JEE Main 2024 (Online) 5th April Evening ShiftQuestion 1
If $y(\theta)=\dfrac{2\cos\theta+\cos2\theta}{\cos3\theta+4\cos2\theta+5\cos\theta+2}$, then at $\theta=\dfrac{\pi}{2}$, $y''+y'+y$ is equal to :
Differentiation and its Application
Solution
Concept. In terms of $c=\cos\theta$, numerator and denominator share the factor $2c^2+2c-1$; so $y=\frac{1}{2(1+\cos\theta)}$. Formulas used. (1) $\cos2\theta=2c^2-1$, $\cos3\theta=4c^3-3c$Step 1 - Simplify. By (1), the numerator is $2c^2+2c-1$ and the denominator is $$4c^3+8c^2+2c-2=2(c+1)(2c^2+2c-1).$$ So $y=\dfrac{1}{2(1+\cos\theta)}$. Step 2 - Derivatives. $$y'=\frac{\sin\theta}{2(1+\cos\theta)^2},\qquad y''=\frac{\cos\theta}{2(1+\cos\theta)^2}+\frac{\sin^2\theta}{(1+\cos\theta)^3}.$$ Step 3 - At $\theta=\frac\pi2$. $y=\frac12$, $y'=\frac12$, $y''=0+1=1$. Step 4 - Value. $y''+y'+y=2$. Answer: (D)