Question

The integral values of ${}^{\prime }{a}^{\prime }$ for which the equation ${\cos }^{2}x-(a^2+a+5)|\cos x|+(a^3+3a^2+2a+6)=0$ has real solution(s)

• A. $-3$
• B. $-2$
• C. $-1$
• D. $0$

Answer 1

The correct option is A $-3$

$Let$ $\left|\cos x\right|=t$

$\therefore t^2-(a^2+a+5)t+(a^3+3a^2+2a+6)=0$

$For$ $a=-3$

$t^2-11=0$

$t=0$ Or $t=11$

$\therefore \cos x=0$ or $\cos x=11$

$But$ $\cos x\le 1$

$\cos x=0$

$For$ $a=-2$

$t^2-7t+6=0$

$(t-1)(t-6)=0$

$t=1$ Or $t=0$

$\therefore \cos x=1$ Or $\cos x=6$

$But$ $\cos x\le 1$

$\therefore \cos x=1$

$Similarlycheckingforresttwooptions,thecaseisnotpossible$

$SobothoptionAandBaretrue.$