Q.12. The sum of all the integral values of p such that the equation , , has at least one solution,

Q: What is the correct answer for JEE Main Mathematics Q12?

The correct answer is C. Sol. Let $f(x) = 3\sin^2x + 12\cos x - 3$.We need to find the range of $f(x)$ for $x \in R$.Substitute $\sin^2x = 1 - \cos^2x$:$f(x) = 3(1 - \cos^2x) + 12\cos x - 3$$f(x) = 3 - 3\cos^2x + 12\cos x - 3$$f(x) = -3\cos^2x + 12\cos x$Let $t = \cos x$. Since $x \in R$, $t \in [-1, 1]$.So we need to find the range of $g(t) = -3t^2 + 12t$ for $t \in [-1, 1]$.This is a quadratic function opening downwards

Q: Q.12. The sum of all the integral values of p such that the equation , , has at least one solution, is :

The correct answer is C. Sol. Let $f(x) = 3\sin^2x + 12\cos x - 3$.We need to find the range of $f(x)$ for $x \in R$.Substitute $\sin^2x = 1 - \cos^2x$:$f(x) = 3(1 - \cos^2x) + 12\cos x - 3$$f(x) = 3 - 3\cos^2x + 12\cos x - 3$$f(x) = -3\cos^2x + 12\cos x$Let $t = \cos x$. Since $x \in R$, $t \in [-1, 1]$.So we need to find the range of $g(t) = -3t^2 + 12t$ for $t \in [-1, 1]$.This is a quadratic function opening downwards

Q: What are the options for this JEE Main Mathematics question?

A. -54 · B. -60 · C. -75 · D. -84

Question

Q.12. The sum of all the integral values of p such that the equation , , has at least one solution, is :

FirstInTest · Accepted Answer

Sol. Let $f(x) = 3\sin^2x + 12\cos x - 3$.We need to find the range of $f(x)$ for $x \in R$.Substitute $\sin^2x = 1 - \cos^2x$:$f(x) = 3(1 - \cos^2x) + 12\cos x - 3$$f(x) = 3 - 3\cos^2x + 12\cos x - 3$$f(x) = -3\cos^2x + 12\cos x$Let $t = \cos x$. Since $x \in R$, $t \in [-1, 1]$.So we need to find the range of $g(t) = -3t^2 + 12t$ for $t \in [-1, 1]$.This is a quadratic function opening downwards. The vertex occurs at $t = -\frac{12}{2(-3)} = -\frac{12}{-6} = 2$.Since the vertex $t=2$ is outside the interval $[-1, 1]$, the maximum and minimum values will occur at the endpoints of the interval.At $t = -1$: $g(-1) = -3(-1)^2 + 12(-1) = -3 - 12 = -15$.At $t = 1$: $g(1) = -3(1)^2 + 12(1) = -3 + 12 = 9$.So, the range of $g(t)$ for $t \in [-1, 1]$ is $[-15, 9]$.For the equation $f(x) = p$ to have at least one solution, $p$ must be in the range of $f(x)$.So, $p \in [-15, 9]$.We need to find the sum of all integral values of $p$.The integral values of $p$ are $-15, -14, \dots, 0, \dots, 8, 9$.This is an arithmetic progression.Sum $= \sum_{p=-15}^{9} p = (-15) + (-14) + \dots + (-1) + 0 + 1 + \dots + 9$.The terms from $-9$ to $9$ cancel out.So the sum is $(-15) + (-14) + (-13) + (-12) + (-11) + (-10)$$= -(15+14+13+12+11+10)$$= -(75) = -75$.

Answer

-54

Answer

-60

Answer

-75

Answer

-84

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