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JEE Main 2009
Exam
JEE Main
Year
2009
Subject
Mathematics
Chapter
Continuity and Differentiability

JEE Main 2009 · Mathematics · Continuity and Differentiability

Let f(x)=x|x| and g(x)= x . Statement-1 : gof is differentiable at...

Let f(x)=x∣x∣f(x)=x|x| and g(x)=sin⁡xg(x)=\sin x. Statement-1 : gof is differentiable at x=0x=0 and its derivative is continuous at that point. Statement-2 : gof is twice differentiable at x=0x=0.

Options

  1. AStatement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1
  2. BStatement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1
  3. CStatement-1 is true, Statement-2 is false
  4. DStatement-1 is false, Statement-2 is true

Correct answer

C. Statement-1 is true, Statement-2 is false

Step-by-step solution

f(x)=x∣x∣ and g(x)=sin⁡xg∘f(x)=sin⁡(x∣x∣)={−sin⁡x2,x<0sin⁡x2,x≥0∴ (gof )′(x)={−2xcos⁡x2,x<02xcos⁡x2,x≥0\begin{aligned}& f(x)=x|x| \text{ and } g(x)=\sin x \\ & g {}^{\circ} f(x)=\sin (x|x|)=\left\{\begin{array}{cc} -\sin x² & , x < 0 \\ \sin x² & , x \geq 0 \end{array}\right. \\ & \therefore \text{ (gof })^{\prime}(x)=\left\{\begin{array}{cc} -2 x \cos x² & , x < 0 \\ 2 x \cos x² & , x \geq 0 \end{array}\right. \end{aligned}Clearly, L(gof )' (0)=0=(0)=0=\mathrm{R}((\text{ gof })′(0)∴)^{\prime}(0)\therefore gof is differentiable at x=0x=0 and also its derivative is continuous at x=0x=0  Now, (gof)" (x)={−2cos⁡x2+4x2sin⁡x2,x<02cos⁡x2−4x2sin⁡x2,x≥0∴L( gof )′′(0)=−2 and R( gof )′′(0)=2∴L( gof )′′(0)≠R( gof )′′(0)\begin{aligned} \text{ Now, (gof)" }(\mathrm{x})=\left\{\begin{array}{cc} -2 \cos \mathrm{x}^2+4 \mathrm{x}^2 \sin \mathrm{x}^2 & , \mathrm{x}< 0 \\ 2 \cos \mathrm{x}^2-4 \mathrm{x}^2 \sin \mathrm{x}^2 & , \mathrm{x} \geq0 \end{array}\right. \\ & \therefore \mathrm{L}(\text{ gof })^{\prime \prime}(0)=-2 \text{ and } \mathrm{R}(\text{ gof })^{\prime \prime}(0)=2 \\ & \therefore \mathrm{L}(\text{ gof })^{\prime \prime}(0) \neq \mathrm{R}(\text{ gof })^{\prime \prime}(0) \end{aligned}\therefore g o f(x)isnottwicedifferentiableatis not twice differentiable atx=0$.
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