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)

Add (1) and (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) = x^{2} + x + 1, then find B.

Solution:

A = {-2, -1, 0, 1, 2} and f : A → B, f(x) = x^{2} + 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) = 0^{2} + 0 + 1

= 0 + 0 + 1

= 1

f(1) = 1^{2} + 1 + 1

= 1 + 1 + 1

= 3

f(2) = 2^{2} + 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) = x^{2}, g(x) = x^{3}, which of the following are surjections?

(i) f : A → A

(ii) g : A → A

Solution:

(i) ∵ A = {x | -1 ≤ x ≤ 1} and f(x) = x^{2}

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

⇒ g : A → A

Let y ∈ A. Then g(x) = y

⇒ x^{3} = 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 x_{1}, x_{2} ∈ R

∵ f(x_{1}) = f(x_{2})

⇒ \(\frac{2 x_{1}+1}{3}=\frac{2 x_{2}+1}{3}\)

⇒ 2x_{1} + 1 = 2x_{2} + 1

⇒ 2x_{1} = 2x_{2}

⇒ x_{1} = x_{2}

∵ f(x_{1}) = f(x_{2}) ⇒ x_{1} = x_{2}, ∀ x_{1}, x_{2} ∈ 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 x_{1}, x_{2} ∈ R

∵ f(x_{1}) = f(x_{2})

⇒ \(2^{x_{1}}=2^{x_{2}}\)

⇒ x_{1} = x_{2}

∴ f(x_{1}) = f(x_{2}) ⇒ x_{1} = x_{2} ∀ x_{1}, x_{2} ∈ R

∴ f(x) = 2x, f : R → (0, ∞) is injection

If y ∈ (0, ∞) and y = 2x ⇒ x = log_{2} (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) = log_{e}x

Solution:

Let x_{1}, x_{2} e (0, ∞)

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

⇒ \(\log _{e}\left(x_{1}\right)=\log _{e}\left(x_{2}\right)\)

⇒ x_{1} = x_{2}

∵ f(x_{1}) = f(x_{2})

⇒ x_{1} = x_{2} ∀ x_{1}, x_{2} ∈ (0, ∞)

∴ f(x) is injection.

Let y ∈ R.

y = log_{e}x ⇒ x = e^{y}

Then f(x) = log_{e}x

= log_{e}(e^{y})

= y . log_{e}e

= y(1)

= y

∴ f is a surjection.

∴ f is a bijection.

(iv) f : [0, ∞) → [0, ∞) defined by f(x) = x^{2}.

Solution:

Let x_{1}, x_{2} ∈ [0, ∞) (i.e.,) domain of f.

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

⇒ \(x_{1}^{2}=x_{2}^{2}\)

⇒ x_{1} = x_{2}

∵ x_{1}, x_{2} ≥ 0

∴ f(x) = x^{2}, f : {0, ∞) → {0, ∞) is injection

Let y ∈ (0, ∞), co-domain of f

Let y = x^{2} ⇒ x = √y, ∵ y ≥ 0

Then f(x) = x^{2}

= \((\sqrt{y})^{2}\)

= y

∴ f is surjection.

Hence f is a bijection.

(v) f : R → [0, ∞) defined by f(x) = x^{2}.

Solution:

Let x_{1}, x_{2} ∈ R.

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

⇒ \(x_{1}^{2}=x_{2}^{2}\)

⇒ x_{1} = ±x_{2}, ∵ x_{1}, x_{2} ∈ R

Hence f is not injection

Let y ∈ [0, ∞)

y = x^{2}

⇒ x = ±√y, where y ∈ [0, ∞)

Then f(x) = x^{2}

= \((\sqrt{y})^{2}\)

= y

∴ f is surjection

Hence f is not a bijection

(vi) f : R → R defined by f(x) = x^{2}.

Solution:

Let x_{1}, x_{2} ∈ R, (domain of f)

∴ f(x_{1}) = f(x_{2})

⇒ \(x_{1}^{2}=x_{2}^{2}\)

⇒ x_{1} = ±x_{2}, ∵ x_{1}, x_{2} ∈ 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