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Limits Continuity and Differentiability question

2011 · Shift 0 · Q30
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  5. /2011 · Shift 0 · Q30

Limits Continuity and Differentiability question

2011 · Shift 0 · Q30

JEE MainMathematicsLimits Continuity and DifferentiabilityMCQ+4 / −1
The value of ppp and qqq for which the function f(x)={sin⁡(p+1)x+sin⁡xx,x<0q,x=0x+x2−xx3/2,x>0f\left( x \right) = \left\{ {\begin{matrix} {{{\sin (p + 1)x + \sin x} \over x}} & {,x \lt 0} \\ q & {,x = 0} \\ {{{\sqrt {x + {x^2}} - \sqrt x } \over {{x^{3/2}}}}} & {,x \gt 0} \\ \end{matrix} } \right.f(x)=⎩⎨⎧​xsin(p+1)x+sinx​qx3/2x+x2​−x​​​,x<0,x=0,x>0​ is continuous for all xxx in R, are
  1. A
    p=52p ={5 \over 2}p=25​, q=12q = {1 \over 2}q=21​
  2. B
    p=−32p =-{3 \over 2}p=−23​, q=12q = {1 \over 2}q=21​
  3. C
    p=12p ={1 \over 2}p=21​, q=32q = {3 \over 2}q=23​
  4. D
    p=12p ={1 \over 2}p=21​, q=−32q = -{3 \over 2}q=−23​
View written solutionFree

Correct answer: B

  1. Continuity needs to be checked only at x=0x=0x=0.

    For x<0x<0x<0 and x>0x>0x>0, each branch is well-defined and continuous on its own domain. So we require lim⁡x→0−f(x)=f(0)=lim⁡x→0+f(x)=q.\lim_{x\to 0^-} f(x)=f(0)=\lim_{x\to 0^+} f(x)=q.limx→0−​f(x)=f(0)=limx→0+​f(x)=q.

  2. Left-hand limit

    For x<0x<0x<0, f(x)=sin⁡((p+1)x)+sin⁡xx.f(x)=\frac{\sin((p+1)x)+\sin x}{x}.f(x)=xsin((p+1)x)+sinx​.

    Using sin⁡(ax)∼ax\sin(ax)\sim axsin(ax)∼ax as x→0x\to 0x→0,

    =\lim_{x\to 0}\left(\frac{\sin((p+1)x)}{x}+\frac{\sin x}{x}\right) =(p+1)+1=p+2.$$ So, $$\lim_{x\to 0^-} f(x)=p+2.$$

  3. Right-hand limit

    For x>0x>0x>0, f(x)=x+x2−xx3/2.f(x)=\frac{\sqrt{x+x^2}-\sqrt{x}}{x^{3/2}}.f(x)=x3/2x+x2​−x​​.

    Factor x\sqrt{x}x​ from the first root: x+x2=x(1+x)=x1+x.\sqrt{x+x^2}=\sqrt{x(1+x)}=\sqrt{x}\sqrt{1+x}.x+x2​=x(1+x)​=x​1+x​.

    Hence f(x)=x(1+x−1)x3/2=1+x−1x.f(x)=\frac{\sqrt{x}(\sqrt{1+x}-1)}{x^{3/2}}=\frac{\sqrt{1+x}-1}{x}.f(x)=x3/2x​(1+x​−1)​=x1+x​−1​.

    Rationalize:

    =\frac{(1+x)-1}{x(\sqrt{1+x}+1)} =\frac{1}{\sqrt{1+x}+1}.$$ Therefore, $$\lim_{x\to 0^+} f(x)=\frac{1}{1+1}=\frac{1}{2}.$$

  4. Continuity at x=0x=0x=0

    We need p+2=q=12.p+2=q=\frac{1}{2}.p+2=q=21​.

    Thus, q=12,p=12−2=−32.q=\frac{1}{2},\qquad p=\frac{1}{2}-2=-\frac{3}{2}.q=21​,p=21​−2=−23​.

  5. Match with options

    The correct pair is p=−32,q=12.p=-\frac{3}{2},\quad q=\frac{1}{2}.p=−23​,q=21​.

    This is Option B.

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