NIMCET 2024 — Mathematics PYQ

Step 1: Analyze the first piece for $x\le 2$

Let $f_1(x)=-x^3+3x^2+1$ defined on $(-\infty ,2]$.

To find the critical points, find its derivative with respect to $x$:

$$f_1^{\prime }(x)=-3x^2+6x$$

Set $f_1^{\prime }(x)=0$:

$$-3x(x-2)=0⟹x=0\text{ or }x=2$$

Now, apply the second derivative test to classify these critical points:

$$f_1^{\prime \prime }(x)=-6x+6$$

• At $x=0$: f_1''(0) = 6 > 0 \implies x = 0 is a local minimum.

  The value is $f(0)=-(0{)}^3+3(0{)}^2+1=1$.

• At $x=1$: Since $f_1^{\prime }(1)=3\ne 0$, $x=1$ is not a critical point, making Statement A false.

• At $x=2$: f_1''(2) = -6(2) + 6 = -6 < 0 \implies x = 2 is a local maximum for this polynomial segment.

  The value is $f(2)=-(2{)}^3+3(2{)}^2+1=-8+12+1=5$.

As $x\to -\infty$, $f_1(x)\to +\infty$. Since the function goes up to infinity on the left, there is no global maximum for this function.

Step 2: Check continuity and behavior at the boundary $x=2$

Let us check if the function is continuous at the junction point $x=2$:

• Left-hand limit / value at $x=2$:

  $$\underset{x\to 2^{-}}{\lim }f(x)=f(2)=5$$

• Right-hand limit near $x=2$:

  $$\underset{x\to 2^{+}}{\lim }f(x)=\cos (2)$$

Since $\cos (2)$ lies between $-1$ and $1$, the right-hand limit ($\approx -0.416$) is strictly less than the left-hand value ($5$).

Since $f(2)=5$ is strictly greater than all values immediately to its left (because it is a local maximum for the first segment) and strictly greater than all values immediately to its right (\cos(x) \le 1 < 5), $x=2$ is a local maximum for the overall function $f(x)$.

Step 3: Analyze the remaining options

• Checking $x=\pi$ (Statement C): For 2 < x \le 4, $f(x)=\cos (x)$. At $x=\pi$, $\cos (\pi )=-1$, which is the absolute minimum value for a cosine function. Therefore, $x=\pi$ is a local minimum, not a maximum. Statement C is false.

• Checking Global Maximum (Statement B & D): As established in Step 1, ${\lim }_{x\to -\infty }f(x)=+\infty$, which means the function values grow infinitely large. Thus, a global maximum does not exist, confirming that the local maximum at $x=2$ is not a global maximum.

Conclusion

The function $f(x)$ has a local maximum at $x=2$, which is not the global maximum.

Therefore, the correct option is B.