Question

if f(x)=x2 -1/x2, then value of f(x)+f(1/x) is​

Answer 1

11th

Maths

Relations and Functions

Algebra of Real Functions

If f(x) = x^3 - x^2 + x + 1...

MATHS

Asked on December 27, 2019 byDivyanshu Adil

If f(x)=x3−x2+x+1, then the value of 2f(1)+f(−1) will be

A

5

B

2

C

0

D

-2

EASY

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ANSWER

Consider the given function.

f(x)=x3−x2+x+1

Put x=1

f(1)=1−1+1+1

f(1)=2

Put x=−1

f(−1)=−1−1−1+1

f(−1)=−2

Since,

2f(1)+f(−1)

=22−2

=0

Hence, this is the answer.

Answer 2

SOLUTION

GIVEN

$\displaystyle \sf{ f(x) = {x}^{2} - \frac{1}{ {x}^{2} } }$

TO DETERMINE

$\displaystyle \sf{ f(x) + f \bigg( \frac{1}{x} \bigg) }$

EVALUATION

Here it is given that

$\displaystyle \sf{ f(x) = {x}^{2} - \frac{1}{ {x}^{2} } }$

Now

$\displaystyle \sf{ f \bigg( \frac{1}{x} \bigg) = {\bigg( \frac{1}{x} \bigg) }^{2} - \frac{1}{ {\bigg( \frac{1}{x} \bigg) }^{2}} }$

$\displaystyle \sf{ \implies \: f \bigg( \frac{1}{x} \bigg) = \frac{1}{ {x}^{2} } - \frac{1}{ \frac{1}{ {x}^{2} } } }$

$\displaystyle \sf{ \implies \: f \bigg( \frac{1}{x} \bigg) = \frac{1}{ {x}^{2} } - {x}^{2} }$

Thus we have

$\displaystyle \sf{ f(x) + f \bigg( \frac{1}{x} \bigg) }$

$\displaystyle \sf{ = {x}^{2} - \frac{1}{ {x}^{2} } + \frac{1}{ {x}^{2} } - {x}^{2} }$

$= 0$

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