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NTA Abhyas 2020
Exam
NTA Abhyas
Year
2020
Subject
Mathematics
Chapter
Continuity and Differentiability

NTA Abhyas 2020 · Mathematics · Continuity and Differentiability

If f(x) = \ cases - 2 x & π ≤ x ≤ (3π)/(2) pcosx + q & (3π)/(2) < x...

If f\left(x\right) = \left\{\begin{cases}- 2\sin x & \pi \le x \le \dfrac{3\pi}{2} \\ pcosx + q & \dfrac{3\pi}{2} < x < 2\pi \\ \cos x - 2 & 2\pi \le x \le \dfrac{5\pi}{2} \\ rsinx & \dfrac{5\pi}{2} < x \le 3\pi\end{cases}\right) is continuous in\left[\pi, 3\pi\right],thenthenumberofpointsatwhich, then the number of points at whichf\left(x\right) is not differentiable is/are

Correct answer

2

Step-by-step solution

At 3π2,\dfrac{3\pi}{2}, −2(−1)=p(0)+q⇒q=2- 2\left(- 1\right) = p\left(0\right) + q ⇒ q = 2
At 2π,2\pi, p+q=1−2⇒p=–3p + q = 1 - 2 ⇒ p = – 3
At 5π2,\dfrac{5\pi}{2}, 0−2=r(1)⇒r=−20 - 2 = r\left(1\right) ⇒ r = - 2
{f}^{'}\left(x\right) = \left\{−2cos⁡x;π≤x<3π2ND;x=3π23sin⁡x;3π2<x<2π0;x=2π−sin⁡x;2π<x<5π2ND;x=5π2<x<3π−2cos⁡x;5π2<x<3π\begin{array}{ll}- 2\cos x; \pi \le x < \dfrac{3\pi}{2} \\ ND; x = \dfrac{3\pi}{2} \\ 3\sin x; \dfrac{3\pi}{2} < x < 2\pi \\ 0; x = 2\pi \\ - \sin x; 2\pi < x < \dfrac{5\pi}{2} \\ ND; x = \dfrac{5\pi}{2} < x < 3\pi \\ - 2\cos x; \dfrac{5\pi}{2} < x < 3\pi\end{array}\right)<br />⇒ f(x)\left(x\right) is not differentiable at 3π2 and 5π2\dfrac{3\pi}{2} \text{ and } \dfrac{5\pi}{2}
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