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 res

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 $A, B, C$ are the angles of a triangle,then $\sin 2A - \sin 2B + \sin 2C =$

In a triangle,if $\tan A + \tan B + \tan C = 6$ and $\tan A \tan B = 2$,then the values of $\tan A, \tan B,$ and $\tan C$ are:

Given that $\cos \left( \frac{\alpha - \beta}{2} \right) = 2\cos \left( \frac{\alpha + \beta}{2} \right)$,then $\tan \frac{\alpha}{2} \tan \frac{\beta}{2}$ is equal to

If $A + B + C = \pi ,$ then $\tan ^2 \frac{A}{2} + \tan ^2 \frac{B}{2} + \tan ^2 \frac{C}{2}$ is always

If $A+B+C=\pi$ and $\cos B=\cos A \cos C$,then $\tan A \tan C=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module