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JEE Advanced 2008Paper 1
Exam
JEE Advanced
Year
2008
Paper
Paper 1
Subject
Mathematics
Chapter
Continuity and Differentiability

JEE Advanced 2008 · Paper 1 · Mathematics · Continuity and Differentiability

Let f and g be real valued functions defined on interval (-1,1)...

Let ff and gg be real valued functions defined on interval (−1,1)(-1,1) such that g′′(x)g^{\prime \prime}(x) is continuous, g(0)≠0,g′(0)=0,g′′(0)≠0g(0) \neq 0, g^{\prime}(0)=0, g^{\prime \prime}(0) \neq 0 and f(x)=g(x)sin⁡xf(x)=g(x) \sin x. Statement 1lim⁡x→0[g(x)cos⁡x−g(0)cosec⁡x]=f′′(0)1 \lim _{x \rightarrow 0}[g(x) \cos x-g(0) \operatorname{cosec} x]=f^{\prime \prime}(0). Statement 2f′(0)=g(0)2 f^{\prime}(0)=g(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

B. Statement 1 is true, Statement 2 is true, Statement 2 is not a correct explanation for Statement 1.

Step-by-step solution

We have, lim⁡x→0g(x)cos⁡x−g(0)sin⁡x=lim⁡x→0g′(x)cos⁡x−g(x)sin⁡xcos⁡x=0\begin{aligned}& \lim _{x \rightarrow 0} \frac{g(x) \cos x-g(0)}{\sin x}\\ & =\lim _{x \rightarrow 0} \frac{g^{\prime}(x) \cos x-g(x) \sin x}{\cos x}=0 \end{aligned}Since, f(x)=g(x)sin⁡xf(x)=g(x) \sin x⇒f′(x)=g′(x)sin⁡x+g(x)cos⁡x⇒f′′(x)=g′′(x)sin⁡x+2g′(x)cos⁡x−g(x)sin⁡x⇒f′′(0)=0\begin{aligned}& \Rightarrow \quad f^{\prime}(x)=g^{\prime}(x) \sin x+g(x) \cos x \\ & \Rightarrow \quad f^{\prime \prime}(x)=g^{\prime \prime}(x) \sin x+2 g^{\prime}(x) \cos x-g(x) \sin x \\ & \Rightarrow \quad f^{\prime \prime}(0)=0 \end{aligned}Thus,Thus,\lim _{x \rightarrow 0}[g(x) \cos x-g(0) \operatorname{cosec} x]=0=f^{\prime \prime}(0)\RightarrowStatement1istrue.Statement2.Statement 1 is true. Statement 2.f^{\prime}(x)=g^{\prime}(x) \sin x+g(x) \cos x$ \Rightarrow \quad f^{\prime}(0)=g(0) Statement 2 is not a correct explanation of Statement 1.
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