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Limits Continuity and Differentiability question

2021 · 17 Mar · Shift 1 · Q24
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  5. /2021 · 17 Mar · Shift 1 · Q24

Limits Continuity and Differentiability question

2021 · 17 Mar · Shift 1 · Q24

JEE MainMathematicsLimits Continuity and DifferentiabilityMCQ+4 / −1
The value of lim⁡x→0+cos⁡−1(x−[x]2).sin⁡−1(x−[x]2)x−x3\mathop {\lim }\limits_{x \to {0^ + }} {{{{\cos }^{ - 1}}(x - {{[x]}^2}).{{\sin }^{ - 1}}(x - {{[x]}^2})} \over {x - {x^3}}}x→0+lim​x−x3cos−1(x−[x]2).sin−1(x−[x]2)​, where [ x ] denotes the greatest integer ≤\le≤ x is :
  1. A
    π\piπ
  2. B
    π4{\pi \over 4}4π​
  3. C
    π2{\pi \over 2}2π​
  4. D
    0
View written solutionFree

Correct answer: C

  1. Interpret the greatest integer term near x=0+x=0^+x=0+

For x→0+x \to 0^+x→0+, we have 0<x<10<x<10<x<1. Hence [x]=0.[x]=0.[x]=0. So, x−[x]2=x−0=x.x-[x]^2=x-0=x.x−[x]2=x−0=x. Therefore the given limit becomes lim⁡x→0+cos⁡−1(x) sin⁡−1(x)x−x3.\lim_{x\to 0^+}\frac{\cos^{-1}(x)\,\sin^{-1}(x)}{x-x^3}.limx→0+​x−x3cos−1(x)sin−1(x)​.

  1. Use the identity connecting inverse sine and inverse cosine

For x∈[−1,1]x\in[-1,1]x∈[−1,1], sin⁡−1(x)+cos⁡−1(x)=π2.\sin^{-1}(x)+\cos^{-1}(x)=\frac{\pi}{2}.sin−1(x)+cos−1(x)=2π​. Thus, cos⁡−1(x)=π2−sin⁡−1(x).\cos^{-1}(x)=\frac{\pi}{2}-\sin^{-1}(x).cos−1(x)=2π​−sin−1(x). So the numerator is cos⁡−1(x)sin⁡−1(x)=(π2−sin⁡−1(x))sin⁡−1(x).\cos^{-1}(x)\sin^{-1}(x)=\left(\frac{\pi}{2}-\sin^{-1}(x)\right)\sin^{-1}(x).cos−1(x)sin−1(x)=(2π​−sin−1(x))sin−1(x).

  1. Behavior as x→0+x\to 0^+x→0+

We know sin⁡−1(x)→0as x→0,\sin^{-1}(x)\to 0 \quad \text{as } x\to 0,sin−1(x)→0as x→0, and hence cos⁡−1(x)→π2.\cos^{-1}(x)\to \frac{\pi}{2}.cos−1(x)→2π​. Also, x−x3=x(1−x2).x-x^3=x(1-x^2).x−x3=x(1−x2). As x→0x\to 0x→0, 1−x2→11-x^2\to 11−x2→1, so denominator behaves like xxx.

Thus the limit is

=\left(\lim_{x\to 0^+}\cos^{-1}(x)\right) \left(\lim_{x\to 0^+}\frac{\sin^{-1}(x)}{x}\right) \left(\lim_{x\to 0^+}\frac{1}{1-x^2}\right).$$ Now, $$\lim_{x\to 0^+}\cos^{-1}(x)=\frac{\pi}{2},$$ $$\lim_{x\to 0^+}\frac{\sin^{-1}(x)}{x}=1,$$ $$\lim_{x\to 0^+}\frac{1}{1-x^2}=1.$$ Therefore, $$\text{Limit}=\frac{\pi}{2}\cdot 1\cdot 1=\frac{\pi}{2}.$$ 4. **Check options** - A: $\pi$ ❌ - B: $\dfrac{\pi}{4}$ ❌ - C: $\dfrac{\pi}{2}$ ✅ - D: $0$ ❌ Hence the correct option is **C**.
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