JEE Main 2026 — Functions Question with Solution

Given equation: $3{\sin }^{2}x+12\cos x-3=p$

Substituting ${\sin }^{2}x=1-{\cos }^{2}x$:

$3(1-{\cos }^{2}x)+12\cos x-3=p$

$-3{\cos }^{2}x+12\cos x=p$

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

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

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

Differentiating with respect to $t$:

$f^{\prime }(t)=-6t+12=6(2-t)$

For $t\in [-1,1]$, $f^{\prime }(t)>0$, which means $f(t)$ is strictly increasing in the interval $[-1,1]$.

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

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

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

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

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

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

The sum of these integral values is:

${\sum }_{k=-15}^{9}k=(-15)+(-14)+(-13)+(-12)+(-11)+(-10)+{\sum }_{k=-9}^{9}k$

Since ${\sum }_{k=-9}^{9}k=0$, the sum simplifies to:

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

Answer: $-75$