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Limits Continuity and Differentiability question

2020 · 5 Sep · Shift 1 · Q35
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  5. /2020 · 5 Sep · Shift 1 · Q35

Limits Continuity and Differentiability question

2020 · 5 Sep · Shift 1 · Q35

JEE MainMathematicsLimits Continuity and DifferentiabilityMCQ+4 / −1
If the function f(x)={k1(x−π)2−1,x≤πk2cos⁡x,x>πf\left( x \right) = \left\{ {\begin{matrix} {{k_1}{{\left( {x - \pi } \right)}^2} - 1,} & {x \le \pi } \\ {{k_2}\cos x,} & {x \gt \pi } \\ \end{matrix} } \right.f(x)={k1​(x−π)2−1,k2​cosx,​x≤πx>π​ is twice differentiable, then the ordered pair (k1, k2) is equal to :
  1. A
    (12,−1)\left( {{1 \over 2},-1} \right)(21​,−1)
  2. B
    (1, 1)
  3. C
    (1, 0)
  4. D
    (12,1)\left( {{1 \over 2},1} \right)(21​,1)
View written solutionFree

Correct answer: D

  1. We need the piecewise function
f(x)={k1(x−π)2−1,x≤πk2cos⁡x,x>πf(x)= \begin{cases} k_1(x-\pi)^2-1, & x\le \pi \\ k_2\cos x, & x>\pi \end{cases}f(x)={k1​(x−π)2−1,k2​cosx,​x≤πx>π​

to be twice differentiable at x=πx=\pix=π.

That requires:

  1. continuity at x=πx=\pix=π
  2. first derivative continuity at x=πx=\pix=π
  3. second derivative continuity at x=πx=\pix=π
  1. Continuity at x=πx=\pix=π

Left value:

f(π)=k1(π−π)2−1=−1f(\pi)=k_1(\pi-\pi)^2-1=-1f(π)=k1​(π−π)2−1=−1

Right-hand limit:

lim⁡x→π+f(x)=lim⁡x→π+k2cos⁡x=k2cos⁡π=−k2\lim_{x\to \pi^+} f(x)=\lim_{x\to \pi^+} k_2\cos x=k_2\cos \pi=-k_2x→π+lim​f(x)=x→π+lim​k2​cosx=k2​cosπ=−k2​

For continuity,

−1=−k2  ⟹  k2=1-1=-k_2 \implies k_2=1−1=−k2​⟹k2​=1
  1. First derivative continuity at x=πx=\pix=π

For x<πx<\pix<π,

f′(x)=2k1(x−π)f'(x)=2k_1(x-\pi)f′(x)=2k1​(x−π)

So,

f−′(π)=lim⁡x→π−2k1(x−π)=0f'_-(\pi)=\lim_{x\to \pi^-}2k_1(x-\pi)=0f−′​(π)=x→π−lim​2k1​(x−π)=0

For x>πx>\pix>π,

f′(x)=−k2sin⁡xf'(x)= -k_2\sin xf′(x)=−k2​sinx

Thus,

f+′(π)=−k2sin⁡π=0f'_+(\pi)= -k_2\sin \pi=0f+′​(π)=−k2​sinπ=0

So first derivatives match automatically:

f−′(π)=f+′(π)=0f'_-(\pi)=f'_+(\pi)=0f−′​(π)=f+′​(π)=0

No new condition arises.

  1. Second derivative continuity at x=πx=\pix=π

For x<πx<\pix<π,

f′′(x)=2k1f''(x)=2k_1f′′(x)=2k1​

Hence,

f−′′(π)=2k1f''_-(\pi)=2k_1f−′′​(π)=2k1​

For x>πx>\pix>π,

f′′(x)=−k2cos⁡xf''(x)= -k_2\cos xf′′(x)=−k2​cosx

Hence,

f+′′(π)=−k2cos⁡π=−k2(−1)=k2f''_+(\pi)= -k_2\cos \pi = -k_2(-1)=k_2f+′′​(π)=−k2​cosπ=−k2​(−1)=k2​

For twice differentiability,

2k1=k22k_1=k_22k1​=k2​

Since k2=1k_2=1k2​=1,

2k1=1  ⟹  k1=122k_1=1 \implies k_1=\frac122k1​=1⟹k1​=21​
  1. Therefore,
(k1,k2)=(12,1)(k_1,k_2)=\left(\frac12,1\right)(k1​,k2​)=(21​,1)

This matches Option D.

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