F-g-R- R are functions such that fx-3x-sin - x 2 -gx- x 3 -2x-sin - x 2 The value of d dx f -1 g -1 x x-12 is equal to

The correct option is A 2 30+x Given that, g:R → R f( x ) = 3x sin( π x2) g( x ) = x^3 + 2x sin( π x2) ∴df^ 1dx( g'( x ))_x = 12 ...

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Byju's Answer

Standard XII

Mathematics

Composite Function

f,g:R→ R are ...

Question

$f, g : R \to R$ are functions such that $f (x) = 3 x - sin (\frac{π x}{2}), g (x) = x^{3} + 2 x - sin (\frac{π x}{2})$
The value of $\frac{d}{d x} f^{- 1} {(g^{- 1} (x))}_{x = 12}$ is equal to

\frac{2}{30 + x}

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\frac{2}{30 - x}

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\frac{2}{3 (28 - π)}

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\frac{2}{3 (28 + π)}

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Solution

The correct option is A $\frac{2}{30 + x}$
Given that,
$g : R \to R$

$f (x) = 3 x - sin (\frac{π x}{2})$

$g (x) = x^{3} + 2 x - sin (\frac{π x}{2})$

$∴ \frac{d f^{- 1}}{d x} {(g^{'} (x))}_{x = 12} = \frac{2}{30 + x}$

Hence, the option $(A)$ is correct.

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f,g:R→R are functions such that f(x)=3x−sin(πx2),g(x)=x3+2x−sin(πx2)The value of ddxf−1(g−1(x))x=12 is equal to

The correct option is A 230+xGiven that,g:R→Rf(x)=3x−sin(πx2)g(x)=x3+2x−sin(πx2)∴df−1dx(g′(x))x=12=230+xHence, the option (A) is correct.

$f, g : R \to R$ are functions such that $f (x) = 3 x - sin (\frac{π x}{2}), g (x) = x^{3} + 2 x - sin (\frac{π x}{2})$
The value of $\frac{d}{d x} f^{- 1} {(g^{- 1} (x))}_{x = 12}$ is equal to

The correct option is A $\frac{2}{30 + x}$
Given that,
$g : R \to R$

$f (x) = 3 x - sin (\frac{π x}{2})$

$g (x) = x^{3} + 2 x - sin (\frac{π x}{2})$

$∴ \frac{d f^{- 1}}{d x} {(g^{'} (x))}_{x = 12} = \frac{2}{30 + x}$

Hence, the option $(A)$ is correct.