways.
Unit’s place digit can be selected in 3 ways.
∴ Total number of three-digit numbers formed = 3 × 4 × 3 = 36

Case II: Number of four-digit numbers formed from 2, 3, 4, 5, 6.
Since repetition of digits is not allowed,
1000’s place digit can be selected in 5 ways.
100’s place digit can be selected in 4 ways.
10’s place digit can be selected in 3 ways.
Unit’s place digit can be selected in 2 ways.
∴ Total number of four-digit numbers formed = 5 × 4 × 3 × 2 = 120

Case III: Number of five-digit numbers formed from 2, 3, 4, 5, 6.
Similarly, since repetition of digits is not allowed,
Total number of five digit numbers formed = 5 × 4 × 3 × 2 × 1 = 120.
∴ Total number of numbers that exceed 400 = 36 + 120 + 120 = 276

Question 12.
How many numbers formed with the digits 0, 1, 2, 5, 7, 8 will fall between 13 and 1000 if digits can be repeated?
Solution:
Case I: 2-digit numbers more than 13, less than 20, formed from the digits 0, 1, 2, 5, 7, 8.
10’s place digit is 1.
∴ it can be selected in 1 way only.
Unit’s place can be filled by any one of the numbers 5, 7, 8.
∴ Unit’s place digit can be selected in 3 ways.
∴ Total number of such numbers = 1 × 3 = 3.

Case II: 2-digit numbers more than 20 formed from 0, 1, 2, 5, 7, 8.
10’s place can be filled by any one of the numbers 2, 5, 7, 8.
∴ 10’s place digit can be selected in 4 ways.
Since repetition is allowed, the unit’s place can be filled by one of the remaining 6 digits.
∴ Unit’s place digit can be selected in 6 ways.
∴ Total number of such numbers = 4 × 6 = 24.

Case III: 3-digit numbers formed from 0, 1, 2, 5, 7, 8.
Similarly, since repetition of digits is allowed, the total number of such numbers = 5 × 6 × 6 = 180.
All cases are mutually exclusive.
∴ Total number of required numbers = 3 + 24 + 180 = 207

Question 13.
A school has three gates and four staircases from the first floor to the second floor. How many ways does a student have to go from outside the school to his classroom on the second floor?
Solution:
A student can go inside the school from outside in 3 ways and from the first floor to the second floor in 4 ways.
∴ A number of ways to choose gates = 3.
The number of ways to choose a staircase = 4.
By using the fundamental principle of multiplication,
number of ways in which a student has to go from outside the school to his classroom = 4 × 3 = 12

Question 14.
How many five-digit numbers formed using the digit 0, 1, 2, 3, 4, 5 are divisible by 5 if digits are not repeated?
Solution:
Here, repetition of digits is not allowed.
For a number to be divisible by 5,
unit’s place digit should be 0 or 5.
Case I: when unit’s place is 0
Unit’s place digit can be selected in 1 way.
10’s place digit can be selected in 5 ways.
100’s place digit can be selected in 4 ways.
1000’s place digit can be selected in 3 ways.
10000’s place digit can be selected in 2 ways.
∴ Total number of numbers = 1 × 5 × 4 × 3 × 2 = 120.

Case II: when the unit’s place is 5
Unit’s place digit can be selected in 1 way.
10000’s place should be a non-zero number.
∴ It can be selected in 4 ways.
1000’s place digit can be selected in 4 ways.
100’s place digit can be selected in 3 ways.
10’s place digit can be selected in 2 ways.
∴ Total number of numbers = 1 × 4 × 4 × 3 × 2 = 96
∴ Total number of required numbers = 120 + 96 = 216