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BITSAT 2023Paper 1
Exam
BITSAT
Year
2023
Paper
Paper 1
Subject
Mathematics
Chapter
Differentiation

BITSAT 2023 · Paper 1 · Mathematics · Differentiation

If y= ((1+ 2 θ)/(1- 2 θ)) , then (d y)/(d θ) at θ=(3 π)/(4) is :

If y=(1+cos⁡2θ1−cos⁡2θ)y=\sqrt{\left(\frac{1+\cos 2 \theta}{1-\cos 2 \theta}\right)}, then dydθ\frac{d y}{d \theta} at θ=3π4\theta=\frac{3 \pi}{4} is :

Options

  1. A-2
  2. B2
  3. C±2\pm 2
  4. DNone of these

Correct answer

A. -2

Step-by-step solution

y=1+cos⁡2θ1−cos⁡2θy=\sqrt{\frac{1+\cos 2 \theta}{1-\cos 2 \theta}}
⇒y=2cos⁡2θ2sin⁡2θ=cot⁡2θ\Rightarrow y=\sqrt{\frac{2 \cos ^2 \theta}{2 \sin ^2 \theta}}=\sqrt{\cot ^2 \theta}
⇒y=cot⁡θ\Rightarrow y=\cot \theta
Differentiate w.r.t. ' θ\theta ', we get: dydθ=−cosec⁡2θ\frac{d y}{d \theta}=-\operatorname{cosec}^2 \theta
Now, \left(\frac{\mathrm{dy}}{\mathrm{d} \theta}\right)_{\theta=\frac{3 \pi}{4}}=-\operatorname{cosec}^2 \left(\frac{3 \pi}{4}\right)
=−cosec⁡2(π−π4)=−cosec⁡2π4=−2=-\operatorname{cosec}^2\left(\pi-\frac{\pi}{4}\right)=-\operatorname{cosec}^2 \frac{\pi}{4}=-2
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