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Limits Continuity and Differentiability question

2019 · 10 Apr · Shift 1 · Q31
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Limits Continuity and Differentiability question

2019 · 10 Apr · Shift 1 · Q31

JEE MainMathematicsLimits Continuity and DifferentiabilityMCQ+4 / −1
If f(x)={sin⁡(p+1)x+sin⁡xx,x<0q,x=0x+x2−xx32,x>0f(x) = \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^{\frac{3}{2}}}}}} & {,x \gt 0} \\ \end{matrix} } \right.f(x)=⎩⎨⎧​xsin(p+1)x+sinx​qx23​x+x2​−x​​​,x<0,x=0,x>0​ is continuous at x = 0, then the ordered pair (p, q) is equal to
  1. A
    (−32,−12)\left( { - {3 \over 2}, - {1 \over 2}} \right)(−23​,−21​)
  2. B
    (−12,32)\left( { - {1 \over 2},{3 \over 2}} \right)(−21​,23​)
  3. C
    (−32,12)\left( { - {3 \over 2}, {1 \over 2}} \right)(−23​,21​)
  4. D
    (52,12)\left( { {5 \over 2}, {1 \over 2}} \right)(25​,21​)
View written solutionFree

Correct answer: C

  1. Condition for continuity at x=0x=0x=0

For continuity at x=0x=0x=0, we need

lim⁡x→0−f(x)=f(0)=lim⁡x→0+f(x).\lim_{x\to 0^-} f(x)=f(0)=\lim_{x\to 0^+} f(x).x→0−lim​f(x)=f(0)=x→0+lim​f(x).

Since f(0)=qf(0)=qf(0)=q, we compute the left-hand and right-hand limits.

  1. Left-hand limit x→0−x\to 0^-x→0−

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 lim⁡t→0sin⁡tt=1\lim_{t\to 0}\frac{\sin t}{t}=1limt→0​tsint​=1,

lim⁡x→0−sin⁡((p+1)x)x=lim⁡x→0(p+1)⋅sin⁡((p+1)x)(p+1)x=p+1,\lim_{x\to 0^-}\frac{\sin((p+1)x)}{x} =\lim_{x\to 0}(p+1)\cdot \frac{\sin((p+1)x)}{(p+1)x}=p+1,x→0−lim​xsin((p+1)x)​=x→0lim​(p+1)⋅(p+1)xsin((p+1)x)​=p+1,

and

lim⁡x→0−sin⁡xx=1.\lim_{x\to 0^-}\frac{\sin x}{x}=1.x→0−lim​xsinx​=1.

Therefore,

lim⁡x→0−f(x)=(p+1)+1=p+2.\lim_{x\to 0^-} f(x)= (p+1)+1=p+2.x→0−lim​f(x)=(p+1)+1=p+2.
  1. Right-hand limit x→0+x\to 0^+x→0+

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 numerator:

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​.

So,

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​.

Now rationalize:

1+x−1x⋅1+x+11+x+1=(1+x)−1x(1+x+1)=11+x+1.\frac{\sqrt{1+x}-1}{x}\cdot \frac{\sqrt{1+x}+1}{\sqrt{1+x}+1} =\frac{(1+x)-1}{x(\sqrt{1+x}+1)} =\frac{1}{\sqrt{1+x}+1}.x1+x​−1​⋅1+x​+11+x​+1​=x(1+x​+1)(1+x)−1​=1+x​+11​.

Hence,

lim⁡x→0+f(x)=11+1=12.\lim_{x\to 0^+} f(x)=\frac{1}{1+1}=\frac12.x→0+lim​f(x)=1+11​=21​.
  1. Apply continuity

Thus,

p+2=q=12.p+2=q=\frac12.p+2=q=21​.

So,

p=12−2=−32,q=12.p=\frac12-2=-\frac32, \qquad q=\frac12.p=21​−2=−23​,q=21​.

Therefore,

(p,q)=(−32,12).(p,q)=\left(-\frac32,\frac12\right).(p,q)=(−23​,21​).
  1. Check options

This matches Option C:

(−32,12).\left(-\frac32,\frac12\right).(−23​,21​).
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