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JEE Advanced 2009Paper 2
Exam
JEE Advanced
Year
2009
Paper
Paper 2
Subject
Mathematics
Chapter
Application of Derivatives

JEE Advanced 2009 · Paper 2 · Mathematics · Application of Derivatives

For the function f(x)=x (1)/(x), x ≥ 1 ,

For the function f(x)=xcos⁡1x,x≥1f(x)=x \cos \frac{1}{x}, x \geq 1,

Options

  1. Afor atleast one xx in the interval [1,∞)[1, \infty), f(x+2)−f(x)<2f(x+2)-f(x) < 2
  2. Blim⁡x→∞f′(x)=1\lim _{x \rightarrow \infty} f^{\prime}(x)=1
  3. Cfor all xx in the interval [1,∞)[1, \infty), f(x+2)−f(x)>2f(x+2)-f(x)>2
  4. Df′(x)f^{\prime}(x) is strictly decreasing in the interval [1,∞)[1, \infty)

Correct answer

B. lim⁡x→∞f′(x)=1\lim _{x \rightarrow \infty} f^{\prime}(x)=1

Step-by-step solution

Given, f(x)=xcos⁡1x,x≥1f(x)=x \cos \frac{1}{x}, x \geq 1⇒f′(x)=1xsin⁡1x+cos⁡1x⇒f′′(x)=−1x3cos⁡(1x)\begin{aligned} & \Rightarrow \quad f^{\prime}(x)=\frac{1}{x} \sin \frac{1}{x}+\cos \frac{1}{x} \\ & \Rightarrow \quad f^{\prime \prime}(x)=-\frac{1}{x^3} \cos \left(\frac{1}{x}\right) \end{aligned}Now,lim⁡x→∞f′(x)=0+1=1⇒Now, \lim _{x \rightarrow \infty} f^{\prime}(x)=0+1=1 \Rightarrow Option (b) is correct. Now, x∈[1,∞)⇒1x∈(0,1]⇒f′′(x)<0⇒\quad x \in[1, \infty) \Rightarrow \frac{1}{x} \in(0,1] \Rightarrow f^{\prime \prime}(x) < 0 \Rightarrow Option (d) is correct. As f′(1)=sin⁡1+cos⁡1>1f′(x)f^{\prime}(1)=\sin 1+\cos 1>1 f^{\prime}(x) is strictly decreasing and lim⁡x→∞\lim _{x \rightarrow \infty} f^{\prime}(x)=1So,graphofSo, graph off^{\prime}(x)isshownasbelow.Now,inis shown as below. Now, in[x, x+2], x \in[1, \infty), f(x)is continuous and differentiablesobyLMVT,f′(x)=f(x+2)−f(x)2Asdifferentiable so by LMVT, f^{\prime}(x)=\frac{f(x+2)-f(x)}{2} Asf^{\prime}(x)>1forallfor allx \in[1, \infty) ⇒f(x+2)−f(x)2>1⇒f(x+2)−f(x)>2 for  all  x∈[1,∞)\begin{aligned} & \Rightarrow \quad \frac{f(x+2)-f(x)}{2}>1 \\ & \Rightarrow f(x+2)-f(x)>2 \quad \text{\text{ for }\text{ all } } x \in[1, \infty) \end{aligned}
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