Practicing the Intermediate 1st Year Maths 1A Textbook Solutions Inter 1st Year Maths 1A Functions Solutions Exercise 1(a) will help students to clear their doubts quickly.

## Intermediate 1st Year Maths 1A Functions Solutions Exercise 1(a)

I.

Question 1.
If the function f is defined by

then find the values of
(i) f(3)
(ii) f(0)
(iii) f(-1.5)
(iv) f(2) + f(-2)
(v) f(-5)
Solution:
(i) f(3)
For x > 1, f(x) = x + 2
∴ f(3) = 3 + 2 = 5

(ii) f(0)
For -1 ≤ x ≤ 1, f(x) = 2
∴ f(0) = 2

(iii) f(-1.5)
For -3 < x < -1, f(x) = x – 1
∴ f(-1.5) = -1.5 – 1 = -2.5

(iv) f(2) + f(-2) For x > 1, f(x) = x + 2
∴ f(2) = 2 + 2 = 4
For -3 < x < -1, f(x) = x – 1
∴ f(-2)= -2 – 1 = -3
f(2) + f(-2) = 4 + (-3) = 1

(v) f(-5) is not defined, since domain of x is {X/X ∈ (-3, ∞)}

Question 2.
If f: R{0}R is defined by f(x) = $$x^{3}-\frac{1}{x^{3}}$$; then show that f(x) + $$f\left(\frac{1}{x}\right)$$ = 0.
Solution:
Given f(x) = $$x^{3}-\frac{1}{x^{3}}$$ ……(i)
Now $$f\left(\frac{1}{x}\right)=\left(\frac{1}{x}\right)^{3}-\frac{1}{\left(\frac{1}{x}\right)^{3}}=\frac{1}{x^{3}}-x^{3}$$ ……(2)
$$f(x)+f\left(\frac{1}{x}\right)=\left(x^{3}-\frac{1}{x^{3}}\right)+\left(\frac{1}{x^{3}}-x^{3}\right)$$ = 0
∴ f(x) + $$f\left(\frac{1}{x}\right)$$ = 0

Question 3.
If f : R → R is defined by f(x) = $$\frac{1-x^{2}}{1+x^{2}}$$, then show that f(tan θ) = cos 2θ.
Solution:
Given f(x) = $$\frac{1-x^{2}}{1+x^{2}}$$

∴ f(tan θ) = cos 2θ

Question 4.
If f : R\{±1} → R is defined by f(x) = $$\log \left|\frac{1+x}{1-x}\right|$$, then show that $$f\left(\frac{2 x}{1+x^{2}}\right)$$ = 2f(x)
Solution:
f : R\{±1} → R and f(x) = $$\log \left|\frac{1+x}{1-x}\right|$$

∴ $$f\left(\frac{2 x}{1+x^{2}}\right)$$ = 2f(x)

Question 5.
If A = {-2, -1, 0, 1, 2} and f : A → B is a surjection defined by f(x) = x2 + x + 1, then find B.
Solution:
A = {-2, -1, 0, 1, 2} and f : A → B, f(x) = x2 + x + 1
f : A → B is a surjection
f(-2) = (-2)2 + (-2) + 1
= 4 – 2 + 1
= 3
f(-1) = (-1)2 + (-1) + 1
= 1 – 1 + 1
= 1
f(0) = 02 + 0 + 1
= 0 + 0 + 1
= 1
f(1) = 12 + 1 + 1
= 1 + 1 + 1
= 3
f(2) = 22 + 2 + 1
= 4 + 2 + 1
= 7
∴ B = f(A) = {3, 1, 7}

Question 6.
If A = {1, 2, 3, 4} and f : A → R is a function defined by f(x) = $$\frac{x^{2}-x+1}{x+1}$$, then find the range of f.
Solution:
A= {1, 2, 3, 4}

∴ Range of f = f(A) = $$\left\{\frac{1}{2}, 1, \frac{7}{4}, \frac{13}{5}\right\}$$

Question 7.
If f(x + y) = f(xy) ∀ x, y ∈ R then prove that f is a constant function.
Solution:
Given f(x + y) = f(x y), x, y ∈ R
take x = y = 0
⇒ f(0) = f(0) ………(1)
Let x = 1, y = 0
⇒ f(1) = f(0) ……..(2)
Let x = 1, y = 1
⇒ f(2) = f(1) ………(3)
from (1), (2), (3)
f(0) = f(1) = f(2)
⇒ f(0) = f(2)
Similarly f(3) = f(0)
f(4) = f(0)
and so on
f(n) = f(0)
∴ f is a constant function

II.

Question 1.
If A = {x | -1 ≤ x ≤ 1}, f(x) = x2, g(x) = x3, which of the following are surjections?
(i) f : A → A
(ii) g : A → A
Solution:
(i) ∵ A = {x | -1 ≤ x ≤ 1} and f(x) = x2
This implies f(x) is a function from A to A
(i.e.,) f : A → A
Now let y ∈ A
If f(x) = y then x2 = y
x = √y
So, if y = -1 then x = √-1 ∉ A
∴ f : A → A is not a surjection.

(ii) ∵ A = {x | -1 ≤ x ≤ 1} and g(x) = x3
⇒ g : A → A
Let y ∈ A. Then g(x) = y
⇒ x3 = y
⇒ x = $$(y)^{1 / 3}$$ ∈ A
So if y = -1 then x = -1 ∈ A
y = 0, then x = 0 ∈ A
y = 1, then x = 1 ∈ A
∴ g : A → A is a surjections.

Question 2.
Which of the following are injections or surjections or bijections? Justify your answers.
(i) f : R → R defined by f(x) = $$\frac{2 x+1}{3}$$
Solution:
f(x) = $$\frac{2 x+1}{3}$$
Let x1, x2 ∈ R
∵ f(x1) = f(x2)
⇒ $$\frac{2 x_{1}+1}{3}=\frac{2 x_{2}+1}{3}$$
⇒ 2x1 + 1 = 2x2 + 1
⇒ 2x1 = 2x2
⇒ x1 = x2
∵ f(x1) = f(x2) ⇒ x1 = x2, ∀ x1, x2 ∈ R
So f(x) = $$\frac{2 x+1}{3}$$, f : R → R is an injection
If y ∈ R (co-domain) then y = $$\frac{2 x+1}{3}$$
⇒ x = $$\frac{3 y-1}{2}$$
Then f(x) = $$\frac{2 x+1}{3}=\frac{2\left(\frac{3 y-1}{2}\right)+1}{3}=y$$
∴ f is a surjection
∴ f : R → R defined by f(x) = $$\frac{2 x+1}{3}$$ is a bijection

(ii) f : R → (0, ∞) defined by f(x) = 2x
Solution:
Let x1, x2 ∈ R
∵ f(x1) = f(x2)
⇒ $$2^{x_{1}}=2^{x_{2}}$$
⇒ x1 = x2
∴ f(x1) = f(x2) ⇒ x1 = x2 ∀ x1, x2 ∈ R
∴ f(x) = 2x, f : R → (0, ∞) is injection
If y ∈ (0, ∞) and y = 2x ⇒ x = log2 (y)
Then f(x) = 2x
= $$2^{\log _{2}(y)}$$
= y
∴ f is a surjection
Hence f is a bijection.

(iii) f : (0, ∞) → R defined by f(x) = logex
Solution:
Let x1, x2 e (0, ∞)
f(x1) = f(x2)
⇒ $$\log _{e}\left(x_{1}\right)=\log _{e}\left(x_{2}\right)$$
⇒ x1 = x2
∵ f(x1) = f(x2)
⇒ x1 = x2 ∀ x1, x2 ∈ (0, ∞)
∴ f(x) is injection.
Let y ∈ R.
y = logex ⇒ x = ey
Then f(x) = logex
= loge(ey)
= y . logee
= y(1)
= y
∴ f is a surjection.
∴ f is a bijection.

(iv) f : [0, ∞) → [0, ∞) defined by f(x) = x2.
Solution:
Let x1, x2 ∈ [0, ∞) (i.e.,) domain of f.
Now f(x1) = f(x2)
⇒ $$x_{1}^{2}=x_{2}^{2}$$
⇒ x1 = x2
∵ x1, x2 ≥ 0
∴ f(x) = x2, f : {0, ∞) → {0, ∞) is injection
Let y ∈ (0, ∞), co-domain of f
Let y = x2 ⇒ x = √y, ∵ y ≥ 0
Then f(x) = x2
= $$(\sqrt{y})^{2}$$
= y
∴ f is surjection.
Hence f is a bijection.

(v) f : R → [0, ∞) defined by f(x) = x2.
Solution:
Let x1, x2 ∈ R.
f(x1) = f(x2)
⇒ $$x_{1}^{2}=x_{2}^{2}$$
⇒ x1 = ±x2, ∵ x1, x2 ∈ R
Hence f is not injection
Let y ∈ [0, ∞)
y = x2
⇒ x = ±√y, where y ∈ [0, ∞)
Then f(x) = x2
= $$(\sqrt{y})^{2}$$
= y
∴ f is surjection
Hence f is not a bijection

(vi) f : R → R defined by f(x) = x2.
Solution:
Let x1, x2 ∈ R, (domain of f)
∴ f(x1) = f(x2)
⇒ $$x_{1}^{2}=x_{2}^{2}$$
⇒ x1 = ±x2, ∵ x1, x2 ∈ R
∴ f(x) is not injection
For elements that belong to (-∞, 0) codomain of f has no pre-image in f.
∴ f is not a surjection
Hence f is neither injection nor surjection.

Question 3.
Is g = {(1, 1) (2, 3) (3, 5) (4, 7)} is a function from A = {1, 2, 3, 4} to B = {1, 3, 5, 7}. If this is given by the formula g(x) = ax + b, then find a and b.
Solution:
A = {1, 2, 3, 4}; B = {1, 3, 5, 7}
g : {(1, 1), (2, 3), (3, 5), (4, 7)}
∵ g(1) = 1, g(2) = 3, g(3) = 5, g(4) = 7
So for each element a ∈ A, there exists a unique b ∈ B such (a, b) ∈ g
∴ g : A → B is a function
Given g(x) = ax + b, ∀ x ∈ A
g(1) = (a) + b = 1
⇒ a + b = 1 ……..(1)
g(2) = 2a + b = 3
⇒ 2a + b = 3 …….(2)
Solve (1) and (2)
a = 2, b = -1

Question 4.
If the function f : R → R defined by f(x) = $$\frac{3^{x}+3^{-x}}{2}$$, then show that f(x + y) + f(x – y) = 2f(x) f(y).
Solution:
f : R → R and f(x) = $$\frac{3^{x}+3^{-x}}{2}$$

∴ f(x + y) + f(x – y) = 2 f(x).f(y)

Question 5.
If the function f : R → R defined by f(x) = $$\frac{4^{x}}{4^{x}+2}$$, then show that f(1 – x) = 1 – f(x) and hence reduce the value of $$f\left(\frac{1}{4}\right)+2 f\left(\frac{1}{2}\right)+f\left(\frac{3}{4}\right)$$
Solution:

∴ f(1 – x) = 1 – f(x)

Question 6.
If the function f : {-1, 1} → {0, 2), defined by f(x) = ax + b is a surjection, then find a and b.
Solution:
f : {-1, 1} → {0, 2} and f(x) = ax + b is a surjection
Given f(-1) = 0 and f(1) = 2 (or) f(-1) = 2, f(1) = 0
Case (i):
f(-1) = 0 and f(1) = 2
a(-1) + b = 0 ⇒ -a + b = 0 ……..(1)
a(1) + b = 2 ⇒ a + b = 2 ……(2)
Solve eq’s (1) and (2), we get a = 1, b = 1
Case (ii):
f(-1) = 2 and f(1) = 0
a(-1) + b = 2 ⇒ -a + b = 2 ……(3)
a(1) + b = 0 ⇒ a + b = 0 ……….(4)
Solve eq’s (3) and (4), we get a = -1, b = 1
Hence a = ±1 and b = 1

Question 7.
If f(x) = cos (log x), then show that $$f\left(\frac{1}{x}\right) \cdot f\left(\frac{1}{y}\right)-\frac{1}{2}\left[f\left(\frac{x}{y}\right)+f(x y)\right]=0$$
Solution:
Given f(x) = cos(log x)
$$f\left(\frac{1}{x}\right)=\cos \left(\log \left(\frac{1}{x}\right)\right)$$
= cos(log 1 – log x)
= cos(-log x)
= cos (log x) (∵ log 1 = 0)
Similarly
$$f\left(\frac{1}{y}\right)$$ = cos (log y)
$$f\left(\frac{x}{y}\right)=\cos \log \left(\frac{x}{y}\right)$$
= cos (log x – log y)
and f(x y) = cos log (x y) = cos (log x + log y)
$$f\left(\frac{x}{y}\right)$$ + f(x y) = cos (log x – log y) + cos (log x + log y)
= 2 cos (log x) cos (log y)
[∵ cos (A – B) + cos (A + B) = 2 cos A . cos B]
LHS = $$f\left(\frac{1}{x}\right) \cdot f\left(\frac{1}{y}\right)-\frac{1}{2}\left[f\left(\frac{x}{y}\right)+f(x y)\right]$$
= cos (log x) cos (log y) – $$\frac{1}{2}$$ [2 cos (log x) cos (log y)]
= 0