If y(θ)=(2 θ+ 2 θ)/( 3 θ+4 2 θ+5 θ+2) , then at θ=(π)/(2), y^ +y^ +y is equal to :

Correct answer: C. 2 aligned & y=(2 θ+2 ^2 θ-1)/(4 ^3 θ-3 θ+8 ^2 θ-4+5 θ+2) & y=((2 ^2 θ+2 θ-1))/((2 ^2 θ+2 θ-1)(2 θ+2)) & y=(1)/(2)((1)/(1+ θ)) & θ=(π)/(2) y=(1)/(2) & y^ =(1)/(2)((-1)/((1+ θ)^2) ×(- θ)) & θ=(π)/(2) y=(1)/(2) aligned aligned y^ & =(1)/(2)[( θ(1+ θ)^2- θ(2)(1+ θ)(- θ))/((1+ θ)^4)] & θ=(π)/(2) y=1 aligned

JEE Main 20245 AprilEvening Shift

Exam: JEE Main
Year: 2024
Date: 5 April
Shift: Evening Shift
Subject: Mathematics
Chapter: Differentiation

JEE Main 2024 · 5 April · Evening Shift · Mathematics · Differentiation

If y(θ)=(2 θ+ 2 θ)/( 3 θ+4 2 θ+5 θ+2) , then at θ=(π)/(2), y^ +y^...

y(\theta)=\frac{2 \cos \theta+\cos 2 \theta}{\cos 3 \theta+4 \cos 2 \theta+5 \cos \theta+2}

, then at

\theta=\frac{\pi}{2}, y^{\prime \prime}+y^{\prime}+y

is equal to :

Options

A $\frac{1}{2}$
B1
C2
D $\frac{3}{2}$

Correct answer

C. 2

Step-by-step solution

\begin{aligned} & y=\frac{2 \cos \theta+2 \cos ^2 \theta-1}{4 \cos ^3 \theta-3 \cos \theta+8 \cos ^2 \theta-4+5 \cos \theta+2} \\ & y=\frac{\left(2 \cos ^2 \theta+2 \cos \theta-1\right)}{\left(2 \cos ^2 \theta+2 \cos \theta-1\right)(2 \cos \theta+2)} \\ & y=\frac{1}{2}\left(\frac{1}{1+\cos \theta}\right) \\ & \Rightarrow \theta=\frac{\pi}{2} \quad y=\frac{1}{2} \\ & y^{\prime}=\frac{1}{2}\left(\frac{-1}{(1+\cos \theta)^2} \times(-\sin \theta)\right) \\ & \Rightarrow \theta=\frac{\pi}{2} \quad y=\frac{1}{2}\end{aligned}

\begin{aligned} y^{\prime \prime} & =\frac{1}{2}\left[\frac{\cos \theta(1+\cos \theta)^2-\sin \theta(2)(1+\cos \theta)(-\sin \theta)}{(1+\cos \theta)^4}\right] \\ & \Rightarrow \theta=\frac{\pi}{2} \quad y=1\end{aligned}

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