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) = 2x^{2} + 3, g(x) = 5x – 2, then find

(i) fog

(ii) gof

(iii) fof

(iv) gog

Solution:

f(x) = 2x^{2} + 3, g(x) = 5x – 2

(i) (fog) (x) = f(g(x))

= f(5x – 2)

= 2(5x – 2)^{2} + 3

= 2(25x^{2} – 20x + 4) + 3

= 50x^{2} – 40x + 11

(ii) (gof) (x) = g(f(x))

= g(2x^{2} + 3)

= 5(2x + 3) – 2

= 10x^{2} + 13

(iii) (fof) (x) = f(f(x))

= f(2x^{2} + 3)

= 2(2x^{2} + 3)^{2} + 3

= 2(4x^{4} + 12x^{2} + 9) + 3

= 8x^{4} + 24x^{2} + 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) = x^{3} + 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) = x^{3} + 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) = 5x^{2}

(ii) f(x) = 8

(iii) f(x) = \(\frac{6 x-7}{3}\)

(iv) f(x) = \(\sqrt{4 x+5}\)

(v) f(x) = 9x^{3} + 8

(vi) f(x) =

Solution:

(i) f(x) = 5x^{2} = y (say)

For two values (x_{1} and x_{2}) 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(x_{1}) = f(x_{2})

∴ \(\frac{6 x_{1}-7}{3}=\frac{6 x_{2}-7}{3}\)

∴ x_{1} = x_{2}

∴ 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(x_{1}) = f(x_{2})

∴ \(\sqrt{4 x_{1}+5}=\sqrt{4 x_{2}+5}\)

∴ x_{1} = x_{2}

∴ f is a one-one function.

f(x) = \(\sqrt{4 x+5}\) = y, (say) y ≥ 0

Squaring on both sides, we get

y^{2} = 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) 9x^{3} + 8

Let f(x_{1}) = f(x_{2})

∴ \(9 x_{1}^{3}+8=9 x_{2}^{3}+8\)

∴ x_{1} = x_{2}

∴ f is a one-one function.

∴ f(x) = 9x^{3} + 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 x_{1}, x_{2} the values of the function are equal.

i.e. f(x_{1}) = f(x_{2})

∴ 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) = x^{2} + 3, x ≤ 2

= 5x + 7, x > 2

(i) f(3) = 5(3) + 7

= 15 + 7

= 22

(ii) f(2) = 2^{2} + 3

= 4 + 3

= 7

(iii) f(0) = 0^{2} + 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

= x^{2}, 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) = 5^{2} = 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) x^{2} + 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) x^{2} + 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