Students must practice these TS Intermediate Maths 1A Solutions Chapter 1 Functions Ex 1(a) to find a better approach to solving the problems.

## TS Inter 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)

Answer:

(i) f(3), For x > 1; f(x) = x + 2

f(3) = 3 + 2 = 5

(ii) f(0), For – 1 ≤ x ≤ 1; 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 such domain of ‘f’ is {x / x > – 3].

Question 2.

If f : R {0} → R defined by f(x) = x^{3} – \(\frac{1}{x^3}\), then show that f(x) + f\(\left(\frac{1}{x}\right)\) = 0.

Answer:

Given f(x) = x^{3} – \(\frac{1}{x^3}\)

Question 3.

If f: R → R defined by f(x) = \(\frac{1-x^2}{1+x^2}\), then show that f(tan θ) = cos 2θ

Answer:

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

= 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).

Answer:

Given f: R – (±1) → R defined by f(x) = log\(\left|\frac{1+x}{1-x}\right|\)

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. (May 2014)

Answer:

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

∴ f(-2) = (-2)^{2} + (-2) + 1=3,

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

f(0) = 0^{2} + 0 + 1 = 1

f(1) =1^{2} + 1 + 1 = 3

f(2) = 2^{2} + 2 + 1 = 7

∴ B = f(A) = (1, 3, 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.

Answer:

Given A = {1, 2, 3, 4} and f(x) = \(\frac{x^2-x+1}{x+1}\)

Question 7.

If f (x + y) = f (xy) ∀ x, y ∈ R, then prove that f is a constant function.

Answer:

Given f (x + y) = f(x y) ∀ x, y ∈ R Suppose x = y = 0 then

f(0 + 0) = f(0 x 0)

⇒ f(0) = f(0) ………………..(1)

Suppose x = 1, y = 0 then then f (1 + 0) = f(1 x 0)

⇒ f(D = f (0) ……………(2)

Suppose x = 1, y = 1 then f (1 + 1) = f(1 x 1)

⇒ f(2) = f(1) …………….. (3)

f(0) = f(1) = f(2)

= f(0) = f(2)

Similarly f(3) = f(0), f(4) = f(0) …………. f(n) = f(0)

∴ f is a constant function.

II.

Question 1.

If A = {x / – 1 ≤ x ≤ 11, f(x) = x^{2}, g(x) = x^{3} Which of the following are surjections

(i) f : A → A

(ii) g : A → A.

Answer:

i) Given A {x / – 1 ≤ x ≤ 1}, f(x) = x^{2}

and f : A → A

Suppose y ∈ A

then x^{2} = y ⇒ x = ± √y

If x = √y and if y = – 1 then x = √-1 ∈ A

f : A → A is not a surjection.

ii) Given A = {x/-1 ≤ x ≤ 1), g(x) = x^{3}

and g : A → A

Suppose ye A then x^{2} = y ⇒ x = \(\sqrt[3]{y}\) ∈ A

If y = -1 then x = -1 ∈ A

y = 0 then x = 0 ∈ A

y = 1 then x = 1 ∈ A

g : A → A is a surjection.

Question 2.

Which of the following are injections or surjections or Bisections ? Justify your answers.

i) f : R → R defined by f(x) = \(\frac{2 x+1}{3}\)

Answer:

Given f(x) = \(\frac{2 x+1}{3}\)

Let a_{1}, a_{2} ∈ R

∴ f(a_{1}) = f(a_{2})

⇒ \(\frac{2 \mathrm{a}_1+1}{3}=\frac{2 \mathrm{a}_2+1}{3}\)

⇒ 2a_{1} + 1 = 2a_{2} + 1

⇒ a_{1} = a_{2}

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

f(x) = \(\frac{2 x+1}{3}\) is an injection.

Suppose y ∈ R (codomain of f) then

y = \(\frac{2 x+1}{3}\) ⇒ x = \(\frac{3 y-1}{2}\)

Then f(x) = f\(\left(\frac{3 y-1}{2}\right)=\frac{\frac{2(3 y-1)}{2}+1}{3}\) = 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) = 2^{x}

Answer:

Let a_{1}, a_{2} ∈ R then f(a_{1}) = f(a_{2})

⇒ 2^{a1} = 2^{a2}

⇒ a_{1} = a_{2} ∀ a_{1}, a_{2} ∈ R

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

Let y ∈ (0, ∞) and y = 2^{x} ⇒ x = log_{2} y

then f(x) = 2^{x} = 2 log2^{y} = y

∴ f is a surjection.

Since f is injection and surjection, f is a bijection.

iii) f : (0, ∞) → R defined by f(x) = log_{e}x.

Answer:

Let x_{1}, x_{2} ∈ (0, ∞)and = log_{e}x. then f(x_{1}) = f(x_{2})

⇒ log_{e}x_{1} = log_{e}x_{2} ⇒ x_{1} = x_{2}

∴ f(x_{1}) = f(x_{2})

x_{1} = x_{2} and f is injection.

Let y ∈ R then y = log_{e}x ⇒ x = e^{y}

f(x) = log_{e}x = log_{e}e^{y} = y and f is a surjection.

Since f is both injective and surjective, f is a bijection.

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

Answer:

Let x_{1}, x_{1} ∈ [0, ∞) given f(x) = x^{2}

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

⇒ x_{1} = x_{2}

x_{1} = x_{2} (∵ x_{1}, x_{2} > 0)

f(x) = x^{2},

∴ f: [0, ∞) → [0, ∞) is an injection.

Let y ∈ [0, ∞)then y = x^{2} ⇒ x = √y (∵ y > 0)

f(x) = x^{2} = (√y)^{2} = y

and f is a surjection

∴ f is a bijection.

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

Answer:

Let x_{1} x_{2} ∈ R and f(x) = x^{2}

∴ f(x_{1}) = f(x_{2})

⇒ x_{1}^{2} = x_{2}^{2}

⇒ x_{1} = ±x_{2} (∵ x_{1}, x_{2} ∈ R)

f is not an injection

Let y ∈ [0, ∞] then y = x^{2} ⇒ x = ±√y

where y ∈ [0, ∞] then f(x) = x^{2} = (√y )^{2} = y.

∴ f is a surjection.

Since f is not injective and only surjective, we say that f is not a bijection.

vi) f : R → R defined by f(x) = x^{2}

Answer:

Let x_{1} x_{2} ∈ R then 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 an injection.

Let y ∈ R then y = x^{2}

⇒ x = ±√y

For elements that belong to (-∞, 0).

codomain R of f has no pre-image in f.

∴ f is not a surjection.

Hence f is not a bijection.

Question 3.

If g = 1(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.

Answer:

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

Hence for an element a ∈ A f ∃ b ∈ B such that g : A → B is a function.

Given g(x) = ax + b ∀ x ∈ A

g(1) = a + b = 1

g(2) = 2a + b = 3

solving 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) = 2 f(x) f(y).

Answer:

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\(\left(\frac{1}{4}\right)\) + 2f\(\left(\frac{1}{2}\right)\) + f\(\left(\frac{3}{4}\right)\).

Answer:

Question 6.

If the function f : {-1, 1} → {0, 2} defined by f(x) = ax + b is a suijection, then find a and b.

Answer:

Since f: {-1, 1} → {0, 2} and f(x) = ax + b is a surjection.

Given f (-1) = 0, f (1) = 2 (or) f (-1) = 2, f (1)=0

Case I : f (-1) = 0, f (1) = 2

∴ – a + b = 0, a + b = 2

Solving b =1 , a = 1

Case II : f (-1) = 2, and f (1) = 0

then – a + b = 2 and a + b = 0

Solving b = 1, a = -1

Hence a = + 1 and b = 1

Question 7.

If f(x) = cos (log x), then show that f\(\left(\frac{1}{x}\right)\) f\(\left(\frac{1}{y}\right)\) – \(\left(\frac{1}{2}\right)\)[f\(\left(\frac{x}{y}\right)\) + f(xy)] = 0

Answer:

Given f(x) = cos(log x)

then f\(\left(\frac{1}{x}\right)\) = cos(log\(\left(\frac{1}{x}\right)\))

= cos(-log x) = cos(log x) (∵ log 1 = 0)

Similarly f\(\left(\frac{1}{x}\right)\) = cos(log y)

f\(\left(\frac{x}{y}\right)\) = cos(log\(\left(\frac{x}{y}\right)\)) = cos(log x – log y)

f(xy) = cos (log xy) = 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

f\(\left(\frac{1}{x}\right)\) f\(\left(\frac{1}{y}\right)\) – \(\left(\frac{1}{2}\right)\)[f\(\left(\frac{x}{y}\right)\) + f(xy)] = cos (log x) cos (log y) – \(\frac{1}{2}\) [2cos (log x) cos (logy)]

= 0