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  • Directions : Questions are Assertion- Reason type questions. Each of these questions contains two statements : Statement- 1 (Assertion) and Statement - 2 (Reason). Each of these

Directions : Questions are Assertion- Reason type questions. Each of these questions contains two statements :

Statement- 1 (Assertion) and Statement - 2 (Reason).

Each of these questions also has four alternative choices, only one of which is the correct answer. You have to select the correct choice.

Question : Let f(x)=x\left | x \right |\; and\; g(x)=sin\, x.

Statement- 1 :  gof is differentiable at x=0 and its derivative is continuous at that point.

Statement- 2 :  gof is twice differentiable at x=0.

Option: 1

Statement-1 is true,Statement-2 is true; Statement-2 is not a correct explanation for Statement-1


Option: 2

Statement- 1 is true, Statement-2 is false


Option: 3

Statement-1 is false, Statement-2 is true


Option: 4

Statement-1 is true, Statement-2 is true; Statement-2 is correct explanation for Statement-1


Answers (1)

\\ f (x) = x/x/ \\ g (x) = \sin x \\ f(x) = \begin{Bmatrix} -x^{2} & x< 0\\ x^{2} & x> 0 \end{matrix}

\\ gof (x) = g [fx] = g (-x^{2}) = \sin (-x^{2}) = \sin -x^{2} and \sin x^{2}

\therefore gof (n) = \begin{Bmatrix} -\sin x ^{2} & x< 0 \\ \sin x ^{2} & x\geqslant 0 \end{matrix}

\therefore [gof (n)]^{1} = \begin{Bmatrix} -2x\cos x ^{2} & x< 0 \\ 2\cos x ^{2} & x\geqslant 0 \end{matrix}

0= at \ both \ 0^{+} \and \ 0^{-}

Now \\ {[gof (x)]}'' = \left\{\begin{matrix} -2 [1.cosx^{2}+ x (-\sin x^{2})\times 2 x] & x< 0 \\ -2 [1.cosx^{2}+ x (-\sin x^{2})\times 2 x] & x\geqslant 0 \end{matrix}\right.

= \left\{\begin{matrix} -2 &\\ +2 \end{matrix}\right.

so It is not twice differential at x = 0

Posted by

Sumit Saini

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