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Maths Question 12 – JEE-MAIN 2026

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The sum of all the integral values of p such that the equation 3sin2⁡x+12cos⁡x−3=p, x∈R, has at least one solution, is:

Simplify the given equation using the fundamental trigonometric identity to express it in terms of a single trigonometric function.

Step 1: Rewrite the equation in terms of a single trigonometric function✦ Active

The given equation is 3sin2⁡x+12cos⁡x−3=p. Use the identity sin2⁡x=1−cos2⁡x to express the equation solely in terms of cos⁡x.

p=3(1−cos2⁡x)+12cos⁡x−3p=3−3cos2⁡x+12cos⁡x−3p=−3cos2⁡x+12cos⁡x
Step 2: Determine the range of the quadratic function○ Expand

Let y=cos⁡x. Since x∈R, the range of y is [−1,1]. We need to find the range of the quadratic function g(y)=−3y2+12y for y∈[−1,1]. The vertex of this downward-opening parabola is at y=−122(−3)=2, which is outside the interval [−1,1]. Since the parabola opens downwards, the function is increasing on [−1,1] (as y<2). Therefore, the minimum and maximum values occur at the endpoints of the interval.

Minimum value at y=−1:g(−1)=−3(−1)2+12(−1)=−3−12=−15Maximum value at y=1:g(1)=−3(1)2+12(1)=−3+12=9

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

Step 3: Calculate the sum of all integral values of p○ Expand

The integral values of p in the interval [−15,9] are −15,−14,…,0,…,8,9. The sum of these integers is calculated as follows:

Sum=∑k=−159k=(−15)+(−14)+⋯+(−1)+0+1+⋯+8+9Sum=−(15×(15+1)2)+(9×(9+1)2)Sum=−(15×162)+(9×102)Sum=−120+45=−75

The sum of all integral values of p is −75.

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