Application of derivatives — JEE Main practice

1. Why: Complete the square inside $f$ so the argument is centred at $a-1$, matching the given condition $f'(a-1)=0$.

$$g(x)=f\big((\tan x-1)^2+(a-1)\big)$$

2. Why: Differentiate using the chain rule.

$$g'(x)=f'\big((\tan x-1)^2+a-1\big)\cdot 2(\tan x-1)\sec^2x$$

3. Why: Since $f''(x)>0$ everywhere, $f'$ is a strictly increasing function, so we can compare $f'$ at different points just by comparing their arguments.

4. Why: On $0<x<\frac{\pi}{4}$, we have $0<\tan x<1$, so $(\tan x-1)^2\in(0,1)$ and the argument of $f'$ exceeds $a-1$.

$$f'\big((\tan x-1)^2+a-1\big)>f'(a-1)=0$$

Also $(\tan x-1)<0$ here, so

$$g'(x)=(+)(-)(+)<0$$

So $g$ is decreasing on $\left(0,\frac{\pi}{4}\right)$ — statement (I) is False.

5. Why: A similar argument on $\left(\frac{\pi}{4},\frac{\pi}{2}\right)$, where $\tan x>1$, flips the sign of $(\tan x-1)$ to positive while $f'(\cdot)>0$ still holds, giving $g'(x)>0$. So $g$ is increasing there — statement (II) is False.

Hence, neither (I) nor (II) is true — the answer is Option B.

1. Why: Differentiate $f(x)=x^{2025}-x^{2000}$ to locate critical points in $[0,1]$.

$$f'(x)=2025x^{2024}-2000x^{1999}=x^{1999}(2025x^5-2000)$$

2. Why: Set $f'(x)=0$ for $x$ in the interior of $[0,1]$ to find the candidate minimizer.

$$x^5=\frac{2000}{2025}=\frac{80}{81}$$

3. Why: Express $f(x)$ in terms of this same power so the value at the critical point can be substituted directly.

$$f(x)=\left(x^5\right)^{81}-\left(x^5\right)^{80}=\left(\frac{80}{81}\right)^{81}-\left(\frac{80}{81}\right)^{80}$$

4. Why: Factor out the common term $\left(\frac{80}{81}\right)^{80}$ to simplify.

$$f(x)=\left(\frac{80}{81}\right)^{80}\left(\frac{80}{81}-1\right)=\left(\frac{80}{81}\right)^{80}\left(-\frac{1}{81}\right)$$

5. Why: Rewrite this in the form $(80)^{80}(n)^{-81}$ given in the problem and match coefficients.

$$f(x)_{\min}=(80)^{80}(-81)^{-81}$$

Comparing gives $n=-81$.

Hence, the answer is Option D.

1. Why: Simplify each trig term using periodicity and co-function identities so the whole expression is written purely in $\sin\theta,\cos\theta$.

$$f(\theta)=4\left(\cos^4\theta+\sin^4\theta\right)-2\left(\cos^6\theta+\sin^6\theta\right)$$

2. Why: Use the identities $\sin^4\theta+\cos^4\theta=1-2\sin^2\theta\cos^2\theta$ and $\sin^6\theta+\cos^6\theta=1-3\sin^2\theta\cos^2\theta$.

$$f(\theta)=4(1-2\sin^2\theta\cos^2\theta)-2(1-3\sin^2\theta\cos^2\theta)=2-2\sin^2\theta\cos^2\theta$$

3. Why: Rewrite $\sin^2\theta\cos^2\theta$ in terms of $\sin2\theta$ to get a single simple sinusoid.

$$f(\theta)=2-\frac{\sin^2 2\theta}{2}$$

4. Why: Since $0\le\sin^22\theta\le1$, find the maximum and minimum of $f(\theta)$.

$$\alpha=f_{\max}=2,\qquad \beta=f_{\min}=2-\frac{1}{2}=\frac{3}{2}$$

5. Why: Compute the required combination.

$$\alpha+2\beta=2+3=5$$

Hence, the answer is Option C.