The sum of all the integral values of p such | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Let $u = \cos x$, where $u \in [-1, 1]$.The given equation is $3(1 - u^2) + 12u - 3 = p$.Simplifying this, we get $3 - 3u^2 + 12u - 3 = p$, which

Vedclass · Accepted Answer

-$75$

Vedclass · Answer

(C) Let $u = \cos x$, where $u \in [-1, 1]$.The given equation is $3(1 - u^2) + 12u - 3 = p$.Simplifying this, we get $3 - 3u^2 + 12u - 3 = p$, which results in $-3u^2 + 12u = p$.Let $g(u) = -3u^2 + 12u$. To find the range of $g(u)$ for $u \in [-1, 1]$, we calculate the values at the boundaries and the vertex.The derivative $g'(u) = -6u + 12$. Setting $g'(u) = 0$ gives $u = 2$, which is outside the interval $[-1, 1]$.Thus, the function $g(u)$ is strictly increasing on $[-1, 1]$.At $u = -1$, $g(-1) = -3(-1)^2 + 12(-1) = -3 - 12 = -15$.At $u = 1$, $g(1) = -3(1)^2 + 12(1) = -3 + 12 = 9$.Therefore, the range of $p$ is $[-15, 9]$.The sum of all integers from $-15$ to $9$ is given by the sum of an arithmetic progression: $\frac{n}{2}(a + l) = \frac{25}{2}(-15 + 9) = \frac{25}{2}(-6) = -75$.

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

Similar Questions

If $A+B+C=2S$,then $\sin(S-A) \cos(S-B) - \sin(S-C) \cos S =$

The maximum value of the function $f(x) = 3 \sin x + 4 \cos x$ is

In $\Delta ABC$,the minimum value of $\frac{\sum \cot^2 \frac{A}{2} \cot^2 \frac{B}{2}}{\prod \cot^2 \frac{A}{2}}$ is

The maximum value of $(7 \cos\theta + 24 \sin\theta) \times (7 \sin\theta - 24 \cos\theta)$ for every $\theta \in R$.

Find the maximum and minimum values of the function given by $h(x) = \sin(2x) + 5$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module