If 4 f(x)-f ((1)/(x))= (2 x+ (1)/(x))(2 x- (1)/(x)), then what is f(2) equal to?

Correct answer: D. 4 4 f(x)-f((1)/(x))=(2 x+(1)/(x))(2 x-(1)/(x))4 f(x)-f( 1 x )=4 x^2- 1 x^2 ...(i) Replace x by (1)/(x) we get 4 f((1)/(x))-f(x)=(4)/(x^2)-x ...(ii) from 4 ×( i) + (ii) 16 f(x)-4 f((1)/(x))=16 x^2-(4)/(x^2) aligned & 16 f(x)-4 f((1)/(x))=16 x^2-(4)/(x^2) & 4 f((1)/(x))-f(x)=(4)/(x^2)-x aligned aligned & 15 f(x)=15 x^2 & f(x)=x^2 aligned Now, f(2)=2^2=4 .

NDA 2022

Exam: NDA
Year: 2022
Subject: Mathematics
Chapter: Functions

NDA 2022 · Mathematics · Functions

If 4 f(x)-f ((1)/(x))= (2 x+ (1)/(x))(2 x- (1)/(x)), then what is...

4 f(x)-f \left(\frac{1}{x}\right)= \left(2

\frac{1}{x}\right)\left(2

\frac{1}{x}\right), then what is

f(2)

equal to?

Options

Correct answer

D. 4

Step-by-step solution

4 f(x)-f\left(\frac{1}{x}\right)=\left(2 x+\frac{1}{x}\right)\left(2 x-\frac{1}{x}\right)4 \mathrm{f}(\mathrm{x})-\mathrm{f}\left(\frac{1}{\mathrm{x}}\right)=4 \mathrm{x}^2-\frac{1}{\mathrm{x}^2}

...(i) Replace

x

\frac{1}{x}

we get

4 f\left(\frac{1}{x}\right)-f(x)=\frac{4}{x^2}-x

...(ii) from

4 \times(

i) + (ii)

16 f(x)-4 f\left(\frac{1}{x}\right)=16 x^2-\frac{4}{x^2}

\begin{aligned} & 16 f(x)-4 f\left(\frac{1}{x}\right)=16 x^2-\frac{4}{x^2} \\ & 4 f\left(\frac{1}{x}\right)-f(x)=\frac{4}{x^2}-x\end{aligned}

$\begin{aligned} & 15 \mathrm{f}(\mathrm{x})=15 \mathrm{x}^2 \\ & \therefore \mathrm{f}(\mathrm{x})=\mathrm{x}^2\end{aligned}

Now,

f(2)=2^2=4

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