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JEE Main & Advanced AIEEE Solved Paper-2009

  • question_answer
    Directions: Questions No. 90 are Assertion - Reason type questions. Each of these questions contains two statements: Statement-1 (Assertion) and Statement-2 (Reason). Each of these questions also have four alternative choices, only one of which is the correct answer. You have to select the correct choice. Let\[f(x)=x|x|\]and\[g(x)=sinx\] Statement 1: gof is differentiable at\[x=0\]and its derivative is continuous at that point Statement 2: gof is twice differentiable at\[x=0\]                                                            

    A) (a) Statement-1 is true, Statement-2 is true, Statement-2 is a correct explanation for statement-1

    B) (b) Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for statement-1.

    C) (c) Statement-1 is true, statement-2 is false.

    D) (d) Statement-1 is false, Statement-2 is true

    Correct Answer: C

    Solution :

    \[g(f(x))=\sin (f(x))=\left\{ \begin{matrix} \sin {{x}^{2}}, & x\ge 0  \\  -2x\,\cos {{x}^{2}}, & x<0  \\ \end{matrix} \right.\] \[(g(f(x)))'=\left\{ \begin{matrix} 2x\,\cos {{x}^{2}}, & x\ge 0  \\ -2x\,\cos {{x}^{2}}, & x<0  \\ \end{matrix} \right.\] R.H.D. of\[(g(f(0)))'=\underset{h\to {{0}^{+}}}{\mathop{\lim }}\,\frac{2h{{\cosh }^{2}}}{h}=2\] L.H.D. of\[(g(f(0)))'=\underset{h\to {{0}^{+}}}{\mathop{\lim }}\,\frac{2h{{\cosh }^{2}}}{-h}=-2\] Clearly gof is twice differentiable at\[x=0\]hence it is differentiable at\[x=0\]and its derivative is continuous at\[x=0\].


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