NIMCET 2024 — Mathematics PYQ

To solve this problem, we need to evaluate the greatest integer values inside the function brackets.

Step 1: Find the value of ${\pi }^2$

The value of the mathematical constant $\pi$ is approximately $3.14159$. Let's compute its square:

$${\pi }^2\approx (3.14159{)}^2\approx 9.8696$$

Step 2: Apply the Greatest Integer Function (GIF)

The greatest integer function $[y]$ outputs the largest integer less than or equal to $y$:

• For $[{\pi }^2]$: Since $9.8696$ lies between the integers $9$ and $10$:

  $$[{\pi }^2]=9$$

• For $[-{\pi }^2]$: Since $-9.8696$ lies between the integers $-10$ and $-9$:

  $$[-{\pi }^2]=-10$$

Step 3: Substitute these values back into $f(x)$

Now, rewrite the function expression:

$$f(x)=\cos (9x)+\cos (-10x)$$

Since cosine is an even function ($\cos (-\theta )=\cos \theta$):

$$f(x)=\cos (9x)+\cos (10x)$$

Step 4: Evaluate $f\left(\frac{\pi }{2}\right)$

Substitute $x=\frac{\pi }{2}$ into the simplified function:

$$f\left(\frac{\pi }{2}\right)=\cos \left(9\cdot \frac{\pi }{2}\right)+\cos \left(10\cdot \frac{\pi }{2}\right)$$

$$f\left(\frac{\pi }{2}\right)=\cos \left(\frac{9\pi }{2}\right)+\cos (5\pi )$$

Let's evaluate each term:

• $\cos \left(\frac{9\pi }{2}\right)$: Any odd multiple of $\frac{\pi }{2}$ has a cosine value of $0$.

  $$\cos \left(\frac{9\pi }{2}\right)=0$$

• $\cos (5\pi )$: Any odd multiple of $\pi$ has a cosine value of $-1$.

  $$\cos (5\pi )=-1$$

Adding the two results together:

$$f\left(\frac{\pi }{2}\right)=0+(-1)=-1$$

Correct Answer: A) $-1$