Students get through Maths 1A Important Questions Inter 1st Year Maths 1A Functions Important Questions which are most likely to be asked in the exam.

## Intermediate 1st Year Maths 1A Functions Important Questions

I.

Question 1.

If f: R – {0} → R is defined by f(x) = x + \(\frac{1}{x}\), then prove that (f(x))^{2} = f(x^{2}) + f(1).

Solution:

f : R – {0} → R and

f(x) = x + \(\frac{1}{x}\)

Now f(x^{2}) + f(1) = (x^{2} + \(\frac{1}{x^{2}}\)) + (1 + \(\frac{1}{1}\))

= x^{2} + 2 + \(\frac{1}{x^{2}}\) = (x + \(\frac{1}{x}\))^{2} = (f(x))^{2}

∴ (f(x))^{2} = f(x^{2}) + f(1)

Question 2.

If the function f is defined by

then find the values, if exist, of f(4), f(2.5), f(-2), f(-4), f(0), f(-7). (Mar. ’14)

Solution:

Domain of f is(-∞, -3) ∪ [-2, 2] ∪ (3, ∞)

i) Since f(x) = 3x – 2 for x > 3

f(4) = 3(4) – 2 = 10

ii) 2.5 does not belong to domain off, hence

f(2.5) is not desired.

iii) ∵ f(x) = x^{2} – 2 for -2 ≤ x ≤ 2

f(-2) = (-2)^{2} – 2 = 4 – 2 = 2

iv) ∵ f(x) = 2x + 1 for x < -3

f(-4) = 2(-4) + 1 = – 8 + 1 = -7

v) ∵ f(x) = x^{2} – 2, for -2 ≤ x ≤ 2

f(0)= (0)^{2} – 2 = 0 – 2 = -2

vi) ∵ f(x) = 2x + 1 for x < -3

f(-7) = 2(-7) + 1 = -14 + 1 = -13

Question 3.

If A = {0, \(\frac{\pi}{6}\), \(\frac{\pi}{4}\), \(\frac{\pi}{3}\), \(\frac{\pi}{2}\)} and f : A → B is a surjection defined by f(x) = cos x, then find B. (A.P. Mar. ’16, ’11; May ’11)

Solution:

∵ f : A → B is a sujection defined by

f(x) = cos x

B = Range of f = f(A)

= {f(0), f(\(\frac{\pi}{6}\)), f(\(\frac{\pi}{4}\)), f(\(\frac{\pi}{3}\)), f(\(\frac{\pi}{2}\))}

= {cos 0, cos \(\frac{\pi}{6}\), cos \(\frac{\pi}{4}\), cos \(\frac{\pi}{3}\), cos \(\frac{\pi}{2}\)}

= {1, \(\frac{\sqrt{3}}{2}\), \(\frac{1}{\sqrt{2}}\), \(\frac{1}{2}\), 0}

Question 4.

Determine whether the function f: R → R defined by f(x) = \(\frac{e^{|x|}-e^{-x}}{e^{x}+e^{-x}}\) is an injection or a surjection or a bijection.

Solution:

f(x) = \(\frac{e^{|x|}-e^{-x}}{e^{x}+e^{-x}}\)

f(0) = \(\frac{e^{0}-e^{-0}}{e^{0}+e^{-0}}\) = \(\frac{1-1}{1+1}\) = 0

f(-1) = \(\frac{e^{1}-e^{1}}{e^{-1}+e^{1}}\) = 0

∵ f(0) = f(-1)

⇒ f is not an injection.

Let y = f(x) = \(\frac{e^{|x|}-e^{-x}}{e^{x}+e^{-x}}\)

When y = 1, there is no x ∈ R such that f(x) = 1

⇒ f is not a surjection

if there is such x ∈ R then

\(\frac{e^{|x|}-e^{-x}}{e^{x}+e^{-x}}\) = 1

⇒ e^{|x|} – e^{-x} = e^{x} + e^{-x}, clearly x ≠ 0

for x > 0, this equation gives

e^{x} – e^{-x} = e^{x} + e^{-x} ⇒ -e^{-x} = e^{-x} which is not possible

for x < 0, this equation gives

e^{-x} – e^{-x} = e^{x} + e^{-x}

⇒ e^{-x} = e^{x} which is also not possible.

Question 5.

Determine whether the function f: R → R defined by

is an injection or a surjection or a bijection

Solution:

∵ f(x’) = x for x > 2

⇒ f(3) = 3

∵ f(x) = 5x – 2, for x ≤ 2

⇒ f(1) = 5(1) – 2 = 3

∵ 1 and 3 have same f-image.

Hence f is not an injection.

Let y ∈ R then y > 2 (or) y ≤ 2.

If y > 2 take x = y ∈ R so that

f(x) = x = y

If y ≤ 2 take x = \(\frac{y+2}{5}\) ∈ R and

x = \(\frac{y+2}{5}\) < 1

∴ f(x) = 5x – 2 = 5\(\left[\frac{y+2}{5}\right]\) – 2 = y

∴ f is a surjection

∵ f is not an injection ⇒ It is not a bijection.

Question 6.

Find the domain of definition of the function y(x), given by the equation 2^{x} + 2^{y} = 2.

Solution:

2^{x} = 2 – 2^{y} < 2 (∵ 2^{y} > 0)

⇒ log_{2} 2^{x} < log_{2}2

⇒ x < 1

∴ Domain = (-∞, 1)

Question 7.

It f: R → R is defined as f(x + y) = f(x)+ f(y) ∀ x, y ∈R and f(1) = 7, then find \(\sum_{r=1}^{n} f(r)\).

Solution:

Consider f(2) = f(1 + 1) = f (1) + f (1) = 2f(1)

f(3) =f(2 + 1) = f(2) + f(1) = 3f(1)

Similarly f(r) = rf(1)

∴ \(\sum_{r=1}^{n} f(r)\) = f(1) + f(2) + ….. + f(n)

= f(1) + 2f(1)+ ….. + nf(1)

= f(1) (1 + 2 + ….. + n)

= \(\frac{7 n(n+1)}{2}\)

Question 8.

If f(x) = \(\frac{\cos ^{2} x+\sin ^{4} x}{\sin ^{2} x+\cos ^{4} x}\) ∀ x ∈ R then show that f(2012) = 1.)

Solution:

f(x) = \(\frac{\cos ^{2} x+\sin ^{4} x}{\sin ^{2} x+\cos ^{4} x}\)

= \(\frac{1-\sin ^{2} x+\sin ^{4} x}{1-\cos ^{2} x+\sin ^{4} x}\)

Question 9.

If f : R → R, g : R → R are defined by f(x) = 4x – 1 and g(x) = x^{2} + 2 then find

i) (gof)(x)

ii) (gof)\(\left(\frac{a+1}{4}\right)\)

iii) (fof)(x)

iv) go(fof)(0) (Mar. ’05)

Solution:

Given f : R → R and g : R → R and

f(x) = 4x – 1 and g(x) = x^{2} + 2

(i) (gof) (x) = g (f(x))

= g (4x – 1), ∵ f(x) = 4x – 1

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

= 16x^{2} – 8x + 1 + 2

= 16x^{2} – 8x + 3

(ii) (gof)\(\left(\frac{a+1}{4}\right)\) = \(g\left(f\left(\frac{a+1}{4}\right)\right)\)

= \(g\left(4\left(\frac{a+1}{4}\right)-1\right)\)

= g(a)

= a^{2} + 2

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

= f(4x – 1), ∵ f(x) = 4x – 1

= 4(4x – 1) – 1

= 16x – 4 – 1 = 16x – 5

(iv) (fof)(0) = f(f(0))

= f(4 × 0 – 1)

= f(-1)

= 4(-1) – 1 = -5

Now go(fof)(0)

= g(-5) = (-5)^{2} + 2 = 27

Question 10.

If f: [0, 3] → [0, 3] is defined by

then show that f[0,3] ⊆ [0, 3] and find fof.

Solution:

0 ≤ x ≤ 2 ⇒ 1 ≤ 1 + x ≤ 3 ——— (1)

2 < x ≤ 3 ⇒ -3 ≤ -x ≤ -2

⇒ 3 – 3 ≤ 3 – x ≤ 3 – 2

⇒ 0 ≤ 3 – x < 1 ——- (2)

from (1) and (2).

f[0, 3] ⊆ [0, 3]

When 0 ≤ x ≤ 1, we have

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

= f(1 + x) = 1 + (1 + x) = 2 + x

[∵ 1 ≤ 1 + x ≤ 2]

When 1 < x ≤ 2, we have

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

= f(1 + x)

= 3 – (1 + x)

= 2 – x, [∵ 2 < 1 + x ≤ 3]

When 2 < x ≤ 3, we have

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

= f(3 – x)

= 1 + (3 – x)

= 4 – x, [∵ 0 ≤ 3 – x < 1]

Question 11.

If f, g : R → R are defined

and

then find (fog)(π) + (gof)(e).

Solution:

(fog)(π) = f(g(π)) = f(0) = 0

(gof)(e) = g(f(e)) = g (1) = -1

∴ (fog)(π) + (gof) (e) = -1.

Question 12.

Let A = {1, 2, 3), B = {a, b, c}, C = (p, q, r}

If f: A → B, g: B → C are defined by

f = {(1, a), (2, c), (3, b)},

g = {(a, q), (b, r), (c, p)} then

show that f^{-1}og^{-1} = (gof)^{-1}

Solution:

Given that

f = {(1, a), (2, c), (3, b)} and

g = {(a, q), (b, r), (c, p)}

then g f = {(1, q), (2, p), (3, r)}

⇒ (gof)^{-1} = {(q, 1), (p, 2), (r, 3)}

⇒ g^{-1} = {(q, a), (r, b), (p, c)} and

f^{-1} = {(a, 1), (c, 2), (b, 3)}

f^{-1} g^{-1} = {(q, 1), (r, 3), (p, 2)}

⇒ (gof)^{-1} = f^{-1}og^{-1}

Question 13.

If f: Q → Q, is defined by f(x) = 5x + 4 for all x ∈ Q, show that f is a bijection and find f^{-1}.(A.P Mar. ’16, May ’12, ’05)

Solution:

Let x_{1}, x_{2} ∈ Q

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

⇒ 5x_{1} + 4 = 5x_{2} + 4

⇒ 5x_{1} = 5x_{2}

⇒ x_{1} = x_{2}

∴ f is an injection.

Let y ∈ Q then x = \(\frac{y-4}{5}\) ∈ Q and

f(x) = f\(\left(\frac{y-4}{5}\right)\) = 5\(\left(\frac{y-4}{5}\right)\) + 4 = y

∴ f is a surjection.

Hence f is a bijection

∴ f^{-1} : Q → Q is also bijection

We have (fof^{-1})(x) = I(x)

⇒ f(f^{-1}(x)) = x, ∵ f(x) = 5x + 4

⇒ 5f^{-1}(x) + 4 = x

⇒ f^{-1}(x) =\(\frac{x-4}{5}\) for ∀ x ∈ Q

Question 14.

Find the domains of the following real valued functions.

i) f(x) = \(\frac{1}{6 x-x^{2}-5}\)

Solution:

f(x) = \(\frac{1}{6 x-x^{2}-5}\) = \(\frac{1}{(x-1)(5-x)}\) ∈ R

⇔ (x – 1) (5 – x) ≠ 0

⇔ x ≠ 1, 5

∴ Domain of f is R – {1, 5}

ii) f(x) = \(\frac{1}{\sqrt{x^{2}-a^{2}}}\), (a > 0) (A.P.) (Mar. ’15)

Solution:

f(x) = \(\frac{1}{\sqrt{x^{2}-a^{2}}}\) ∈ R

⇔ x^{2} – a^{2} > 0

⇔ (x + a)(x – a) > 0

⇔ x ∈ (-∞, -a) ∪ (a, ∞)

∴ Domain of f is (-∞, -a) ∪ (a, ∞) = R – [-a, a]

iii) f(x) = \(\sqrt{(x+2)(x-3)}\)

Solution:

f(x) = \(\sqrt{(x+2)(x-3)}\) ∈ R

⇔ (x + 2)(x – 3) ≥ 0

⇔ x ∈ (-∞, -2) ∪ [3, ∞)

∴ Domain of f is

(-∞, -2] ∪ [3, ∞) = R – (-2, 3)

iv) f(x) = \(\sqrt{(x-\alpha)(\beta-x)}\) (0 < α < β)

Solution:

f(x) = \(\sqrt{(x-\alpha)(\beta-x)}\) ∈R

⇔ (x – α) (β – α) ≥ 0

⇔ α ≤ x ≤ β ; (∵ α < β)

⇔ x ∈ [α, β]

∴ Domain of f is [α, β]

v) f(x) = \(\sqrt{2-x}\) + \(\sqrt{1+x}\)

Solution:

f(x) = \(\sqrt{2-x}\) + \(\sqrt{1+x}\) + x ∈ R

⇔ 2 – x ≥ 0 and 1 + x ≥ 0

⇔ 2 ≥ x and x ≥ -1

⇔ -1 ≤ x ≤ 2

⇔ x ∈ [-1, 2]

∴ Domain of f is [-1, 2].

vi) f(x) = \(\sqrt{x^{2}-1}\) + \(\frac{1}{\sqrt{x^{2}-3 x+2}}\)

Solution:

f(x) = \(\sqrt{x^{2}-1}\) + \(\frac{1}{\sqrt{x^{2}-3 x+2}}\) ∈ R

⇔ x^{2} – 1 ≥ 0 and x^{2} – 3x + 2 > 0

⇔ (x + 1)(x – 1) ≥ 0 and (x – 1)(x – 2) > 0

⇔ x ∈ (-∞, -1] ∪ [1, ∞) and x ∈ (-∞, 1)u(2, ∞)

⇔ x ∈ (R – (-1, 1)) ∩ (R – [1, 2])

⇔ x ∈ R – {(-1, 1) ∪ [1,2]}

⇔ x ∈ R – (-1, 2]

⇔ x ∈ (-∞, -1] ∪ (2, ∞)

∴ Domain of f is (-∞, -1) ∪ (2, ∞) = R – (-1, 2]

vii) f(x) = \(\frac{1}{\sqrt{|x|-x}}\)

Solution:

f(x) = \(\frac{1}{\sqrt{|x|-x}}\) ∈ R

⇔ |x| – x > 0

⇔ |x| > x

⇔ x ∈ (-∞, 0)

∴ Domain of f is (-∞, 0)

viii) f(x) = \(\sqrt{|x|-x}\)

Solution:

\(\sqrt{|x|-x}\) ∈ R

⇔ |x| – x ≥ 0

⇔ |x| ≥ x

⇔ x ∈ R

∴ Domain of f is R

Question 15.

If f = ((4, 5), (5, 6), (6, -4)} and g = ((4, -4), (6, 5), (8, 5)} then find

(i) f + g

(ii) f – g

(iii) 2f + 4g

(iv) f + 4

(v) fg

(vi) \(\frac{\mathbf{f}}{\mathbf{g}}\)

(vii) |f|

(viii) \(\sqrt{f}\)

(ix) f^{2}

(x) f^{3}

Solution:

Given that f = ((4, 5), (5, 6), (6, -4)}

g {(4, —4), (6, 5), (8, 5)}

Domain of f = {4, 5, 6} = A

Domain of g = (4, 6, 8} = B

Domain of f ± g = A ∩ B = {4, 6}

i) f + g = {4, 5 + (-4), (6, -4 + 5)}

= {(4, 1), (6, 1)}

ii) f – g = {(4, 5 – (-4)), (6, -4, -5)}

= {(4, 9), (6, -9)}

iii) Domain of 2f = A = {4, 5, 6}

Domain of 4g = B = {4, 6, 8}

Domain of 2f + 4g = A ∩ B = {4, 6}

∴ 2f = {(4, 10), (5, 12), (6, -8)}

4g = {(4, —16), (6, 20), (8, 20)}

∴ 2f + 4g = {(4, 10 + (-16), 6, -8 + 20)}

= {(4, -6), (6, 12)}

iv) Domain of f + 4 = A= {4, 5, 6}

f + 4 ={4, 5 + 4), (5, 6 +4), (6, -4 + 4)}

= ((4, 9), (5, 10), (6, 0)}

v) Domain of fg = A ∩ B = {4, 6}

fg = {(4, (5) (-4), (6, (-4) (5))}

= {(4, -20), (6, -20)}

vi) Domain of \(\frac{f}{g}\) = {4, 6}

∴ \(\frac{f}{g}\) = {(4, \(\frac{-5}{4}\)), (6, \(\frac{-4}{5}\))}

vii) Domain of |f| = {4, 5, 6}

∴ |f| = {(4, |5|), (5, |6|), (6, |-4|)}

= {(4, 5), (5, 6), (6, 4))

viii) Domain of \(\sqrt{f}\) = {4, 5}

∴ \(\sqrt{f}\) = {(4, \(\sqrt{5}\)), (5, \(\sqrt{6}\))}

ix) Domain of f^{2} = {4, 5, 6} = A

∴ f^{2} = ((4, (5)^{2}) (5, (6)^{2}, (6, (-4)^{2})}

f^{2} = {(4, 25), (5, 36), (6, 16)}

x) Domain of f^{3} = A = {4, 5, 6}

∴ f^{3} = {(4, 5^{3}), (5, 6^{3}), (6, (-4)^{3}}

= {(4, 125) (5, 216) (6, -64)}

Question 16.

Find the domains and ranges of the following real valued functions.

i) f(x) = \(\frac{2+x}{2-x}\)

ii) f(x) = \(\frac{x}{1+x^{2}}\)

iii) f(x) = \(\sqrt{9-x^{2}}\) (A.P.)(Mar. ’15)

Solution:

i) f(x) = \(\frac{2+x}{2-x}\) ∈ R

⇔ 2 – x ≠ 0 x ⇔ x ∈ R -{2}

∴ Domain of f is R – {2}

Let f(x) = \(\frac{y}{1}\) = \(\frac{2+x}{2-x}\).

Apply componendo and dividendo rule

⇒ \(\frac{y+1}{y-1}\) = \(\frac{(2+x)+(2-x)}{(2+x)-(2-x)}\)

⇒ \(\frac{y+1}{y-1}\) = \(\frac{4}{2 x}\)

⇒ x = \(\frac{2(y-1)}{y+1}\)

Clearly x is not defined for y + 1 = 0

(i.e.,) y = -1

∴ Range of f is R – {-1}.

ii) f(x) = \(\frac{x}{1+x^{2}}\)

Solution:

f(x) = \(\frac{x}{1+x^{2}}\) ∈ R

∵ ∀ x ∈ R, x^{2} + 1 ≠ 0

Domain of f is R

Let f(x) = y = \(\frac{x}{1+x^{2}}\)

⇒ x^{2}y – x + y = 0

⇒ x = \(\frac{-(-1) \pm \sqrt{1-4 y^{2}}}{y}\) is a real number.

iff 1 – 4y^{2} ≥ 0; y ≠ 0

⇔ (1 – 2y)(1 + 2y) ≥ 0 and y ≠ 0

⇔ y ∈ \(\left[-\frac{1}{2} ; \frac{1}{2}\right]\) – {0}

Also x = 0 ⇒ y = 0

∴ Range of f = \(\left[-\frac{1}{2}, \frac{1}{2}\right]\)

iii) f(x) = \(\sqrt{9-x^{2}}\)

Solution:

f(x) = \(\sqrt{9-x^{2}}\) ∈ R

⇔ 9 – y^{2} ≥ 0

⇔ x ∈ [-3, 3]

∴ Domain of f is [-3, 3]

Let f(x) = y = \(\sqrt{9-x^{2}}\)

⇒ x = \(\sqrt{9-y^{2}}\) ∈ R

⇔ 9 – y^{2} ≥ 0 ⇔ (3 + y)(3 – y) ≥ 0

∴ -3 ≤ y ≤ 3

But f(x) attains only non negative values

∴ Range of f = [0, 3].

Question 17.

If f(x) = x^{2} and g(x) = |x|, find the following functions.

i) f + g

ii) f – g

iii) fg

iv) 2f

v) f^{2}

vi) f + 3

Solution:

Given f(x) = x^{2}

Domain of f = Domain of g = R

Hence domain of all the functions (i) to (vi) is R

iv) 2f(x) = 2 f(x) = 2x^{2}

v) f^{2}(x) = (f(x))^{2} = (x^{2})^{2} = x^{4}

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

Question 18.

Determine whether the following functions are even or odd.

i) f(x) = a^{x} – a^{-x} + sin x

Solution:

Given f(x) = a^{x} – a^{-x} + sin x

∴ f(- x) = a^{-x} – a^{x} + sin (-x)

= a^{-x} – a^{x} – sin x

= – (a^{x} – a^{x} – sin x) = – f(x)

∴ f(x) is an odd function.

ii) f(x) = x\(\left(\frac{e^{x}-1}{e^{x}+1}\right)\)

Solution:

f(x) = x\(\left(\frac{e^{x}-1}{e^{x}+1}\right)\)

∴ f is an even function.

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

Solution:

Given f(x) = log (x + \(\sqrt{x^{2}+1}\))

Then f(-x) = log (-x + \(\sqrt{(-x)^{2}+1}\))

= log (\(\sqrt{x^{2}+1}\) – x)

∴ f is an odd function.

Question 19.

Find the domains of the following real valued functions.

i) f(x) = \(\frac{1}{\sqrt{[x]^{2}-[x]-2}}\)

Solution:

f(x) = \(\frac{1}{\sqrt{[x]^{2}-[x]-2}}\) ∈ R

⇔ [x]^{2} – [x] – 2 > 0

⇔ ([x] + 1) ([x] – 2) > 0

⇔ [x] < -1, (or) [x] > 2

But [x] < -1 ⇒ [x] = -2, -3, -4, ……..

⇒ x < -1 [x] > 2 ⇒ [x] = 3, 4, 5, ……

⇒ x ≥ 3

∴ Domain of f = (-∞, -1) ∪ [3, ∞]

= R – [-1, 3)

ii) f(x) = log (x – [x])

f(x) = log (x – [x]) ∈ R

⇔ x – [x] > 0

⇔ x > [x]

⇔ x is a non integer

∴ Domain of f is R – Z

iii) f(x) = \(\sqrt{\log _{10}\left(\frac{3-x}{x}\right)}\)

Solution:

f(x) = \(\sqrt{\log _{10}\left(\frac{3-x}{x}\right)}\) ∈ R

⇔ log_{10}\(\left(\frac{3-x}{x}\right)\) ≥ 0 and \(\frac{3-x}{x}\) > 0

⇔ \(\frac{3-x}{x}\) ≥ 10° = 1 and 3 – x > 0, x > 0

⇔ 3 – x ≥ x and 0 < x < 3

⇔ x ≤ \(\frac{3}{2}\) and 0 < x < 3

⇔ x ∈ (-∞, \(\frac{3}{2}\)] ∩ (0, 3) = (0, \(\frac{3}{2}\)]

∴ Domain of f is (0, \(\frac{3}{2}\)]

iv) f(x) = \(\frac{\sqrt{3+x}+\sqrt{3-x}}{x}\)

Solution:

f(x) = \(\frac{\sqrt{3+x}+\sqrt{3-x}}{x}\) ∈ R

⇔ 3 + x ≥ 0, 3 – x ≥ 0 and x ≠ 0

⇔ x ≥ -3, x ≤ 3 and x ≠ 0

⇔ -3 ≤ x ≤ 3, and x ≠ 0

⇔ x ∈ [-3, 3] and x ≠ 0

⇔ x ∈ [-3, 3] – {0}

∴ Domain of f is [-3, 3] – {0}

Question 20.

If f : A → B and g : B → C are two injective functions then the mapping gof : A → C is an injection.

Solution:

f : A → B and g: B → C are one—one.

∴ gof : A → C

To prove that g o f is one — one function

Let a_{1}, a_{2} ∈ A ∴ f(a_{1}), f(a_{2}) ∈ B and g(f(a_{1})), g(f(a_{2})) ∈ C i.e., (gof) (a_{1}), gof(a_{2}) ∈ C

Now (gof) (a_{1}) = gof (a_{2})

⇒ g(f(a_{1})) = g(f(a_{2}))

⇒ f(a_{1}) = f(a_{2}) (∵ g is one -one)

⇒ a_{1} = a_{2} (∵ f is one-one)

Hence gof: A → C is a one-one function.

Note : The converse of the above theorem is not true.

If f : A → B, g: B → C and go f is one-one then both f and g need not be one-one.

For, consider

A = {1, 2}, B = {p, q, r), C = {s, t}

Let f = {(1, p), (2, q)} and

g = {(p, s), (q, t), (r, t)}

Now gof = {(1, s), (2, t)} ⇒ gof is one—one

from A to C

Observe that g: B → C is not one—one.

Question 21.

If f : A → B and g : B → C are two onto (surjective) functions then the mapping gof : A → C is a surjection. (May. ’08)

Solution:

f : A → B and g : B → C are onto.

∴ gof : A → C

To prove that gof is onto. Let c be any element of C.

Since g : B → C is an onto function there exists an element b ∈ B such that g(b) = c

Since f : A → B is an onto function, there exists an element a ∈ A such that f(a) = b

Now g(b) = c ⇒ g [f(a)] c = (gof) (a) = c

Thus for any element c ∈ C there is an element a ∈ A such that (gof) (a) = c

∴ gof : A → C is a surjection

Question 22.

If f : A → B and g : B → C are two bijective functions then the mapping gof : A → C is a bijection. (Mar. ’16, May ’12)

Solution:

f and g are injections gof: A → C is an injection.

f and g are surjections= gof : A → C is a surjection.

Hence it follows that if f and g are bijections,

gof is also a bijection.

Note : The converse of the above theorem is not true.

Question 23.

If f : A → B and g : B → C are such that gof is a surjection, then g is necessarily a surjection.

Solution:

Let c ∈ C. Since gof is a surjection, from A to

C there exists an element a ∈ A such that (gof) (a) = c, i.e., g(f(a)) = c

Since g: B → C and f(a) ∈ B, ∀ c ∈ C there exists an element belonging to B.

Hence g is a surjection.

Question 24.

If f : A → B and g : B → C and h : C → D are functions then ho(gof) = (hog)of (Mar. ’12, ’08)

Solution:

f : A → B and g : B → C gof : A → C

Now gof : A → C and h : C → D

⇒ ho(gof) : A → D

Similarly (hog) of : A → D

Thus ho (gof) and (hog) of both exist and have the same domain A and co-domain D.

Let a be an element of A.

Now [ho(gof)](a) = h[(gof)(a)] = h[g(f(a))]

= (hog) [f(a)] = [(hog)of] (a)

∴ ho(gof) = (hog)of

Note : Thus composition of mappings is associative.

Question 25.

Let f : A → B, I_{A} and I_{B} are identity functions on A and B respectively. Then for I_{A} = f = I_{B} of Mar.’12, ’08

Solution:

∵ I_{A} : A → A and f : A → B are functions foI_{A} is a function from A to B. Hence foI_{A} and f are definded on same domain A.

i) Let a ∈ A then (foI_{A})(a) = f(I_{A}(a))

= f(a)

[∵ I_{A}(a) = a, ∀ a ∈ A]

∴ foI_{A} = f ————- (1)

ii) ∵ f : A → B, I_{B} : B → B are functions, I_{B} of is a function from A to B

∴ The functions I_{B} of and f are defined on the same domain A.

Let a ∈ A, then (I_{B} of)(a)

= I_{B}(f (a)) = f(a)

∵ f : A → B, we have f(a) ∈ B

∴ I_{B} of = f ———– (2)

From(1) & (2) foI_{A} = I_{Bo}f = f

Question 26.

Let A and B be two non-empty sets. If f : A → B is a bijection, then f^{-1} : B → A is also a bijection.

Solution:

f : A → B is a bijection.

∴ f^{-1} : B → A is a unique function.

(i) To prove that f^{-1} is one—one:

Let b_{1} and b_{2} be any two different elements of B. i.e., b_{1} ≠ b_{2}.

Then, we have to prove that

f^{-1}(b_{1}) ≠ F^{-1}(b_{2})

Let f^{-1}(b_{1}) = a_{1} and f^{-1}(b_{2}) = a_{2}

such that a_{1}, a_{2} ∈ A

Then b_{1} = f(a_{1}) and b_{2} = f(a_{2})

Now b_{1} ≠ b_{2} ⇒ f(a_{1}) ≠ f(a_{2})

⇒ a_{1} ≠ a_{2} (∵ f is a bijection)

⇒ f^{-1}(b_{1}) ≠ f^{-1}(b_{2})

∴ f^{-1} is one—one.

(ii) To prove that f^{-1} is onto:

Let ‘a’ be an element of A.

Then there exists an element b ∈ B such that f(a) = b (or) f^{-1}(b) = a

(or) a = f^{-1}(b)

Thus ‘a’ is the f^{-1} – image of the element b ∈ B

Hence f^{-1} is onto.

∴ f : B → A is a bijection.

Question 27.

If f : A → B is a bijection, then f^{-1}of = I_{A} and fof^{-1} = I_{B}

(A.P) (Mar. 15,12, ’07; May ’07, ‘06)

Solution:

f : A → B is a bijection ⇒ f^{-1} : B → A is also a bijection.

By definition, fof^{-1} : B → B and f^{-1}of : A → A are bijection

Also I_{A} : A → A and I_{B} : B → B

To prove that f^{-1}of = I_{A}

Let a ∈ A

Since f : A → B there exists a unique element

b ∈ B such that f(a) = b

a = f^{-1} (b) (∵ f is a bijection)

∴ (f^{-1}of) (a) = f^{-1}[f(a)]

= f^{-1}(b) = a = I_{A} (a)

f^{-1} of = I_{A}

Similarly it can be shown that fof^{-1} = I_{B}

Question 28.

If f : A → B and g : B → A are two functions such that gof = I_{A} and fog = I_{B} then g = f^{-1}.

Solution:

(i) To prove that f is one—one:

Let a_{1}, a_{2} ∈ A.

Since f: A → B, f(a_{1}), f(a_{2}) ∈ B

Now f(a_{1}) = f(a_{2})

⇒ g[f(a_{1})] = g[f(a_{2})]

⇒ (gof)(a_{1}) = gof(a_{2})

⇒ I_{A} (a_{1}) = I_{A} (a_{2})

∴ a_{1} = a_{2} = ∴ f is one — one

(ii) To prove that f is onto :

Let b be an element of B

∴ b = I_{B} fog(b)

⇒ b = f{(g(b)} ⇒ f{g(b)} = b

i.e., there exists a pre-image g(b) ∈ A for b, under the mapping f. ∴ f is onto

Thus f is one-one, onto and hence

f^{-1} : B → A exists and is also one — one and onto.

(iii) To prove g = f^{-1}:

Now g : B → A and f^{-1} : B → A

Let a ∈ A and b be the f – image of a, where b ∈ B

∴ f(a) = b ⇒ a =f^{-1}(b)

Now g(b) = g[f(a)] (gof) (a)

= I_{A}(a) = a = f^{-1}(b)

∴ g = f^{-1}

Question 29.

If f : A → B and g : B → C are bijective functions, then (gof)^{-1} = f^{-1}og^{-1}(AP) (Mar. ‘16’14, ‘11; May ‘11)

Solution:

f : A → B, g : B → C are bijections

⇒ gof : A → C is a bijection

Also g^{-1} : C → B and f^{-1} : B → A are bijections

⇒ f^{-1}og^{-1} : C → A is a bijection.

Let c be any element of C.

Then ∃ an element b ∈ B such that g(b) = c

⇒ b = g^{-1}(c)

Also ∃ an element a A such that f(a) = b

⇒ a = f^{-1}(b)

Now (gof) (a) = g(f(a) = g(b) = c

⇒ a = f^{-1}(b)

Now (gof) (a) = g(f(a) = g(b) = c

⇒ a = (gof)^{-1} (c) ⇒ (gof) (c) = a

Also (f^{-1}og^{-1}) (c) ⇒ f^{-1} (g^{-1} (c)) = f^{-1} (b) = a

∴ From (1) and (2);

(gof)^{-1} (c) = (f^{-1}og^{-1}(c))

⇒ (gof)^{-1} = f^{-1}og^{-1}