Bhagya

Balbharati Maharashtra State Board 12th Commerce Maths Solution Book Pdf Chapter 1 Mathematical Logic Ex 1.3 Questions and Answers.

Maharashtra State Board 12th Commerce Maths Solutions Chapter 1 Mathematical Logic Ex 1.3

Question 1.
Write the negation of each of the following statements:
(i) All men are animals.
Solution:
Some men are not animals.

(ii) 3 is a natural number.
Solution:
-3 is not a natural number.

(iii) It is false that Nagpur is the capital of Maharashtra.
Solution:
Nagpur is the capital of Maharashtra.

(iv) 2 + 3 ≠ 5.
Solution:
2 + 3 = 5.

Question 2.
Write the truth value of the negation of each of the following statements:
(i) √5 is an irrational number.
Solution:
Let p : √5 is an irrational number.
The truth value of p is T.
Therefore, the truth value of ~p is F.

(ii) London is in England.
Solution:
Let p : London is in England.
The truth value of p is T.
Therefore, the truth value of ~p is F.

(iii) For every x ∈ N, x + 3 < 8.
Solution:
Let p : For every x ∈ N, x + 3 < 8.
The truth value of p is F.
Therefore, the truth value of ~p is T.

Balbharati Maharashtra State Board 12th Commerce Maths Solution Book Pdf Chapter 1 Mathematical Logic Ex 1.2 Questions and Answers.

Maharashtra State Board 12th Commerce Maths Solutions Chapter 1 Mathematical Logic Ex 1.2

Question 1.
Express the following statements in symbolic form:
(i) e is a vowel or 2 + 3 = 5.
Solution:
Let p : e is a vowel.
q: 2 + 3 = 5.
Then the symbolic form of the given statement is p ∨ q.

(ii) Mango is a fruit but potato is a vegetable.
Solution:
Let p : Mango is a fruit.
q : Potato is a vegetable.
Then the symbolic form of the given statement is p ∧ q.

(iii) Milk is white or grass is green.
Solution:
Let p : Milk is white.
q : Grass is green.
Then the symbolic form of the given statement is p ∨ q.

(iv) I like playing but not singing.
Solution:
Let p : I like playing.
q : I am not singing.
Then the symbolic form of the given statement is p ∧ q.

(v) Even though it is cloudy, it is still raining.
Solution:
The given statement is equivalent to:
It is cloudy and it is still raining.
Let p : It is cloudy.
q : It is still raining.
Then the symbolic form of the given statement is p ∧ q.

Question 2.
Write the truth values of the following statements:
(I) Earth is a planet and Moon is a star.
Solution:
Let p : Earth is a planet.
q : Moon is a star.
Then the symbolic form of the given statement is p ∧ q.
The truth values of p and q are T and F respectively.
∴ the truth value of p ∧ q is F. …[T ∧ F ≡ F]

(ii) 16 is an even number and 8 is a perfect square.
Solution:
Let p : 16 is an even number.
q : 8 is a perfect square.
Then the symbolic form of the given statement is p ∧ q.
The truth values of p and q are T and F respectively.
∴ the truth value of p ∧ q is F. ….[T ∧ F ≡ F]

(iii) A quadratic equation has two distinct roots or 6 has three prime factors.
Solution:
Let p : A quadratic equation has two distinct roots.
q : 6 has three prime factors.
Then the symbolic form of the given statement is p ∨ q.
The truth values of both p and q are F.
∴ the truth value of p ∨ q is F. …..[F ∨ F ≡ F]

(iv) The Himalayas are the highest mountains but they are part of India in the northeast.
Solution:
Let p : the Himalayas are the highest mountains.
q : They are part of India in the northeast.
Then the symbolic form of the given statement is p ∧ q.
The truth values of both p and q are T.
∴ the truth value of p ∧ q is T. …..[T ∧ T ≡ T]

Balbharati Maharashtra State Board 12th Commerce Maths Solution Book Pdf Chapter 1 Mathematical Logic Ex 1.1 Questions and Answers.

Maharashtra State Board 12th Commerce Maths Solutions Chapter 1 Mathematical Logic Ex 1.1

State which of the following sentences are statements. Justify your answer. In case of statements, write down the truth value:

Question (i).
A triangle has ‘ n’ sides.
Solution:
It is a statement that is false, hence its truth value is ‘F’.

Question (ii).
The sum of interior angles of a triangle is 180°.
Solution:
It is a statement that is true, hence its truth value is ‘T’.

Question (iii).
You are amazing!
Solution:
It is an exclamatory sentence, hence it is not a statement.

Question (iv).
Please grant me a loan.
Solution:
It is an imperative sentence, hence it is not a statement.

Question (v).
√-4 is an irrational number.
Solution:
It is a statement that is false, hence its truth value is ‘F’.

Question (vi).
x² – 6x + 8 = 0 implies x = -4 or x = -2.
Solution:
It is a statement that is false, hence its truth value is ‘F’.

Question (vii).
He is an actor.
Solution:
It is an open sentence, hence it is not a statement.

Question (viii).
Did you eat lunch yet?
Solution:
It is an interrogative sentence, hence it is not a statement.

Question (ix).
Have a cup of cappuccino.
Solution:
It is an imperative sentence, hence it is not a statement.

Question (x).
(x + y)² = x² + 2xy + y² for all x, y ∈ R.
Solution:
It is a mathematical identity that is true, hence its truth value is ‘T’.

Question (xi).
Every real number is a complex number.
Solution:
It is a statement that is true, hence its truth value is ‘T.

Question (xii).
1 is a prime number.
Solution:
It is a statement that is false, hence its truth value is ‘F’.

Question (xiii).
With the sunset, the day ends.
Solution:
It is a statement that is true, hence its truth value is ‘T’.

Question (xiv).
1! = 0.
Solution:
It is a statement that is false, hence its truth value is

Question (xv).
3 + 5 > 11.
Solution:
It is a statement that is false, hence its truth value is ‘F’.

Question (xvi).
The number π is an irrational number.
Solution:
It is a statement that is true, hence its truth value is ‘T’.

Question (xvii).
x² – y² = (x + y)(x – y) for all x, y ∈ R.
Solution:
It is a mathematical identity that is true, hence its truth value is ‘T’.

Question (xviii).
The number 2 is only even a prime number.
Solution:
It is a statement that is true, hence its truth value is ‘T’.

Question (xix).
Two coplanar lines are either parallel or intersecting.
Solution:
It is a statement that is true, hence its truth value is ‘T’.

Question (xx).
The number of arrangements of 7 girls in a row for a photograph is 7!
Solution:
It is a statement that is true, hence its truth value is ‘T’.

Question (xxi).
Give me a compass box.
Solution:
It is an imperative sentence, hence it is not a statement.

Question (xxii).
Bring the motor car here.
Solution:
It is an imperative sentence, hence it is not a statement.

Question (xxiii).
It may rain today.
Solution:
It is an open sentence, hence it is not a statement.

Question (xxiv).
If a + b < 7, where a ≥ 0 and b ≥ 0, then a < 7 and b < 7.
Solution:
It is a statement that is true, hence its truth value is ‘T’.

Question (xxv).
Can you speak English?
Solution:
It is an interrogative sentence, hence it is not a statement.

Balbharti Maharashtra State Board 11th Maths Book Solutions Pdf Chapter 9 Differentiation Miscellaneous Exercise 9 Questions and Answers.

Maharashtra State Board 11th Maths Solutions Chapter 9 Differentiation Miscellaneous Exercise 9

(I) Select the appropriate option from the given alternatives.

Question 1.
If y = \(\frac{x-4}{\sqrt{x+2}}\), then \(\frac{d y}{d x}\) is
(A) \(\frac{1}{x+4}\)
(B) \(\frac{\sqrt{x}}{\left(\sqrt{x+2)^{2}}\right.}\)
(C) \(\frac{1}{2 \sqrt{x}}\)
(D) \(\frac{x}{(\sqrt{x}+2)^{2}}\)
Answer:
(C) \(\frac{1}{2 \sqrt{x}}\)
Hint:

Question 2.
If y = \(\frac{a x+b}{c x+d}\),then \(\frac{d y}{d x}\) =
(A) \(\frac{a b-c d}{(c x+d)^{2}}\)
(B) \(\frac{a x-c}{(c x+d)^{2}}\)
(C) \(\frac{a c-b d}{(c x+d)^{2}}\)
(D) \(\frac{a d-b c}{(c x+d)^{2}}\)
Answer:
(D) \(\frac{a d-b c}{(c x+d)^{2}}\)
Hint:

Question 3.
If y = \(\frac{3 x+5}{4 x+5}\), then \(\frac{d y}{d x}\) =
(A) \(-\frac{15}{(3 x+5)^{2}}\)
(B) \(-\frac{15}{(4 x+5)^{2}}\)
(C) \(-\frac{5}{(4 x+5)^{2}}\)
(D) \(-\frac{13}{(4 x+5)^{2}}\)
Answer:
(C) \(-\frac{5}{(4 x+5)^{2}}\)
Hint:

Question 4.
If y = \(\frac{5 \sin x-2}{4 \sin x+3}\), then \(\frac{d y}{d x}\) =
(A) \(\frac{7 \cos x}{(4 \sin x+3)^{2}}\)
(B) \(\frac{23 \cos x}{(4 \sin x+3)^{2}}\)
(C) \(-\frac{7 \cos x}{(4 \sin x+3)^{2}}\)
(D) \(-\frac{15 \cos x}{(4 \sin x+3)^{2}}\)
Answer:
(B) \(\frac{23 \cos x}{(4 \sin x+3)^{2}}\)
Hint:

Question 5.
Suppose f(x) is the derivative of g(x) and g(x) is the derivative of h(x).
If h(x) = a sin x + b cos x + c, then f(x) + h(x) =
(A) 0
(B) c
(C) -c
(D) -2(a sin x + b cos x)
Answer:
(B) c
Hint:
h(x) = a sin x + b cos x + c
Differentiating w.r.t. x, we get
h'(x) = a cos x – b sin x = g(x) …..[given]
Differentiating w.r.t. x, we get
g'(x) = -a sin x – b cos x = f(x) …..[given]
∴ f(x) + h(x) = -a sin x – b cos x + a sin x + b cos x + c
∴ f(x) + h(x) = c

Question 6.
If f(x) = 2x + 6, for 0 ≤ x ≤ 2
= ax² + bx, for 2 < x ≤ 4
is differentiable at x = 2, then the values of a and b are
(A) a = \(-\frac{3}{2}\), b = 3
(B) a = \(\frac{3}{2}\), b = 8
(C) a = \(\frac{1}{2}\), b = 8
(D) a = \(-\frac{3}{2}\), b = 8
Answer:
(D) a = \(-\frac{3}{2}\), b = 8
Hint:
f(x) = 2x + 6, 0 ≤ x ≤ 2
= ax² + bx, 2 < x ≤ 4
Lf'(2) = 2, Rf'(2) = 4a + b
Since f is differentiable at x = 2,
Lf'(2) = Rf'(2)
∴ 2 = 4a + b …..(i)
f is continuous at x = 2.
∴ \(\lim _{x \rightarrow 2^{+}} f(x)=f(2)=\lim _{x \rightarrow 2^{-}} f(x)\)
∴ 4a + 2b = 2(2) + 6
∴ 4a + 2b = 10
∴ 2a + b = 5 …..(ii)
Solving (i) and (ii), we get
a = \(-\frac{3}{2}\), b = 8

Question 7.
If f(x) = x² + sin x + 1, for x ≤ 0
= x² – 2x + 1, for x ≤ 0, then
(A) f is continuous at x = 0, but not differentiable at x = 0
(B) f is neither continuous nor differentiable at x = 0
(C) f is not continuous at x = 0, but differentiable at x = 0
(D) f is both continuous and differentiable at x = 0
Answer:
(A) f is continuous at x = 0, but not differentiable at x = 0
Hint:

Question 8.
If f(x) = \(\frac{x^{50}}{50}+\frac{x^{49}}{49}+\frac{x^{48}}{48}+\ldots .+\frac{x^{2}}{2}+x+1\), then f'(1) =
(A) 48
(B) 49
(C) 50
(D) 51
Answer:
(C) 50
Hint:

(II).

Question 1.
Determine whether the following function is differentiable at x = 3 where,
f(x) = x² + 2, for x ≥ 3
= 6x – 7, for x < 3.
Solution:
f(x) = x² + 2, x ≥ 3
= 6x – 7, x < 3
Differentiability at x = 3


Here, Lf'(3) = Rf'(3)
∴ f is differentiable at x = 3.

Question 2.
Find the values of p and q that make function f(x) differentiable everywhere on R.
f(x) = 3 – x, for x < 1
= px² + qx, for x ≥ 1.
Solution:
f(x) is differentiable everywhere on R.
∴ f(x) is differentiable at x = 1.
∴ f(x) is continuous at x = 1.

f(x) is differentiable at x = 1.
∴ Lf'(1) = Rf'(1)
∴ -1 = 2p + q …..(ii)
Subtracting (i) from (ii), we get
p = -3
Substituting p = -3 in (i), we get
p + q = 2
∴ -3 + q = 2
∴ q = 5

Question 3.
Determine the values of p and q that make the function f(x) differentiable on R where
f(x) = px³, for x < 2
= x² + q, for x ≥ 2
Solution:
f(x) is differentiable on R.
∴ f(x) is differentiable at x = 2.
∴ f(x) is continuous at x = 2.
Continuity at x = 2:
f(x) is continuous at x = 2.


f(x) is differentiable at x = 2.
∴ Lf'(2) = Rf'(2)
∴ 12p = 4
∴ p = \(\frac{1}{3}\)
Substituting p = \(\frac{1}{3}\) in (i), we get
8(\(\frac{1}{3}\) – q = 4
∴ q = \(\frac{8}{3}\) – 4 = \(\frac{-4}{3}\)

Question 4.
Determine all real values of p and q that ensure the function
f(x) = px + q, for x ≤ 1
= tan(\(\frac{\pi x}{4}\)), for 1 < x < 2
is differentiable at x = 1.
Solution:
f(x) is differentiable at x = 1.
∴ f(x) is continuous at x = 1.
Continuity at x= 1:
f(x) is continuous at x = 1.

Question 5.
Discuss whether the function f(x) = |x + 1| + |x – 1| is differentiable ∀ x ∈ R.
Solution:


Here, Lf'(1) ≠ Rf'(1)
∴ f is not differentiable at x = 1.
∴ f is not differentiable at x = -1 and x = 1.
∴ f is not differentiable ∀ x ∈ R.

Question 6.
Test whether the function
f(x) = 2x – 3, for x ≥ 2
= x – 1, for x < 2
is differentiable at x = 2.
Solution:

Question 7.
Test whether the function
f(x) = x² + 1, for x ≥ 2
= 2x + 1, for x < 2
is differentiable at x = 2.
Solution:

Question 8.
Test whether the function
f(x) = 5x – 3x², for x ≥ 1
= 3 – x, for x < 1
is differentiable at x = 1.
Solution:

Here, Lf'(1) = Rf'(1)
∴ f(x) is differentiable at x = 1.

Question 9.
If f(2) = 4, f'(2) = 1, then find \(\lim _{x \rightarrow 2}\left[\frac{x f(2)-2 f(x)}{x-2}\right]\)
Solution:

Question 10.
If y = \(\frac{\mathbf{e}^{x}}{\sqrt{x}}\), find \(\frac{d y}{d x}\) when x = 1.
Solution:

Balbharti Maharashtra State Board 11th Maths Book Solutions Pdf Chapter 9 Differentiation Ex 9.2 Questions and Answers.

Maharashtra State Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.2

(I) Differentiate the following w.r.t. x

Question 1.
y = \(x^{\frac{4}{3}}+e^{x}-\sin x\)
Solution:

Question 2.
y = √x + tan x – x³
Solution:

Question 3.
y = log x – cosec x + \(5^{x}-\frac{3}{x^{\frac{3}{2}}}\)
Solution:

Question 4.
y = \(x^{\frac{7}{3}}+5 x^{\frac{4}{5}}-\frac{5}{x^{\frac{2}{5}}}\)
Solution:

Question 5.
y = 7^x + x⁷ – \(\frac{2}{3}\) x√x – log x + 7⁷
Solution:

Question 6.
y = 3 cot x – 5e^x + 3 log x – \(\frac{4}{x^{\frac{3}{4}}}\)
Solution:

(II) Diffrentiate the following w.r.t. x

Question 1.
y = x⁵ tan x
Solution:

Question 2.
y = x³ log x
Solution:

Question 3.
y = (x² + 2)² sin x
Solution:

Question 4.
y = e^x log x
Solution:

Question 5.
y = \(x^{\frac{3}{2}} e^{x} \log x\)
Solution:

Question 6.
y = \(\log e^{x^{3}} \log x^{3}\)
Solution:

(III) Diffrentiate the following w.r.t. x

Question 1.
y = x²√x + x⁴ log x
Solution:

Question 2.
y = e^x sec x – \(x^{\frac{5}{3}}\) log x
Solution:

Question 3.
y = x⁴ + x√x cos x – x² e^x
Solution:

Question 4.
y = (x³ – 2) tan x – x cos x + 7^x . x⁷
Solution:

Question 5.
y = sin x log x + e^x cos x – e^x √x
Solution:

Question 6.
y = e^x tan x + cos x log x – √x 5^x
Solution:

(IV) Diffrentiate the following w.r.t.x.

Question 1.
y = \(\frac{x^{2}+3}{x^{2}-5}\)
Solution:

Question 2.
y = \(\frac{\sqrt{x}+5}{\sqrt{x}-5}\)
Solution:

Question 3.
y = \(\frac{x e^{x}}{x+e^{x}}\)
Solution:

Question 4.
y = \(\frac{x \log x}{x+\log x}\)
Solution:
y = \(\frac{x \log x}{x+\log x}\)

Question 5.
y = \(\frac{x^{2} \sin x}{x+\cos x}\)
Solution:

Question 6.
y = \(\frac{5 e^{x}-4}{3 e^{x}-2}\)
Solution:

(V).

Question 1.
If f(x) is a quadratic polynomial such that f(0) = 3, f'(2) = 2 and f'(3) = 12, then find f(x).
Solution:
Let f(x) = ax² + bx + c …..(i)
∴ f(0) = a(0)² + b(0) + c
∴ f(0) = c
But, f(0) = 3 …..(given)
∴ c = 3 …..(ii)
Differentiating (i) w.r.t. x, we get
f'(x) = 2ax + b
∴ f'(2) = 2a(2) + b
∴ f'(2) = 4a + b
But, f'(2) = 2 …..(given)
∴ 4a + b = 2 …..(iii)
Also, f'(3) = 2a(3) + b
∴ f'(3) = 6a + b
But, f'(3) = 12 …..(given)
∴ 6a + b = 12 …..(iv)
equation (iv) – equation (iii), we get
2a = 10
∴ a = 5
Substituting a = 5 in (iii), we get
4(5) + b = 2
∴ b = -18
∴ a = 5, b = -18, c = 3
∴ f(x) = 5x² – 18x + 3

Check:
If f(0) = 3, f'(2) = 2 and f'(3) = 12, then our answer is correct.
f(x) = 5x² – 18x + 3 and f'(x) = 10x – 18
f(0) = 5(0)² – 18(0) + 3 = 3
f'(2) = 10(2) – 18 = 2
f'(3) = 10(3) – 18 = 12
Thus, our answer is correct.

Question 2.
If f(x) = a sin x – b cos x, f'(\(\frac{\pi}{4}\)) = √2 and f'(\(\frac{\pi}{6}\)) = 2, then find f(x).
Solution:
f(x) = a sin x – b cos x
Differentiating w.r.t. x, we get
f'(x) = a cos x – b (- sin x)
∴ f'(x) = a cos x + b sin x


Now, f(x) = a sin x – b cos x
∴ f(x) = (√3 + 1) sin x + (√3 – 1) cos x

VI. Fill in the blanks. (Activity Problems)

Question 1.
y = e^x . tan x
Diff. w.r.t. x

Solution:

Question 2.
y = \(\frac{\sin x}{x^{2}+2}\)
diff. w.r.t. x

Solution:

Question 3.
y = (3x² + 5) cos x
Diff. w.r.t. x

Solution:

Question 4.
Differentiate tan x and sec x w.r.t. x using the formulae for differentiation of \(\frac{u}{v}\) and \(\frac{1}{v}\) respectively.
Solution:

Balbharti Maharashtra State Board 11th Maths Book Solutions Pdf Chapter 9 Differentiation Ex 9.1 Questions and Answers.

Maharashtra State Board 11th Maths Solutions Chapter 9 Differentiation Ex 9.1

Question 1.
Find the derivatives of the following w.r.t. x by using the method of the first principle.
(a) x² + 3x – 1
Solution:
Let f(x) = x² + 3x – 1
∴ f(x + h) = (x + h)² + 3(x + h) – 1
= x² + 2xh + h² + 3x + 3h – 1
By first principle, we get

(b) sin(3x)
Solution:
Let f(x) = sin 3x
f(x + h) = sin3(x + h) = sin(3x + 3h)
By first principle, we get

(c) e^2x+1
Solution:

(d) 3^x
Solution:

(e) log(2x + 5)
Solution:
Let f(x) = log(2x + 5)
∴ f(x + h) = log[2(x + h) + 5] = log (2x + 2h + 5)
By first principle, we get

(f) tan(2x + 3)
Solution:
Let f(x) = tan(2x + 3)
∴ f(x + h) = tan[2(x + h) + 3] = tan(2x + 2h + 3)
By first principle, we get

(g) sec(5x – 2)
Solution:
Let f(x) = sec(5x – 2)
f(x + h) = sec[5(x + h) – 2] = sec(5x + 5h – 2)
By first principle, we get

(h) x√x
Solution:

Question 2.
Find the derivatives of the following w.r.t. x. at the points indicated against them by using the method of the first principle.
(i) \(\sqrt{2 x+5}\) at x = 2
Solution:

(ii) tan x at x = \(\frac{\pi}{4}\)
Solution:

(iii) 2^3x+1 at x = 2
Solution:

(iv) log(2x + 1) at x = 2
Solution:
Let f(x) = log(2x + 1)
∴ f(2) = log [2(2) + 1] = log 5 and
f(2 + h) = log [2(2 + h) + 1] = log(2h + 5)
By first principle, we get

(v) e^3x-4 at x = 2
Solution:

(vi) cos x at x = \(\frac{5 \pi}{4}\)
Solution:

Question 3.
Show that the function f is not differentiable at x = -3,
where f(x) = x² + 2 for x < -3
= 2 – 3x for x ≥ -3
Solution:


∴ L f'(-3) ≠ R f'(-3)
∴ f is not differentiable at x = -3.

Question 4.
Show that f(x) = x² is continuous and differentiable at x = 0.
Solution:

Question 5.
Discuss the continuity and differentiability of
(i) f(x) = x |x| at x = 0
Solution:

(ii) f(x) = (2x + 3) |2x + 3| at x = \(-\frac{3}{2}\)
Solution:

Question 6.
Discuss the continuity and differentiability of f(x) at x = 2.
f(x) = [x] if x ∈ [0, 4). [where [ ] is a greatest integer (floor) function]
Solution:
Explanation:
x ∈ [0, 4)
∴ 0 ≤ x < 4
We will plot graph for 0 ≤ x < 4
not for x < 0 and upto x = 4 making on X-axis.

f(x) = [x]
∴ Greatest integer function is discontinuous at all integer values of x and hence not differentiable at all integers.
∴ f is not continuous at x = 2.
∵ f(x) = 1, x < 2
= 2, x ≥ 2
x ∈ neighbourhood of x = 2.
∴ L.H.L. = 1, R.H.L. = 2
∴ f is not continuous at x = 2.
∴ f is not differentiable at x = 2.

Question 7.
Test the continuity and differentiability of
f(x) = 3x + 2 if x > 2
= 12 – x² if x ≤ 2 at x = 2.
Solution:

Question 8.
If f(x) = sin x – cos x if x ≤ \(\frac{\pi}{2}\)
= 2x – π + 1 if x > \(\frac{\pi}{2}\)
Test the continuity and differentiability of f at x = \(\frac{\pi}{2}\).
Solution:

Question 9.
Examine the function
f(x) = x² cos(\(\frac{1}{x}\)), for x ≠ 0
= 0, for x = 0
for continuity and differentiability at x = 0.
Solution:

Balbharti Maharashtra State Board 11th Maths Book Solutions Pdf Chapter 8 Continuity Miscellaneous Exercise 8 Questions and Answers.

Maharashtra State Board 11th Maths Solutions Chapter 8 Continuity Miscellaneous Exercise 8

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

Question 1.
f(x) = \(\frac{2^{\cot x}-1}{\pi-2 x}\), for x ≠ \(\frac{\pi}{2}\)
= log √2, for x = \(\frac{\pi}{2}\)
(A) f is continuous at x = \(\frac{\pi}{2}\)
(B) f has a jump discontinuity at x = \(\frac{\pi}{2}\)
(C) f has a removable discontinuity
(D) \(\lim _{x \rightarrow \frac{\pi}{2}} f(x)=2 \log 3\)
Answer:
(A) f is continuous at x = \(\frac{\pi}{2}\)
Hint:

Question 2.
If f(x) = \(\frac{1-\sqrt{2} \sin x}{\pi-4 x}\), for x ≠ \(\frac{\pi}{4}\) is continuous at x = \(\frac{\pi}{4}\), then f(\(\frac{\pi}{4}\)) =
(A) \(\frac{1}{\sqrt{2}}\)
(B) \(-\frac{1}{\sqrt{2}}\)
(C) \(-\frac{1}{4}\)
(D) \(\frac{1}{4}\)
Answer:
(D) \(\frac{1}{4}\)
Hint:

Question 3.
If f(x) = \(\frac{(\sin 2 x) \tan 5 x}{\left(e^{2 x}-1\right)^{2}}\), for x ≠ 0 is continuous at x = 0, then f(0) is
(A) \(\frac{10}{e^{2}}\)
(B) \(\frac{10}{e^{4}}\)
(C) \(\frac{5}{4}\)
(D) \(\frac{5}{2}\)
Answer:
(D) \(\frac{5}{2}\)
Hint:

Question 4.
f(x) = \(\frac{x^{2}-7 x+10}{x^{2}+2 x-8}\), for x ∈ [-6, -3]
(A) f is discontinuous at x = 2
(B) f is discontinuous at x = -4
(C) f is discontinuous at x = 0
(D) f is discontinuous at x = 2 and x = -4
Answer:
(B) f is discontinuous at x = -4
Hint:
f(x) = \(\frac{x^{2}-7 x+10}{x^{2}+2 x-8}\), for x ∈ [-6, -3]
= \(\frac{x^{2}-7 x+10}{(x+4)(x-2)}\)
Here f(x) is a rational function and is continuous everywhere except at the points Where denominator becomes zero.
Here, denominator becomes zero when x = -4 or x = 2
But x = 2 does not lie in the given interval.
∴ x = -4 is the point of discontinuity.

Question 5.
If f(x) = ax² + bx + 1, for |x – 1| ≥ 3 and
= 4x + 5, for -2 < x < 4
is continuous everywhere then,
(A) a = \(\frac{1}{2}\), b = 3
(B) a = \(-\frac{1}{2}\), b = -3
(C) a = \(-\frac{1}{2}\), b = 3
(D) a = \(\frac{1}{2}\), b = -3
Answer:
(A) a = \(\frac{1}{2}\), b = 3
Hint:
f(x) = ax² + bx + 1, |x – 1| ≥ 3
= 4x + 5; -2 < x < 4
The first interval is
|x – 1| ≥ 3
∴ x – 1 ≥ 3 or x – 1 ≤ -3
∴ x ≥ 4 or x ≤ -2
∴ f(x) is same for x ≤ -2 as well as x ≥ 4.
∴ f(x) is defined as:
f(x) = ax² + bx + 1; x ≤ -2
= 4x + 5; -2 < x < 4
= ax² + bx + 1; x ≥ 4
f(x) is continuous everywhere.
∴ f(x) is continuous at x = -2 and x = 4.
As f(x) is continuous at x = -2,
\(\lim _{x \rightarrow-2^{-}} f(x)=\lim _{x \rightarrow-2^{+}} f(x)\)
∴ \(\lim _{x \rightarrow-2}\left(a x^{2}+b x+1\right)=\lim _{x \rightarrow-2}(4 x+5)\)
∴ a(-2)² + b(-2) + 1 = 4(-2) + 5
∴ 4a – 2b + 1 = -3
∴ 4a – 2b = -4
∴ 2a – b = -2 …..(i)
∵ f(x) is continuous at x = 4,
\(\lim _{x \rightarrow 4^{-}} \mathrm{f}(x)=\lim _{x \rightarrow 4^{+}} \mathrm{f}(x)\)
∴ \(\lim _{x \rightarrow 4}(4 x+5)=\lim _{x \rightarrow 4}\left(a x^{2}+b x+1\right)\)
4(4) + 5 = a(4)² + b(4) + 1
16a + 4b + 1 = 21
16a + 4b = 20
4a + b = 5 …..(ii)
Adding (i) and (ii), we get
6a = 3
∴ a = \(\frac{1}{2}\)
Substituting a = \(\frac{1}{2}\) in (ii), we get
4(\(\frac{1}{2}\)) + b = 5
∴ 2 + b = 5
∴ b = 3
∴ a = \(\frac{1}{2}\), b = 3

Question 6.
f(x) = \(\frac{\left(16^{x}-1\right)\left(9^{x}-1\right)}{\left(27^{x}-1\right)\left(32^{x}-1\right)}\), for x ≠ 0
= k, for x = 0
is continuous at x = 0, then ‘k’ =
(A) \(\frac{8}{3}\)
(B) \(\frac{8}{15}\)
(C) \(-\frac{8}{15}\)
(D) \(\frac{20}{3}\)
Answer:
(B) \(\frac{8}{15}\)
Hint:

Question 7.
f(x) = \(\frac{32^{x}-8^{x}-4^{x}+1}{4^{x}-2^{x+1}+1}\), for x ≠ 0
= k, for x = 0,
is continuous at x = 0, then value of ‘k’ is
(A) 6
(B) 4
(C) (log 2) (log 4)
(D) 3 log 4
Answer:
(A) 6
Hint:

Question 8.
If f(x) = \(\frac{12^{x}-4^{x}-3^{x}+1}{1-\cos 2 x}\), for x ≠ 0 is continuous at x = 0 then the value of f(0) is
(A) \(\frac{\log 12}{2}\)
(B) log 2 . log 3
(C) \(\frac{\log 2 \cdot \log 3}{2}\)
(D) None of these
Answer:
(B) log 2 . log 3
Hint:

Question 9.
If f(x) = \(\left(\frac{4+5 x}{4-7 x}\right)^{\frac{4}{x}}\), for x ≠ 0 and f(0) = k, is continuous at x = 0, then k is
(A) e⁷
(B) e³
(C) e¹²
(D) \(e^{\frac{3}{4}}\)
Answer:
(C) e¹²
Hint:

Question 10.
If f(x) = \(\lfloor x\rfloor\) for x ∈ (-1, 2), then f is discontinuous at
(A) x = -1, 0, 1, 2
(B) x = -1, 0, 1
(C) x = 0, 1
(D) x = 2
Answer:
(C) x = 0, 1
Hint:
f(x) = \(\lfloor x\rfloor\), x ∈ (-1, 2)
This function is discontinuous at all integer values of x between -1 and 2.
∴ f(x) is discontinuous at x = 0 and x = 1.

II. Discuss the continuity of the following functions at the point(s) or on the interval indicated against them.

Question 1.
f(x) = \(\frac{x^{2}-3 x-10}{x-5}\), for 3 ≤ x ≤ 6, x ≠ 5
= 10, for x = 5
= \(\frac{x^{2}-3 x-10}{x-5}\), for 6 < x ≤ 9
Solution:

Question 2.
f(x) = 2x² – 2x + 5, for 0 ≤ x ≤ 2
= \(\frac{1-3 x-x^{2}}{1-x}\), for 2 < x < 4
= \(\frac{x^{2}-25}{x-5}\), for 4 ≤ x ≤ 7 and x ≠ 5
= 7, for x = 5
Solution:
The domain of f(x) is [0, 7].
(i) For 0 ≤ x ≤ 2
f(x) = 2x² – 2x + 5
It is a polynomial function and is Continuous at all point in [0, 2).

(ii) For 2 < x < 4
f(x) = \(\frac{1-3 x-x^{2}}{1-x}\)
It is a rational function and is continuous everwhere except at points where its denominator becomes zero.
Denominator becomes zero at x = 1
But x = 1 does not lie in the interval.
f(x) is continuous at all points in (2, 4).

(iii) For 4 ≤ x ≤ 7, x ≤ 5
f(x) = \(\frac{x^{2}-25}{x-5}\)
It is a rational function and is continuous everywhere except at points where its denominator becomes zero.
Denominator becomes zero at x = 5
But x = 5 does not lie in the interval.
∴ f(x) is continuous at all points in (4, 7] – {5}.

(iv) For continuity at x = 2:

∴ f(x) is continuous at x = 2.

(v) For continuity at x = 4:

∴ f(x) is continuous at x = 4.

(vi) For continuity at x = 5.
f(5) = 7

∴ f(x) is discontinuous at x = 5.
Thus, f(x) is continuous at all points on its domain except at x = 5.

Question 3.
f(x) = \(\frac{\cos 4 x-\cos 9 x}{1-\cos x}\), for x ≠ 0
f(0) = \(\frac{68}{15}\), at x = 0 on \(-\frac{\pi}{2}\) ≤ x ≤ \(\frac{\pi}{2}\)
Solution:
The domain of f(x) is [\(-\frac{\pi}{2}\), \(\frac{\pi}{2}\)]
(i) For [\(-\frac{\pi}{2}\), \(\frac{\pi}{2}\)] – {0}:
f(x) = \(\frac{\cos 4 x-\cos 9 x}{1-\cos x}\)
It is a rational function and is continuous everywhere except at points where its denominator becomes zero.
Denominator becomes zero when cos x = 1,
i.e., x = 0
But x = 0 does not lie in the interval.
∴ f(x) is continuous at all points in [\(-\frac{\pi}{2}\), \(\frac{\pi}{2}\)] – {0}

(ii) For continuity at x = 0:


∴ \(\lim _{x \rightarrow 0} f(x) \neq f(0)\)
∴ f(x) is discontinuous at x = 0.

Question 4.
f(x) = \(\frac{\sin ^{2} \pi x}{3(1-x)^{2}}\), for x ≠ 1
= \(\frac{\pi^{2} \sin ^{2}\left(\frac{\pi x}{2}\right)}{3+4 \cos ^{2}\left(\frac{\pi x}{2}\right)}\), for x = 1, at x = 1
Solution:

Question 5.
f(x) = \(\frac{|x+1|}{2 x^{2}+x-1}\), for x ≠ -1
= 0, for x = -1, at x = -1
Solution:

Question 6.
f(x) = [x + 1] for x ∈ [-2, 2)
Where [*] is greatest integer function.
Solution:
f(x) = [x + 1], x ∈ [-2, 2)
∴ f(x) = -1, x ∈ [-2, -1)
= 0, x ∈ [-1, 0)
= 1, x ∈ [0, 1)
= 2, x ∈ [1, 2)

∴ \(\lim _{x \rightarrow-1^{-}} \mathrm{f}(x)=\lim _{x \rightarrow-1^{+}} \mathrm{f}(x)\)
∴ f(x) is discontinuous at x = -1.
Similarly, f(x) is discontinuous at the points x = 0 and x = 1.

Question 7.
f(x) = 2x² + x + 1, for |x – 3| ≥ 2
= x² + 3, for 1 < x < 5
Solution:
|x – 3| ≥ 2
∴ x – 3 ≥ 2 or x – 3 ≤ -2
∴ x ≥ 5 or x ≤ 1
∴ f(x) = 2x² + x + 1, x ≤ 1
= x² + 3, 1 < x < 5
= 2x² + x + 1, x ≥ 5
Consider the intervals
x < 1 , i.e., (-∞, 1)
1 < x < 5, i.e., (1, 5) x > 5, i.e., (5, ∞)
In all these intervals, f(x) is a polynomial function and hence is continuous at all points.
For continuity at x = 1:

∴ f(x) is discontinuous at x = 5.
∴ f(x) is continuous for all x ∈ R, except at x = 5.

III. Identify discontinuities if any for the following functions as either a jump or a removable discontinuity on their respective domains.

Question 1.
f(x) = x² + x – 3, for x ∈ [-5, -2)
= x² – 5, for x ∈ (-2, 5]
Solution:
f(-2) has not been defined.
\(\lim _{x \rightarrow-2^{-}} f(x)=\lim _{x \rightarrow-2^{-}}\left(x^{2}+x-3\right)\)
= (-2)² + (-2) – 3
= 4 – 2 – 3
= -1
\(\lim _{x \rightarrow-2^{+}} f(x)=\lim _{x \rightarrow-2^{+}}\left(x^{2}-5\right)\)
= (-2)² – 5
= 4 – 5
= -1
∴ \(\lim _{x \rightarrow-2^{-}} f(x)=\lim _{x \rightarrow-2^{+}} f(x)\)
∴ \(\lim _{x \rightarrow-2} f(x) \text { exists. }\)
But f(-2) has not been defined.
∴ f(x) has a removable discontinuity at x = -2.

Question 2.
f(x) = x² + 5x + 1, for 0 ≤ x ≤ 3
= x³ + x + 5, for 3 < x ≤ 6
Solution:
\(\lim _{x \rightarrow 3^{-}} f(x)=\lim _{x \rightarrow 3^{-}}\left(x^{2}+5 x+1\right)\)
= (3)² + 5(3) + 1
= 9 + 15 + 1
= 25
\(\lim _{x \rightarrow 3^{+}} f(x)=\lim _{x \rightarrow 3^{+}}\left(x^{3}+x+5\right)\)
= (3)³ + 3 + 5
= 27 + 3 + 5
= 35
∴ \(\lim _{x \rightarrow 3^{-}} f(x) \neq \lim _{x \rightarrow 3^{+}} f(x)\)
∴ \(\lim _{x \rightarrow 3} f(x)\) does not exist.
∴ f(x) is discontinuous at x = 3.
∴ f(x) has a jump discontinuity at x = 3.

Question 3.
f(x) = \(\frac{x^{2}+x+1}{x+1}\), for x ∈ [0, 3)
= \(\frac{3 x+4}{x^{2}-5}\), for x ∈ [3, 6]
Solution:


∴ f(x) is continuous at x = 3.

IV. Discuss the continuity of the following functions at the point or on the interval indicated against them. If the function is discontinuous, identify the type of discontinuity and state whether the discontinuity is removable. If it has a removable discontinuity, redefine the function so that it becomes continuous.

Question 1.
f(x) = \(\frac{(x+3)\left(x^{2}-6 x+8\right)}{x^{2}-x-12}\)
Solution:
f(x) = \(\frac{(x+3)\left(x^{2}-6 x+8\right)}{x^{2}-x-12}\)
= \(\frac{(x+3)(x-2)(x-4)}{(x-4)(x+3)}\)
∴ f(x) is not defined at x = 4 and x = -3.
∴ The domain of function f = R – {-3, 4}.
For x ≠ -3 and 4,

f(x) is discontinuous at x = 4 and x = -3.
This discontinuity is removable.
∴ f(x) can be redefined as
f(x) = \(\frac{(x+3)\left(x^{2}-6 x+8\right)}{x^{2}-x-12}\), for x ≠ 4, x ≠ -3
= -5, for x ∈ R – {-3, 4}, x = -3
= 2, for x ∈ R – {-3, 4}, x = 4

Question 2.
f(x) = x² + 2x + 5, for x ≤ 3
= x³ – 2x² – 5, for x > 3
Solution:

∴ f(x) is discontinuous at x = 3.
This discontinuity is irremovable.

V. Find k if the following functions are continuous at the points indicated against them.

Question 1.
f(x) = \(\left(\frac{5 x-8}{8-3 x}\right)^{\frac{3}{2 x-4}}\), for x ≠ 2
= k, for x = 2 at x = 2.
Solution:
f(x) is continuous at x = 2. …..(given)

Question 2.
f(x) = \(\frac{45^{x}-9^{x}-5^{x}+1}{\left(k^{x}-1\right)\left(3^{x}-1\right)}\), for x ≠ 0
= \(\frac{2}{3}\), for x = 0, at x = 0
Solution:
f(x) is continuous at x = 0 …..(given)

VI. Find a and b if the following functions are continuous at the points or on the interval indicated against them.

Question 1.
f(x) = \(\frac{4 \tan x+5 \sin x}{a^{x}-1}\), for x < 0
= \(\frac{9}{\log 2}\), for x = 0
= \(\frac{11 x+7 x \cdot \cos x}{b^{x}-1}\), for x < 0
Solution:

Question 2.
f(x) = ax² + bx + 1, for |2x – 3| ≥ 2
= 3x + 2, for \(\frac{1}{2}\) < x < \(\frac{5}{2}\)
Solution:

VII. Find f(a), if f is continuous at x = a where,

Question 1.
f(x) = \(\frac{1+\cos (\pi x)}{\pi(1-x)^{2}}\), for x ≠ 1 and at a = 1.
Solution:

Question 2.
f(x) = \(\frac{1-\cos [7(x-\pi)]}{5(x-\pi)^{2}}\), for x ≠ π and at a = π.
Solution:

VIII. Solve using intermediate value theorem.

Question 1.
Show that 5^x – 6x = 0 has a root in [1, 2].
Solution:
Let f(x) = 5^x – 6x
5^x and 6x are continuous functions for all x ∈ R.
∴ 5^x – 6x is also continuous for all x ∈ R.
i.e., f(x) is continuous for all x ∈ R.
A root of f(x) exists, if f(x) = 0 for at least one value of x.
f(1) = 5¹ – 6(1) = -1 < 0
f(2) = (5)² – 6(2) = 13 > 0
∴ f(1) < 0 and f(2) > 0
∴ By intermediate value theorem, there has to be a point ‘c’ between 1 and 2 such that f(c) = 0.
∴ There is a root of the given equation in [1, 2].

Question 2.
Show that x³ – 5x² + 3x + 6 = 0 has at least two real roots between x = 1 and x = 5.
Solution:
Let f(x) = x³ – 5x² + 3x + 6
f(x) is a polynomial function and hence it is continuous for all x ∈ R.
A root of f(x) exists, if f(x) = 0 for at least one value of x.
Here, we have been asked to show that f(x) has at least two roots between x = 1 and x = 5.
f(1) = (1)³ – 5(1)² + 3(1) + 6
= 5 > 0
f(2) = (2)³ – 5(2)² + 3(2) + 6
= 8 – 20 + 6 + 6
= 0
∴ x = 2 is a root of f(x).
Also, f(3) = (3)³ – 5(3)² + 3(3) + 6
= 27 – 45 + 9 + 6
= -3 < 0
f(4) = (4)³ – 5(4)² + 3(4) + 6
= 64 – 80 + 12 + 6
= 2 > 0
∴ f(3) < 0 and f(4) > 0
∴ By intermediate value theorem, there has to be a point ‘c’ between 3 and 4 such that f(c) = 0.
∴ There are two roots, x = 2 and a root between x = 3 and x = 4.
Thus, there are at least two roots of the given equation between x = 1 and x = 5.

Balbharti Maharashtra State Board 11th Maths Book Solutions Pdf Chapter 8 Continuity Ex 8.1 Questions and Answers.

Maharashtra State Board 11th Maths Solutions Chapter 8 Continuity Ex 8.1

Question 1.
Examine the continuity of
(i) f(x) = x³ + 2x² – x – 2 at x = -2
Solution:
Given, f(x) = x³ + 2x² – x – 2
f(x) is a polynomial function and hence it is continuous for all x ∈ R.
∴ f(x) is continuous at x = -2.

(ii) f(x) = sin x, for x ≤ \(\frac{\pi}{4}\)
= cos x, for x > \(\frac{\pi}{4}\), at x = \(\frac{\pi}{4}\)
Solution:

(iii) f(x) = \(\frac{x^{2}-9}{x-3}\), for x ≠ 3
= 8 for x = 3, at x = 3.
Solution:
f(3) = 8 ….(given)

∴ f(x) is discontinuous at x = 3.

Question 2.
Examine whether the function is continuous at the points indicated against them.
(i) f(x) = x³ – 2x + 1, if x ≤ 2
= 3x – 2, if x > 2, at x = 2.
Solution:

(ii) f(x) = \(\frac{x^{2}+18 x-19}{x-1}\), for x ≠ 1
= 20, for x = 1, at x = 1.
Solution:

(iii) f(x) = \(\frac{x}{\tan 3 x}+2\), for x < 0
= \(\frac{7}{3}\), for x ≥ 0, at x = 0.
Solution:

Question 3.
Find all the points of discontinuities of f(x) = [x] on the interval (-3, 2).
Solution:
f(x) = [x], x ∈ (-3, 2)
i.e., f(x) = -3, x ∈ (-3, -2)
= -2, x ∈ [-2, -1)
= -1, x ∈ [- 1, 0)
= 0, x ∈ [0, 1)
= 1, x ∈ [1, 2)

Similarly, f(x) is discontinuous at the points x = -1, x = 0, x = 1.
Thus all the integer values of x in the interval (-3, 2),
i.e., the points x = -2, x = -1, x = 0 and x = 1 are the required points of discontinuities.

Question 4.
Discuss the continuity of the function f(x) = |2x + 3|, at x = \(\frac{-3}{2}\).
Solution:

Question 5.
Test the continuity of the following functions at the points or intervals indicated against them.


Solution:

Question 6.
Identify discontinuities for the following functions as either a jump or a removable discontinuity.
(i) f(x) = \(\frac{x^{2}-10 x+21}{x-7}\)
Solution:
Given, f(x) = \(\frac{x^{2}-10 x+21}{x-7}\)
It is a rational function and is discontinuous if
x – 7 = 0, i.e., x = 7
∴ f(x) is continuous for all x ∈ R, except at x = 7.
∴ f(7) is not defined.

Thus, \(\lim _{x \rightarrow 7} \mathrm{f}(x)\) exist but f(7) is not defined.
∴ f(x) has a removable discontinuity.

(ii) f(x) = x² + 3x – 2, for x ≤ 4
= 5x + 3, for x > 4.
Solution:
f(x) = x² + 3x – 2, x ≤ 4
= 5x + 3, x > 4
f(x) is a polynomial function for both the intervals.
∴ f(x) is continuous for both the given intervals.
Let us test the continuity at x = 4.

∴ f(x) is discontinuous at x = 4.
∴ f(x) has a jump discontinuity at x = 4.

(iii) f(x) = x² – 3x – 2, for x < -3 = 3 + 8x, for x > -3.
Solution:
f(x) = x² – 3x – 2, x < -3 = 3 + 8x, x > -3
f(x) is a polynomial function for both the intervals.
∴ f(x) is continuous for both the given intervals.
Let us test the continuity at x = -3.

∴ f(x) is discontinuous at x = -3.
∴ f(x) has a jump discontinuity at x = -3

(iv) f(x) = 4 + sin x, for x < π = 3 – cos x for x > π.
Solution:
f(x) = 4 + sin x, x < π = 3 – cos x, x > π
sin x and cos x are continuous for all x ∈ R.
4 and 3 are constant functions.
∴ 4 + sin x and 3 – cos x are continuous for all x ∈ R.
∴ f(x) is continuous for both the given intervals.
Let us test the continuity at x = π.

But f(π) is not defined.
∴ f(x) has a removable discontinuity at x = π.

Question 7.
Show that the following functions have a continuous extension to the point where f(x) is not defined. Also, find the extension.
(i) f(x) = \(\frac{1-\cos 2 x}{\sin x}\), for x ≠ 0.
Solution:
f(x) = \(\frac{1-\cos 2 x}{\sin x}\), for x ≠ 0
Here, f(0) is not defined.
Consider,

But f(0) is not defined.
∴ f(x) has a removable discontinuity at x = 0.
∴ The extension of the original function is
f(x) = \(\frac{1-\cos 2 x}{\sin x}\) for x ≠ 0
= 0 for x = 0
∴ f(x) is continuous at x = 0.

(ii) f(x) = \(\frac{3 \sin ^{2} x+2 \cos x(1-\cos 2 x)}{2\left(1-\cos ^{2} x\right)}\), for x ≠ 0.
Solution:
f(x) = \(\frac{3 \sin ^{2} x+2 \cos x(1-\cos 2 x)}{2\left(1-\cos ^{2} x\right)}\), for x ≠ 0
Here, f(0) is not defined.
Consider,

But f(0) is not defined.
∴ f(x) has a removable discontinuity at x = 0.
∴ The extension of the original function is
f(x) = \(\frac{3 \sin ^{2} x+2 \cos x(1-\cos 2 x)}{2\left(1-\cos ^{2} x\right)}\), x ≠ 0
= \(\frac{7}{2}\), x = 0
∴ f(x) is continuous at x = 0.

(iii) f(x) = \(\frac{x^{2}-1}{x^{3}+1}\), for x ≠ -1
Solution:
f(x) = \(\frac{x^{2}-1}{x^{3}+1}\), for x ≠ -1
Here, f(-1) is not defined.
Consider,

But f(-1) is not defined.
∴ f(x) has a removable discontinuity at x = -1.
∴ The extension of the original function is
f(x) = \(\frac{x^{2}-1}{x^{3}+1}\), x ≠ -1
= \(-\frac{2}{3}\), x = -1
∴ f(x) is continuous at x = \(-\frac{2}{3}\)

Question 8.
Discuss the continuity of the following functions at the points indicated against them.

Solution:

Question 9.
Which of the following functions has a removable discontinuity? If it has a removable discontinuity, redefine the function so that it becomes continuous.


Solution:

Question 10.
(i) If f(x) = \(\frac{\sqrt{2+\sin x}-\sqrt{3}}{\cos ^{2} x}\), for x ≠ \(\frac{\pi}{2}\), is continuous at x = \(\frac{\pi}{2}\) then find f(\(\frac{\pi}{2}\)).
Solution:
f(x) is continuous at x = \(\frac{\pi}{2}\), …..(given)

(ii) If f(x) = \(\frac{\cos ^{2} x-\sin ^{2} x-1}{\sqrt{3 x^{2}+1}-1}\) for x ≠ 0, is continuous at x = 0 then find f(0).
Solution:
f(x) is continuous at x = 0, …..(given)

(iii) If f(x) = \(\frac{4^{x-\pi}+4^{\pi-x}-2}{(x-\pi)^{2}}\) for x ≠ π, is continuous at x = π, then find f(π).
Solution:
f(x) is continuous at x = π, …..(given)

Question 11.
(i) If f(x) = \(\frac{24^{x}-8^{x}-3^{x}+1}{12^{x}-4^{x}-3^{x}+1}\), for x ≠ 0
= k, for x = 0
is continuous at x = 0, then find k.
Solution:
f(x) is continuous at x = 0 …..(given)

(ii) If f(x) = \(\frac{5^{x}+5^{-x}-2}{x^{2}}\), for x ≠ 0
= k, for x = 0
is continuous at x = 0, then find k.
Solution:
f(x) is continuous at x = 0 …..(given)

(iii) If f(x) = \(\frac{\sin 2 x}{5 x}\) – a, for x > 0
= 4 for x = 0
= x² + b – 3, for x < 0
is continuous at x = 0, find a and b.
Solution:
f(x) is continuous at x = 0 ……(given)

(iv) For what values of a and b is the function
f(x) = ax + 2b + 18, for x ≤ 0
= x² + 3a – b, for 0 < x ≤ 2 = 8x – 2, for x > 2,
continuous for every x?
Solution:
f(x) is continuous for every x …..(given)
∴ f(x) is continuous at x = 0 and x = 2.
As f(x) is continuous at x = 0,
\(\lim _{x \rightarrow 0^{-}} \mathrm{f}(x)=\lim _{x \rightarrow 0^{+}} \mathrm{f}(x)\)

∴ (2)² + 3a – b = 8(2) – 2
∴ 4 + 3a – b = 14
∴ 3a – b = 10 …….(ii)
Subtracting (i) from (ii), we get
2a = 4
∴ a = 2
Substituting a = 2 in (i), we get
2 – b = 6
∴ b = -4
∴ a = 2 and b = -4

(v) For what values of a and b is the function
f(x) = \(\frac{x^{2}-4}{x-2}\), for x < 2
= ax² – bx + 3, for 2 ≤ x < 3
= 2x – a + b, for x ≥ 3
continuous for every x on R?
Solution:
f(x) is continuous for every x on R …..(given)
∴ f(x) is continuous at x = 2 and x = 3.
As f(x) is continuous at x = 2,

∴ a(3)² – b(3) + 3 = 2(3) – a + b
∴ 9a – 3b + 3 = 6 – a + b
∴ 10a – 4b = 3 …..(ii)
Multiplying (i) by 2, we get
8a – 4b = 2 ….(iii)
Subtracting (ii) from (iii), we get
-2a = -1
∴ a = \(\frac{1}{2}\)
Substituting a = \(\frac{1}{2}\) in (i), we get
4(\(\frac{1}{2}\)) – 2b = 1
∴ 2 – 2b = 1
∴ 1 = 2b
∴ b = \(\frac{1}{2}\)
∴ a = \(\frac{1}{2}\) and b = \(\frac{1}{2}\)

Question 12.
Discuss the continuity of f on its domain, where
f(x) = |x + 1|, for -3 ≤ x ≤ 2
= |x – 5|, for 2 < x ≤ 7
Solution:

Question 13.
Discuss the continuity of f(x) at x = \(\frac{\pi}{4}\) where,
f(x) = \(\frac{(\sin x+\cos x)^{3}-2 \sqrt{2}}{\sin 2 x-1}\), for x ≠ \(\frac{\pi}{4}\)
= \(\frac{3}{\sqrt{2}}\), for x = \(\frac{\pi}{4}\)
Solution:

Question 14.
Determine the values of p and q such that the following function is continuous on the entire real number line.
f(x) = x + 1, for 1 < x < 3
= x² + px + q, for |x – 2| ≥ 1.
Solution:
|x – 2| ≥ 1
∴ x – 2 ≥ 1 or x – 2 ≤ -1
∴ x ≥ 3 or x ≤ 1
∴ f(x) = x² + px + q for x ≥ 3 as well as x ≤ 1
Thus, f(x) = x² + px + q; x ≤ 1
= x + 1; 1 < x < 3 = x² + px + q; x > 3
f(x) is continuous for all x ∈ R.
∴ f(x) is continuous at x = 1 and x = 3.
As f(x) is continuous at x = 1,

Subtracting (i) from (ii), we get
2p = -6
∴ p = -3
Substituting p = -3 in (i), we get
-3 + q = 1
∴ q = 4
∴ p = -3 and q = 4

Question 15.
Show that there is a root for the equation 2x³ – x – 16 = 0 between 2 and 3.
Solution:
Let f(x) = 2x³ – x – 16
f(x) is a polynomial function and hence it is continuous for all x ∈ R.
A root of f(x) exists, if f(x) = 0 for at least one value of x.
f(2) = 2(2)³ – 2 – 16 = -2 < 0
f(3) = 2(3)³ – 3 – 16 = 35 > 0
∴ f(2) < 0 and f(3) > 0
∴ By intermediate value theorem,
there has to be point ‘c’ between 2 and 3 such that f(c) = 0.
∴ There is a root of the given equation between 2 and 3.

Question 16.
Show that there is a root for the equation x³ – 3x = 0 between 1 and 2.
Solution:
Let f(x) = x³ – 3x
f(x) is a polynomial function and hence it is continuous for all x ∈ R.
A root of f(x) exists, if f(x) = 0 for at least one value of x.
f(1) = (1)³ – 3(1) = -2 < 0
f(2) = (2)³ – 3(2) = 2 > 0
∴ f(1) < 0 and f(2) > 0
∴ By intermediate value theorem,
there has to be point ‘c’ between 1 and 2 such that f(c) = 0.
There is a root of the given equation between 1 and 2.

Question 17.
Let f(x) = ax + b (where a and b are unknown)
= x² + 5 for x ∈ R
Find the values of a and b, so that f(x) is continuous at x = 1.

Solution:
f(x) = x² + 5, x ∈ R
∴ f(1) = 1 + 5 = 6
If f(x) = ax + b is continuous at x = 1, then
f(1) = \(\lim _{x \rightarrow 1}(a x+b)\) = a + b
∴ 6 = a + b where, a, b ∈ R
∴ There are infinitely many values of a and b.

Question 18.
Activity: Suppose f(x) = px + 3 for a ≤ x ≤ b
= 5x² – q for b < x ≤ c
Find the condition on p, q, so that f(x) is continuous on [a, c], by filling in the boxes.

Solution:

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

Maharashtra State Board 11th Maths Solutions Chapter 7 Limits Miscellaneous Exercise 7

I. Select the correct answer from the given alternatives.

Question 1.
\(\lim _{x \rightarrow 2}\left(\frac{x^{4}-16}{x^{2}-5 x+6}\right)=\)
(A) 23
(B) 32
(C) -32
(D) -16
Answer:
(C) -32
Hint:

Question 2.
\(\lim _{x \rightarrow-2}\left(\frac{x^{7}+128}{x^{3}+8}\right)=\)
(A) \(\frac{56}{3}\)
(B) \(\frac{112}{3}\)
(C) \(\frac{121}{3}\)
(D) \(\frac{28}{3}\)
Answer:
(B) \(\frac{112}{3}\)
Hint:

Question 3.
\(\lim _{x \rightarrow 3}\left(\frac{1}{x^{2}-11 x+24}+\frac{1}{x^{2}-x-6}\right)=\)
(A) \(-\frac{2}{25}\)
(B) \(\frac{2}{25}\)
(C) \(\frac{7}{25}\)
(D) \(-\frac{7}{25}\)
Answer:
(A) \(-\frac{2}{25}\)
Hint:

Question 4.
\(\lim _{x \rightarrow 5}\left(\frac{\sqrt{x+4}-3}{\sqrt{3 x-11-2}}\right)=\)
(A) \(\frac{-2}{9}\)
(B) \(\frac{2}{7}\)
(C) \(\frac{5}{9}\)
(D) \(\frac{2}{9}\)
Answer:
(D) \(\frac{2}{9}\)
Hint:

Question 5.
\(\lim _{x \rightarrow \frac{\pi}{3}}\left(\frac{\tan ^{2} x-3}{\sec ^{3} x-8}\right)=\)
(A) 1
(B) \(\frac{1}{2}\)
(C) \(\frac{1}{3}\)
(D) \(\frac{1}{4}\)
Answer:
(C) \(\frac{1}{3}\)
Hint:

Question 6.
\(\lim _{x \rightarrow 0}\left(\frac{5 \sin x-x \cos x}{2 \tan x-3 x^{2}}\right)=\)
(A) 0
(B) 1
(C) 2
(D) 3
Answer:
(C) 2
Hint:

Question 7.
\(\lim _{x \rightarrow \frac{\pi}{2}}\left[\frac{3 \cos x+\cos 3 x}{(2 x-\pi)^{3}}\right]=\)
(A) \(\frac{3}{2}\)
(B) \(\frac{1}{2}\)
(C) \(-\frac{1}{2}\)
(D) \(\frac{1}{4}\)
Answer:
(C) \(-\frac{1}{2}\)
Hint:

Question 8.
\(\lim _{x \rightarrow 0}\left(\frac{15^{x}-3^{x}-5^{x}+1}{\sin ^{2} x}\right)=\)
(A) log 15
(B) log 3 + log 5
(C) log 3 . log 5
(D) 3 log 5
Answer:
(C) log 3 . log 5
Hint:

Question 9.
\(\lim _{x \rightarrow 0}\left(\frac{3+5 x}{3-4 x}\right)^{\frac{1}{x}}=\)
(A) e³
(B) e⁶
(C) e⁹
(D) e^-3
Answer:
(A) e³
Hint:

Question 10.
\(\lim _{x \rightarrow 0}\left[\frac{\log (5+x)-\log (5-x)}{\sin x}\right]=\)
(A) \(\frac{3}{2}\)
(B) \(-\frac{5}{2}\)
(C) \(-\frac{1}{2}\)
(D) \(\frac{2}{5}\)
Answer:
(D) \(\frac{2}{5}\)
Hint:

Question 11.
\(\lim _{x \rightarrow \frac{\pi}{2}}\left(\frac{3^{\cos x}-1}{\frac{\pi}{2}-x}\right)=\)
(A) 1
(B) log 3
(C) \(3^{\frac{\pi}{2}}\)
(D) 3 log 3
Answer:
(B) log 3
Hint:

Question 12.
\(\lim _{x \rightarrow 0}\left[\frac{x \cdot \log (1+3 x)}{\left(e^{3 x}-1\right)^{2}}\right]=\)
(A) \(\frac{1}{\mathrm{e}^{9}}\)
(B) \(\frac{1}{\mathrm{e}^{3}}\)
(C) \(\frac{1}{9}\)
(D) \(\frac{1}{3}\)
Answer:
(D) \(\frac{1}{3}\)
Hint:

Question 13.
\(\lim _{x \rightarrow 0}\left[\frac{\left(3^{\sin x}-1\right)^{3}}{\left(3^{x}-1\right) \cdot \tan x \cdot \log (1+x)}\right]=\)
(A) 3 log 3
(B) 2 log 3
(C) (log 3)²
(D) (log 3)³
Answer:
(C) (log 3)²
Hint:

Question 14.
\(\lim _{x \rightarrow 3}\left[\frac{5^{x-3}-4^{x-3}}{\sin (x-3)}\right]=\)
(A) log 5 – 4
(B) log \(\frac{5}{4}\)
(C) \(\frac{\log 5}{\log 4}\)
(D) \(\frac{\log 5}{4}\)
Answer:
(B) log \(\frac{5}{4}\)
Hint:

Question 15.
\(\lim _{x \rightarrow \infty}\left[\frac{(2 x+3)^{7}(x-5)^{3}}{(2 x-5)^{10}}\right]=\)
(A) \(\frac{3}{8}\)
(B) \(\frac{1}{8}\)
(C) \(\frac{1}{6}\)
(D) \(\frac{1}{4}\)
Answer:
(B) \(\frac{1}{8}\)
Hint:

(II) Evaluate the following.

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

Question 2.
\(\lim _{x \rightarrow 0}[x]\) ([*] is a greatest integer function.)
Solution:

Question 3.
If f(r) = πr² then find \(\lim _{h \rightarrow 0}\left[\frac{f(r+h)-f(r)}{h}\right]\)
Solution:

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

Question 5.
Find the limit of the function, if it exists, at x = 1
\(f(x)=\left\{\begin{array}{lll}
7-4 x & \text { for } & x<1 \\
x^{2}+2 & \text { for } & x \geq 1
\end{array}\right.\)
Solution:

Question 6.
Given that 7x ≤ f(x) ≤ 3x² – 6 for all x. Determine the value of \(\lim _{x \rightarrow 3} f(x)\)
Solution:

Question 7.
\(\lim _{x \rightarrow 0}\left[\frac{\sec x^{2}-1}{x^{4}}\right]\)
Solution:

Question 8.
\(\lim _{x \rightarrow 0}\left[\frac{e^{x}+e^{-x}-2}{x \cdot \tan x}\right]\)
Solution:

Question 9.
\(\lim _{x \rightarrow 0}\left[\frac{x\left(6^{x}-3^{x}\right)}{\cos (6 x)-\cos (4 x)}\right]\)
Solution:

Question 10.
\(\lim _{x \rightarrow 0}\left[\frac{a^{3 x}-a^{2 x}-a^{x}+1}{x \cdot \tan x}\right]\)
Solution:

Question 11.
\(\lim _{x \rightarrow a}\left[\frac{\sin x-\sin a}{x-a}\right]\)
Solution:

Question 12.
\(\lim _{x \rightarrow 2}\left[\frac{\log x-\log 2}{x-2}\right]\)
Solution:

Question 13.
\(\lim _{x \rightarrow 1}\left[\frac{a b^{x}-a^{x} b}{x^{2}-1}\right]\)
Solution:

Question 14.
\(\lim _{x \rightarrow 0}\left[\frac{\left(5^{x}-1\right)^{2}}{\left(2^{x}-1\right) \log (1+x)}\right]\)
Solution:

Question 15.
\(\lim _{x \rightarrow \infty}\left[\frac{(2 x+1)^{2}(7 x-3)^{3}}{(5 x+2)^{5}}\right]\)
Solution:

Question 16.
\(\lim _{x \rightarrow a}\left[\frac{x \cos a-a \cos x}{x-a}\right]\)
Solution:

Question 17.
\(\lim _{x \rightarrow \frac{\pi}{4}}\left[\frac{(\sin x-\cos x)^{2}}{\sqrt{2}-\sin x-\cos x}\right]\)
Solution:

Question 18.
\(\lim _{x \rightarrow 1}\left[\frac{2^{2 x-2}-2^{x}+1}{\sin ^{2}(x-1)}\right]\)
Solution:

Question 19.
\(\lim _{x \rightarrow 1}\left[\frac{4^{x-1}-2^{x}+1}{(x-1)^{2}}\right]\)
Solution:

Question 20.
\(\lim _{x \rightarrow 1}\left[\frac{\sqrt{x}-1}{\log x}\right]\)
Solution:

Question 21.
\(\lim _{x \rightarrow 0}\left(\frac{\sqrt{1-\cos x}}{x}\right)\)
Solution:

Question 22.
\(\lim _{x \rightarrow 1}\left(\frac{x+3 x^{2}+5 x^{3}+\cdots \cdots \cdots \cdots \cdots+(2 n-1) x^{n}-n^{2}}{x-1}\right)\)
Solution:

Question 23.
\(\lim _{x \rightarrow 0} \frac{1}{x^{12}}\left[1-\cos \left(\frac{x^{2}}{2}\right)-\cos \left(\frac{x^{4}}{4}\right)+\cos \left(\frac{x^{2}}{2}\right) \cdot \cos \left(\frac{x^{4}}{4}\right)\right]\)
Solution:

Question 24.
\(\lim _{x \rightarrow \infty}\left(\frac{8 x^{2}+5 x+3}{2 x^{2}-7 x-5}\right)^{\frac{4 x+3}{8 x-1}}\)
Solution:

Balbharti Maharashtra State Board 11th Maths Book Solutions Pdf Chapter 7 Limits Ex 7.7 Questions and Answers.

Maharashtra State Board 11th Maths Solutions Chapter 7 Limits Ex 7.7

I. Evaluate the following:

Question 1.
\(\lim _{x \rightarrow \infty}\left[\frac{a x^{3}+b x^{2}+c x+d}{e x^{3}+f x^{2}+g x+h}\right]\)
Solution:

Question 2.
\(\lim _{x \rightarrow \infty}\left[\frac{x^{3}+3 x+2}{(x+4)(x-6)(x-3)}\right]\)
Solution:

Question 3.
\(\lim _{x \rightarrow \infty}\left[\frac{7 x^{2}+5 x-3}{8 x^{2}-2 x+7}\right]\)
Solution:

II. Evaluate the following:

Question 1.
\(\lim _{x \rightarrow \infty}\left[\frac{7 x^{2}+2 x-3}{\sqrt{x^{4}+x+2}}\right]\)
Solution:

Question 2.
\(\lim _{x \rightarrow \infty}\left[\sqrt{x^{2}+4 x+16}-\sqrt{x^{2}+16}\right]\)
Solution:

Question 3.
\(\lim _{x \rightarrow \infty}\left[\sqrt{x^{4}+4 x^{2}}-x^{2}\right]\)
Solution:

III. Evaluate the following:

Question 1.
\(\lim _{x \rightarrow \infty}\left[\frac{\left(3 x^{2}+4\right)\left(4 x^{2}-6\right)\left(5 x^{2}+2\right)}{4 x^{6}+2 x^{4}-1}\right]\)
Solution:

Question 2.
\(\lim _{x \rightarrow \infty}\left[\frac{(3 x-4)^{3}(4 x+3)^{4}}{(3 x+2)^{7}}\right]\)
Solution:

Question 3.
\(\lim _{x \rightarrow \infty}[\sqrt{x}(\sqrt{x+1}-\sqrt{x})]\)
Solution:

Question 4.
\(\lim _{x \rightarrow \infty}\left[\frac{(2 x-1)^{20}(3 x-1)^{30}}{(2 x+1)^{50}}\right]\)
Solution:

Question 5.
\(\lim _{x \rightarrow \infty}\left[\frac{\sqrt{x^{2}+5}-\sqrt{x^{2}-3}}{\sqrt{x^{2}+3}-\sqrt{x^{2}+1}}\right]\)
Solution: