JEE Main 2026 — Mathematics Algebra

Given equation: $3\sin^2 x + 12\cos x - 3 = p$3sin2x+12cosx−3=p

Substituting $\sin^2 x = 1 - \cos^2 x$sin2x=1−cos2x:

$$3(1 - \cos^2 x) + 12\cos x - 3 = p$$3(1−cos2x)+12cosx−3=p

$$-3\cos^2 x + 12\cos x = p$$−3cos2x+12cosx=p

Let $\cos x = t$cosx=t. Since $x \in \mathbb{R}$x∈R, $t \in [-1, 1]$t∈[−1,1].

The equation becomes $p = -3t^2 + 12t$p=−3t2+12t.

Let $f(t) = -3t^2 + 12t$f(t)=−3t2+12t.

Differentiating with respect to $t$t:

$$f'(t) = -6t + 12 = 6(2 - t)$$f′(t)=−6t+12=6(2−t)

For $t \in [-1, 1]$t∈[−1,1], $f'(t) > 0$f′(t)>0, which means $f(t)$f(t) is strictly increasing in the interval $[-1, 1]$[−1,1].

The minimum value of $f(t)$f(t) occurs at $t = -1$t=−1:

$$f(-1) = -3(-1)^2 + 12(-1) = -15$$f(−1)=−3(−1)2+12(−1)=−15

The maximum value of $f(t)$f(t) occurs at $t = 1$t=1:

$$f(1) = -3(1)^2 + 12(1) = 9$$f(1)=−3(1)2+12(1)=9

Thus, the range of $p$p for which the equation has at least one solution is $[-15, 9]$[−15,9].

The integral values of $p$p are $-15, -14, -13, \dots, 9$−15,−14,−13,…,9.

The sum of these integral values is:

$$\sum_{k=-15}^{9} k = (-15) + (-14) + (-13) + (-12) + (-11) + (-10) + \sum_{k=-9}^{9} k$$k=−15∑9​k=(−15)+(−14)+(−13)+(−12)+(−11)+(−10)+k=−9∑9​k

Since $\displaystyle \sum_{k=-9}^{9} k = 0$k=−9∑9​k=0, the sum simplifies to:

$$-15 - 14 - 13 - 12 - 11 - 10 = -75$$−15−14−13−12−11−10=−75

Answer: $-75$−75