If 2 f x 2-3 f 1- x 2- x 2-1- then the domain of the function f isA- R B- R -0C- R -1-0-1D- R -

The correct option is B ℝ {0}2f(x^2)+3f(1x^2)=x^2 1 ⋯ (1) Replace x^2 with 1x^2, 2f(1x^2)+3f(x^2)=1x^2 1 ⋯ nbsp; (2) On solving (1) and (2), we get ...

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Byju's Answer

Standard XII

Mathematics

Domain

If 2 f x 2+3 ...

Question

If $2 f (x^{2}) + 3 f (\frac{1}{x^{2}}) = x^{2} - 1,$ then the domain of the function $f$ is

R

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R - {0}

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R - {- 1, 0, 1}

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R^{+}

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Solution

The correct option is B $R - {0}$
$2 f (x^{2}) + 3 f (\frac{1}{x^{2}}) = x^{2} - 1 \dots (1)$
Replace $x^{2}$ with $\frac{1}{x^{2}},$
$2 f (\frac{1}{x^{2}}) + 3 f (x^{2}) = \frac{1}{x^{2}} - 1 \dots (2)$

On solving $(1)$ and $(2),$ we get
$f (x^{2}) = \frac{1}{5} [\frac{3}{x^{2}} - 2 x^{2} - 1]$
$\Rightarrow f (x) = \frac{1}{5} [\frac{3}{x} - 2 x - 1]$
$∴$ Domain of $f$ is $R - {0}$

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If 2f(x2)+3f(1x2)=x2−1, then the domain of the function f is

The correct option is B R−{0}2f(x2)+3f(1x2)=x2−1 ⋯(1) Replace x2 with 1x2, 2f(1x2)+3f(x2)=1x2−1 ⋯(2) On solving (1) and (2), we get f(x2)=15[3x2−2x2−1] ⇒f(x)=15[3x−2x−1] ∴ Domain of f is R−{0}

If $2 f (x^{2}) + 3 f (\frac{1}{x^{2}}) = x^{2} - 1,$ then the domain of the function $f$ is