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JEE Main 2026 · 5 Apr · Shift 1, Mathematics question 12

The sum of all the integral values of pp such that the equation 3sin⁡2x+12cos⁡x−3=p3\sin^2x+12\cos x-3=p, x∈Rx\in\mathbb R, has at least one solution, is:

Choose one option
Answer and worked solution

Official answer: C

In terms of c=cos⁡xc=\cos x.
p=3(1−c2)+12c−3=12−3(c−2)2,c∈[−1,1].p=3(1-c^2)+12c-3=12-3(c-2)^2,\qquad c\in[-1,1].

Range. This is increasing on [−1,1][-1,1]: from −15-15 at c=−1c=-1 to 99 at c=1c=1.

Integers −15,…,9-15,\dots,9: the terms from −9-9 to 99 cancel, leaving −15−14−13−12−11−10=−75-15-14-13-12-11-10=-75.

(C)