JEE Main 2026 (Online) 5th April Morning ShiftQuestion 1
The sum of all the integral values of $p$ such that the equation $3\sin^{2}x+12\cos x-3=p$, $x\in\mathbb{R}$, has at least one solution, is :
Trigonometric Ratio & Identities
Solution
Concept. Write the left side as a quadratic in $c=\cos x$ and find its range on $[-1,1]$. Formulas used. (1) $\sin^2x=1-\cos^2x$Step 1 - Rewrite. By (1), with $c=\cos x\in[-1,1]$: $$p=3(1-c^2)+12c-3=-3c^2+12c=12-3(c-2)^2.$$ Step 2 - Range. The vertex $c=2$ lies to the right of $[-1,1]$, so $p$ increases on $[-1,1]$: $$p(-1)=-15,\qquad p(1)=9,\qquad p\in[-15,9].$$ Step 3 - Sum of integers. $-15+(-14)+\cdots+9$: $25$ terms with average $\frac{-15+9}{2}=-3$, so the sum is $-75$. Answer: (C)