NIMCET 2011 — Mathematics PYQ

Recall the triple angle formula for cosine:

$\cos 3x=4{\cos }^{3}x-3\cos x$

Rearranging this, we get:

${\cos }^{3}x=\frac{\cos 3x+3\cos x}{4}$

Substitute this into the integral:

$I={\int }_0^{\frac{\pi }{2}}\{a^2{\cos }^{3}x+a\sin x-20\cos x\}dx$

Step 2: Evaluate the integral

We integrate each term separately from $0$ to $\frac{\pi }{2}$:

1. ${\int }_0^{\frac{\pi }{2}}{\cos }^{3}xdx$: Using Wallis' Formula or substitution ($u=\sin x$), the value is $\frac{2}{3}$.

2. ${\int }_0^{\frac{\pi }{2}}\sin xdx=[-\cos x{]}_0^{\frac{\pi }{2}}=-(0-1)=1$.

3. ${\int }_0^{\frac{\pi }{2}}\cos xdx=[\sin x{]}_0^{\frac{\pi }{2}}=1-0=1$.

Now substitute these values back into the expression:

$a^2(\frac{2}{3})+a(1)-20(1)\le \frac{-a^2}{3}$

Step 3: Solve the inequality

$\frac{2a^2}{3}+a-20\le \frac{-a^2}{3}$

Add $\frac{a^2}{3}$ to both sides:

$\frac{3a^2}{3}+a-20\le 0$

$a^2+a-20\le 0$

Step 4: Factorize the quadratic

$(a+5)(a-4)\le 0$

The roots are $a=-5$ and $a=4$. For the expression to be less than or equal to zero, $a$ must lie in the interval:

$-5\le a\le 4$

Step 5: Consider the constraint

The problem states that $a$ is a positive integer. Therefore:

$a\in \{1,2,3,4\}$

There are four such values.

Correct Option:

(d) four