JEE Main 2019 (Online) 11th January Evening ShiftQuestion 1
Let $\alpha$ and $\beta$ be the roots of the quadratic equation $x^{2}\sin\theta-x(\sin\theta\cos\theta+1)+\cos\theta=0$ $(0<\theta<45^\circ)$, and $\alpha<\beta$. Then $\sum_{n=0}^{\infty}\left(\alpha^{n}+\frac{(-1)^{n}}{\beta^{n}}\right)$ is equal to :
Quadratic Equation · Relation Between Roots and Coefficients
Solution
Concept. The quadratic factorises; one root is below $1$ and the other above $1$, so both series are convergent G.P.s. Formulas used. (1) $\sum_{n=0}^{\infty}r^n=\dfrac{1}{1-r}$ for $|r|<1$Step 1 - Factorise. $$x^2\sin\theta-x(\sin\theta\cos\theta+1)+\cos\theta=(x\sin\theta-1)(x-\cos\theta).$$ The roots are $\cos\theta$ and $\dfrac{1}{\sin\theta}$. Step 2 - Which is smaller. For $0<\theta<45^\circ$: $\cos\theta<1$ and $\frac1{\sin\theta}>\sqrt2$. So $\alpha=\cos\theta$, $\beta=\frac1{\sin\theta}$. Step 3 - The two sums. By (1), $$\sum_{n=0}^{\infty}\alpha^n=\frac{1}{1-\cos\theta},\qquad\sum_{n=0}^{\infty}\frac{(-1)^n}{\beta^n}=\sum_{n=0}^{\infty}(-\sin\theta)^n=\frac{1}{1+\sin\theta}.$$ Step 4 - Total. $\dfrac{1}{1-\cos\theta}+\dfrac{1}{1+\sin\theta}$. Answer: (C)