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 + y^{2} = 9

(iii) x^{2} – 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 + y^{2} = 9

∴ y^{2} = 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) x^{2} – y = 25

∴ y = x^{2} – 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) = m^{2} – 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) = m^{2} – 3m + 1

(i) f(0) = 0^{2} – 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

= x^{2} + 2x + 1 – 3x – 3 + 1

= x^{2} – 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) = 6x^{2} + x – 2

(iv) g(x) = x^{3} – 2x^{2} – 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 – 2x^{2} = 0

∴ x^{2} = 9

∴ x = ±3

(iii) g(x) = 6x^{2} + x – 2

g(x) = 0

∴ 6x^{2} + 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) = x^{3} – 2x^{2} – 5x + 6

= ( x- 1) (x^{2} – 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) = x^{4} + 2x^{2}, g(x) = 11x^{2}

(ii) f(x) = √x – 3, g(x) = 5 – x

Solution:

(i) f(x) = x^{4} + 2x^{2}, g(x) = 11x^{2}

f(x) = g(x)

∴ x^{4} + 2x^{2} = 11x^{2}

∴ x^{4} – 9x^{2} = 0

∴ x^{2} (x^{2} – 9) = 0

∴ x^{2} = 0 or x^{2} – 9 = 0

∴ x = 0 or x^{2} = 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 + x^{2} – 16x

∴ x^{2} – 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) = 7x^{2} + 4x – 1

Solution:

f(x) = 7x^{2} + 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) = -x^{2} + 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 – x^{2} ≥ 0

∴ x^{2} ≤ 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) = s^{2}

(b) perimeter (P) = 4s

∴ s = \(\frac{\mathrm{P}}{4}\)

Area (A) = s^{2} = \(\left(\frac{\mathrm{P}}{4}\right)^{2}\)

∴ 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) = πr^{2}

(ii) Diameter (d) = 2r

∴ r = \(\frac{\mathrm{d}}{2}\)

∴ Area (A) = πr^{2} = \(\frac{\pi \mathrm{d}^{2}}{4}\)

(iii) Circumference (C) = 2πr

∴ r = \(\frac{C}{2 \pi}\)

Area (A) = πr^{2} = \(\pi\left(\frac{\mathrm{C}}{2 \pi}\right)^{2}\)

∴ 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) = x^{2}

Solution:

f: N → N given by f(x) = x^{2}

∴ f is injective.

For every y = x^{2} ∈ N, there does not exist x ∈ N.

Example: 7 ∈ N (codomain) for which there is no x in domain N such that x^{2} = 7

∴ f is not surjective.

(ii) f : Z → Z given by f(x) = x^{2}

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 x^{2} ≥ 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) = x^{2}

Solution:

f : R → R given by f(x) = x^{2}

∴ f is not injective.

f(x) = x^{2} ≥ 0

Therefore all negative integers of codomain are not images under f.

∴ f is not surjective.

(iv) f : N → N given by f(x) = x^{3}

Solution:

f: N → N given by f(x) = x^{3}

∴ 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) = x^{3}

Solution:

f: R → R given by f(x) = x^{3}

∴ 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(x_{1}) = f(x_{2})

Then, x_{1} = x_{2} for all x_{1}, x_{2} …..(i)

g is a one-one function.

Let g(y_{1}) = g(y_{2})

Then, y_{1} = y_{2} for all y_{1}, y_{2} …..(ii)

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

∴ g(f(x_{1})) = g(f(x_{2}))

∴ g(y_{1}) = g(y_{2}),

where y_{1} = f(x_{1}), y_{2} = f(x_{2}) ∈ B

∴ y_{1} = y_{2} …..[From (ii)]

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

∴ x_{1} = x_{2} ….[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) 2^{5} = 32

(ii) 54^{0} = 1

(iii) 23^{1} = 23

(iv) \(9^{\frac{3}{2}}\) = 27

(v) 3^{-4} = \(\frac{1}{81}\)

(vi) 10^{-2} = 0.01

(vii) e^{2} = 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) log_{2} 64 = 6

(ii) \(\log _{5} \frac{1}{25}\) = -2

(iii) log_{10} 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) log_{2} 64 = 6

∴ 64 = 2^{6}, i.e., 2^{6} = 64

Question 19.

Find the domain of

(i) f(x) = ln (x – 5)

(ii) f(x) = log10 (x^{2} – 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) = log_{10} (x^{2} – 5x + 6)

x^{2} – 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 (2^{5} x 3)

= \(\frac{1}{2}\) (log 2^{5} + log 3) …..[∵ log mn = log m + log n]

= \(\frac{1}{2}\) (5 log 2 + log 3) ……[∵ log m^{n} = 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) = ax^{2} – bx + 6 and f(2) = 3 and f(4) = 30, find a and b.

Solulion:

f(x) = ax^{2} – 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) 2log_{10} x = 1 + \(\log _{10}\left(x+\frac{11}{10}\right)\)

Solution:

∴ x^{2} = 10x + 11

∴ x^{2} – 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) log_{2} x + log_{4} x + log_{16} x = \(\frac{21}{4}\)

Solution:

(iv) x + log_{10} (1 + 2^{x}) = x log_{10} 5 + log_{10} 6

Solution:

∴ a + a^{2} = 6

∴ a^{2} + 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_{a}bc, 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: