If - f - is an increasing function from R - R such that f - x -0 and f -1 exists then f -1 x - dx 2 isA -0B- cannot be determined C -0D- -0

The correct option is D lt;0f'(x) gt;0 and f”(x) gt;0 Let g(x)=f^ 1(x) f(g(x))=x f'(g(x)).g'(x)=1 g'(x)=1/f'(g(x)) f”(g(x)).g'(x) ∴ g”(x). lt;0. d ...

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Standard XII

Mathematics

Logarithmic Differentiation

If ' f ' is a...

Question

If ‘f ’ is an increasing function from $R \to R$ such that f'' (x) > 0 and $f^{- 1}$ exists then $\frac{(f^{- 1} (x))}{d x^{2}}$ is

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Solution

The correct option is A <0
$f^{'} (x) > 0 a n d f^{''} (x) > 0 L e t g (x) = f^{- 1} (x) f (g (x)) = x f^{'} (g (x)) . g^{'} (x) = 1 g^{'} (x) = \frac{1}{f^{'} (g (x))} f^{''} (g (x)) . g^{'} (x) ∴ g^{''} (x) . < 0. \frac{d^{2}}{d x^{2}} (f^{- 1} (x) = g w^{''} (x)) . < 0$

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If ‘f ’ is an increasing function from R→R such that f'' (x) > 0 and f−1 exists then (f−1(x))dx2 is

The correct option is A <0f′(x)>0 and f′′(x)>0Let g(x)=f−1(x)f(g(x))=xf′(g(x)).g′(x)=1g′(x)=1f′(g(x))f′′(g(x)).g′(x)∴g′′(x).<0.d2dx2(f−1(x)=gw′′(x)).<0

If ‘f ’ is an increasing function from $R \to R$ such that f'' (x) > 0 and $f^{- 1}$ exists then $\frac{(f^{- 1} (x))}{d x^{2}}$ is

The correct option is A <0
$f^{'} (x) > 0 a n d f^{''} (x) > 0 L e t g (x) = f^{- 1} (x) f (g (x)) = x f^{'} (g (x)) . g^{'} (x) = 1 g^{'} (x) = \frac{1}{f^{'} (g (x))} f^{''} (g (x)) . g^{'} (x) ∴ g^{''} (x) . < 0. \frac{d^{2}}{d x^{2}} (f^{- 1} (x) = g w^{''} (x)) . < 0$