Which of the following statements are correct.

Difficult

A

Suppose f, g and h be three real valued functions defined on R. Let $f (x) = 2 x + |x|, g (x) = \frac{1}{3} (2 x - |x|)$ and $h (x) = f (g (x))$ . Then the function $h (f (x))$ is continuous but not derivable in R.
B

$If f (1) = 3, f' (1) = 2, f'' (1) = 4 and let f^{- 1} (x) = g (x) then g'' (3) is equal to - \frac{1}{2}$
C

If $y = 2 \sin^{- 1} \sqrt{1 - x} + \sin^{- 1} (2 \sqrt{x (1 - x)}) t h e n i t s d e r i v a t i v e i s z e r o$
D

Let $f (x) = \{\begin{matrix} x^{p} \cos (\frac{1}{x}), & x \neq 0 \\ 0, & x = 0 \end{matrix}$ then f(x) will be differentiable at x = 0 if p > 1

Solution

$option 1$
$f (x) = 2 x + |x|$
$= 3 x x > 0 = x x < 0 g (x) = \frac{1}{3} (2 x - |x|) = \frac{1}{3} (x) x > 0 = x x < 0$
$\begin{array}{l} when x > 0 \\ h (x) = f (g (x)) \\ = f (x) \\ x \\ when x < 0 \\ h (x) = f (g (x)) \\ = f (x) \\ = x \\ in both cases h (x) = x \\ hence h (f (x)) = 3 x x \geq 0 \\ x x < 0 \end{array}$

$\begin{array}{l} option 3 : \\ x = \sin^{2} θ, 2 θ \in [0, \frac{π}{2}] \\ y = 2 \sin^{- 1} (\cos θ) + s i n^{- 1} (2 \sin θ \cos θ) \\ = - 2 θ + 2 θ \\ = 0 \end{array}$
$\begin{array}{l} option 4: \\ f (x) = 2^{p} \cos (\frac{1}{x}) x \neq 0 \\ 0 x = 0 \\ when p>1then f (x) is continuous and differtiablehence h (f (x)) is continuous but not differtiable \end{array}$
$\begin{array}{l} option 2: \\ g i v e n f (x) =, f^{'} (1) = 2, f^{''} (1) = 4, a n d \\ f^{- 1} (x) = g (x) \Rightarrow g (3) = 1 \\ since f (x) a n d g (x) are inverse function to each other \\ g^{'} (x) = \frac{1}{f^{'} (g (x))} a n d g^{'} (3) = \frac{1}{2} \\ g^{''} (2) = \frac{- 1}{{[f^{'} (g (x))]}^{2}} f^{''} [g (x)] . g^{'} (x) \\ g^{''} (3) = \frac{- 1}{{[f^{'} (g (3))]}^{2}} f^{''} [g (3)] . g^{'} (3) \\ = \frac{- 1}{4} .4. \frac{1}{2} = \frac{- 1}{2} \end{array}$

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