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Limits Continuity and Differentiability question

2019 · 12 Jan · Shift 2 · Q28
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  5. /2019 · 12 Jan · Shift 2 · Q28

Limits Continuity and Differentiability question

2019 · 12 Jan · Shift 2 · Q28

JEE MainMathematicsLimits Continuity and DifferentiabilityMCQ+4 / −1
lim⁡x→1−π−2sin⁡−1x1−x\mathop {\lim }\limits_{x \to {1^ - }} {{\sqrt \pi - \sqrt {2{{\sin }^{ - 1}}x} } \over {\sqrt {1 - x} }}x→1−lim​1−x​π​−2sin−1x​​ is equal to :
  1. A
    2π\sqrt {{2 \over \pi }}π2​​
  2. B
    12π{1 \over {\sqrt {2\pi } }}2π​1​
  3. C
    π2\sqrt {{\pi \over 2}}2π​​
  4. D
    π\sqrt \piπ​
View written solutionFree

Correct answer: A

  1. We need to evaluate L=lim⁡x→1−π−2sin⁡−1x1−x.L=\lim_{x\to 1^-}\frac{\sqrt{\pi}-\sqrt{2\sin^{-1}x}}{\sqrt{1-x}}.L=limx→1−​1−x​π​−2sin−1x​​.

  2. Observe that as x→1−x\to 1^-x→1−, sin⁡−1x→sin⁡−1(1)=π2,\sin^{-1}x \to \sin^{-1}(1)=\frac{\pi}{2},sin−1x→sin−1(1)=2π​, so 2sin⁡−1x→π.\sqrt{2\sin^{-1}x}\to \sqrt{\pi}.2sin−1x​→π​. Hence the expression is of the indeterminate form 00\frac{0}{0}00​.

  3. Rationalize the numerator: π−2sin⁡−1x1−x⋅π+2sin⁡−1xπ+2sin⁡−1x\frac{\sqrt{\pi}-\sqrt{2\sin^{-1}x}}{\sqrt{1-x}}\cdot \frac{\sqrt{\pi}+\sqrt{2\sin^{-1}x}}{\sqrt{\pi}+\sqrt{2\sin^{-1}x}}1−x​π​−2sin−1x​​⋅π​+2sin−1x​π​+2sin−1x​​

This gives L=lim⁡x→1−π−2sin⁡−1x1−x(π+2sin⁡−1x).L=\lim_{x\to 1^-}\frac{\pi-2\sin^{-1}x}{\sqrt{1-x}\left(\sqrt{\pi}+\sqrt{2\sin^{-1}x}\right)}.L=limx→1−​1−x​(π​+2sin−1x​)π−2sin−1x​.

  1. Use the identity π−2sin⁡−1x=2(π2−sin⁡−1x)=2cos⁡−1x.\pi-2\sin^{-1}x=2\left(\frac{\pi}{2}-\sin^{-1}x\right)=2\cos^{-1}x.π−2sin−1x=2(2π​−sin−1x)=2cos−1x. So, L=lim⁡x→1−2cos⁡−1x1−x(π+2sin⁡−1x).L=\lim_{x\to 1^-}\frac{2\cos^{-1}x}{\sqrt{1-x}\left(\sqrt{\pi}+\sqrt{2\sin^{-1}x}\right)}.L=limx→1−​1−x​(π​+2sin−1x​)2cos−1x​.

  2. Now use the standard limit lim⁡x→1−cos⁡−1x2(1−x)=1,\lim_{x\to 1^-}\frac{\cos^{-1}x}{\sqrt{2(1-x)}}=1,limx→1−​2(1−x)​cos−1x​=1, which implies cos⁡−1x∼2(1−x)as x→1−.\cos^{-1}x \sim \sqrt{2(1-x)}\quad \text{as }x\to 1^-.cos−1x∼2(1−x)​as x→1−.

Therefore, 2cos⁡−1x∼22(1−x).2\cos^{-1}x \sim 2\sqrt{2(1-x)}.2cos−1x∼22(1−x)​.

So, L=lim⁡x→1−22(1−x)1−x(π+2sin⁡−1x).L=\lim_{x\to 1^-}\frac{2\sqrt{2(1-x)}}{\sqrt{1-x}\left(\sqrt{\pi}+\sqrt{2\sin^{-1}x}\right)}.L=limx→1−​1−x​(π​+2sin−1x​)22(1−x)​​.

  1. Cancel 1−x\sqrt{1-x}1−x​: L=lim⁡x→1−22π+2sin⁡−1x.L=\lim_{x\to 1^-}\frac{2\sqrt{2}}{\sqrt{\pi}+\sqrt{2\sin^{-1}x}}.L=limx→1−​π​+2sin−1x​22​​.

As x→1−x\to 1^-x→1−, 2sin⁡−1x→π.\sqrt{2\sin^{-1}x}\to \sqrt{\pi}.2sin−1x​→π​. Hence L=22π+π=222π=2π.L=\frac{2\sqrt{2}}{\sqrt{\pi}+\sqrt{\pi}}=\frac{2\sqrt{2}}{2\sqrt{\pi}}=\sqrt{\frac{2}{\pi}}.L=π​+π​22​​=2π​22​​=π2​​.

  1. Therefore the correct option is 2π.\boxed{\sqrt{\frac{2}{\pi}}}.π2​​​. So, option A\boxed{A}A​ is correct.
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