# Function

Algebra

## Domain of the function $f(x) = \frac{x}{\sqrt{sin(lnx)-cos(lnx)}}$ is

Problem: Domain of the function $\large f(x) = \frac{x}{\sqrt{sin(lnx)-cos(lnx)}}$ is

(A) $( e^{2n\pi} , e^{(3n+1/2)\pi})$

(B) $( e^{(2n+1/4)\pi} , e^{(2n+5/4)\pi})$

(C) $( e^{(2n+1/4)\pi} , e^{(3n-3/4)\pi} )$

(D) none of these

Sol: For domain sin (ln x) > cos (ln x) and x > 0

$2n\pi + \frac{\pi}{4} < lnx < 2n\pi + \frac{5\pi}{4}$ n ∈ I

Hence (B) is the correct answer

## If a function satisfies the condition $f(x+\frac{1}{x}) = x^2 + \frac{1}{x^2} , x \ne 0$ then domain of f (x) is

Problem: If a function satisfies the condition $f(x+\frac{1}{x}) = x^2 + \frac{1}{x^2} , x \ne 0$ then domain of f (x) is

(A) [-2 , 2]

(B) (-∞ , 2] ∪ [2 , ∞ )

(C) (0 , ∞)

(D) none of these

Sol: $f(x+ \frac{1}{x}) = (x+\frac{1}{x})^2 – 2$

⇒ f(y) = y2 –2, where y = x + 1/x

for x > 0, y = x + 1/x  ≥ 2 and for x < 0, y = x + 1/x  ≤ –2

Hene (B) is the correct answer.

## If $f(x) + 2f(1-x) = x^2 + 2 \; \forall x \in R$  , then f(x) is given as

Problem: If $f(x) + 2f(1-x) = x^2 + 2 \; \forall x \in R$  , then f(x) is given as

(A) $\frac{(x-2)^2}{3}$

(B) x2 – 2

(C) 1

(D) None of these

Ans: (A)

Sol. By replacing x with (1 – x) in the given expression

f(1 – x) + 2 f(1 – 1 + x)

= (1 – x)2 + 2

⇒ f(1 – x) + 2 f(x)

= (1 – x)2 + 2

Now f(x) + 2 f(1 – x) – 2(f(1 – x) + 2f(x))

= x2 + 2 – 2((1 – x)2 + 2)

⇒  -3 f(x) = x2 + 2 – 2(3 – 2x + x2)

⇒  3 f(x) = x2 – 4x + 4

⇒  $f(x) = \frac{(x-2)^2}{3}$

## Domain of the function $f(x) = \frac{1}{\sqrt{10_{C_{x-1}}-3 . 10_{C_x}}}$ contains the points

Q: Domain of the function $\Large f(x) = \frac{1}{\sqrt{10_{C_{x-1}}-3 . 10_{C_x}}}$ contains the points

(A) 9, 10, 11

(B) 9, 10, 12

(C) all natural numbers

(D) none of these

Sol. Given function is defined if $\large 10_{C_{x-1}} > 3. 10_{C_x}$

$\large \frac{1}{11-x} > \frac{3}{x}$

⇒ 4x > 33

⇒ x  ≥  9 but x ≤ 10

⇒ x = 9, 10

Hence (D) is the correct answer.

## $f(x) = \frac{cosx}{[\frac{2x}{\pi}] + \frac{1}{2}}$ , where x is not an integral multiple of π

Q: $\large f(x) = \frac{cosx}{[\frac{2x}{\pi}] + \frac{1}{2}}$ , where x is not an integral multiple of π and [.] denotes the greatest integer function is

(A) an odd function

(B) even function

(C) neither odd nor even

(D) none of these

Sol: Clearly $\large [- \frac{2x}{\pi}] + \frac{1}{2} = -([ \frac{2x}{\pi}] + \frac{1}{2})$

⇒ f(x) is an odd function.

Hence (A) is the correct answer.