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JEE Main 2011
Exam
JEE Main
Year
2011
Subject
Mathematics
Chapter
Continuity and Differentiability

JEE Main 2011 · Mathematics · Continuity and Differentiability

The value of p and q for which the function f(x)=\ array cl ( (p+1)...

The value of pp and qq for which the function f(x)=\left\{sin⁡(p+1)x+sin⁡xxx<0q,x=0x+x2−xx3/2,x>0\begin{array}{cl}\frac{\sin (p+1) x+\sin x}{x} & x < 0 \\ q & , x=0 \\ \frac{\sqrt{x+x^2}-\sqrt{x}}{x^{3 / 2}} & , x>0\end{array}\right. is continuous for all x\mathrm{x} in R\mathrm{R}, is

Options

  1. Ap=52,q=12p=\frac{5}{2}, q=\frac{1}{2}
  2. Bp=−32,q=12p=-\frac{3}{2}, q=\frac{1}{2}
  3. Cp=12,q=32p=\frac{1}{2}, q=\frac{3}{2}
  4. Dp=12,q=−32p=\frac{1}{2}, q=-\frac{3}{2}

Correct answer

B. p=−32,q=12p=-\frac{3}{2}, q=\frac{1}{2}

Step-by-step solution

lim⁡x→0sin⁡(p+1)+sin⁡xx=q=lim⁡x→0x+x2−xx3/2\lim _{x \rightarrow 0} \frac{\sin (p+1)+\sin x}{x}=q=\lim _{x \rightarrow 0} \frac{\sqrt{x+x^2}-\sqrt{x}}{x^{3 / 2}} lim⁡x→0(p+1)cos⁡(p+1)x+cos⁡x=q=12\lim _{x \rightarrow 0}(p+1) \cos (p+1) x+\cos x=q=\frac{1}{2} ⇒p+1+1=12⇒p=−32;q=12\Rightarrow p+1+1=\frac{1}{2} \Rightarrow p=-\frac{3}{2} ; q=\frac{1}{2}
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