JEE 2026: Application of Derivatives
Complete step-by-step Understanding
JEE Main 2026 (January)
Q1. Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be a twice differentiable function such that $f^{\prime \prime}(x)\gt 0$ for all $x \in \mathbb{R}$ and $f^{\prime}(a-1)=0$, where $a$ is a real number. Let $g(x)=f\left(\tan ^2 x-2 \tan x+a\right), 0 \lt x \lt \frac{\pi}{2}$.Consider the following two statements:
(I) $g$ is increasing in $\left(0, \frac{\pi}{4}\right)$
(II) $g$ is decreasing in $\left(\frac{\pi}{4}, \frac{\pi}{2}\right)$. Then,
(1) Neither (I) nor (II) is True
(2) Only (I) is True
(3) Both (I) and (II) are True
(4) Only (II) is True
Solution: Let us break it into digestible pieces with graphics.The core idea: We have a composite function $g(x) = f(\tan^2 x – 2\tan x + a)$, and we need to find where $g$ increases or decreases. The key insight comes from three layers working together.
Layer 1 — Understanding $f$:** Since $f”(x) > 0$ everywhere, $f’$ is strictly increasing. And since $f'(a-1) = 0$, the function $f’$ crosses zero exactly at $x = a-1$. This means $f$ has a minimum at $x = a-1$.**Layer 2 — The inner function $h(x) = \tan^2 x – 2\tan x + a$:** Let $t = \tan x$. Then $h = t^2 – 2t + a = (t-1)^2 + (a-1)$. This is a parabola in $t$ with minimum value $a-1$ at $t = 1$.**Key observation:** Since $\tan x$ increases from $0$ to $\infty$ as $x$ goes from $0$ to $\frac{\pi}{2}$, the inner function $h(x)$ **decreases** from $0$ to $\frac{\pi}{4}$ (as $\tan x$ moves from $0$ toward $1$) and **increases** from $\frac{\pi}{4}$ to $\frac{\pi}{2}$ (as $\tan x$ moves from $1$ to $\infty$). The minimum of $h(x)$ is $a-1$, achieved at $x = \frac{\pi}{4}$.
Layer 3 — The crucial sign check for $g'(x)$:
Since $g(x) = f(h(x))$, by the chain rule:
$g'(x) = f'(h(x)) \cdot h'(x)$The minimum value of $h(x)$ is $a – 1$, so $h(x) \geq a – 1$ for all $x$. But we know $f'(a-1) = 0$ and $f’$ is increasing. Therefore:$h(x) \geq a-1 \implies f'(h(x)) \geq f'(a-1) = 0$So $f'(h(x)) \geq 0$ always. The sign of $g'(x)$ is entirely determined by $h'(x)$!Now let’s see the complete picture of $g(x)$:Verdict on the two statements:
– Statement (I): “$g$ is increasing on $\left(0, \frac{\pi}{4}\right)$” → **FALSE**. $g$ is actually *decreasing* there.
– Statement (II): “$g$ is decreasing on $\left(\frac{\pi}{4}, \frac{\pi}{2}\right)$” → **FALSE**. $g$ is actually *increasing* there.
Answer: Option (1) — Neither (I) nor (II) is True.
—The 3-step mental model to remember:
1. $f” > 0$ means $f’$ is a rising curve that hits zero at $x = a-1$, so $f'(x) \geq 0$ whenever $x \geq a-1$. 2. $h(x) = (\tan x – 1)^2 + (a-1)$ has its minimum value $a-1$ at $x = \frac{\pi}{4}$, so $h(x) \geq a-1$ always — meaning $f'(h(x)) \geq 0$ always. 3. Therefore $g'(x) = \underbrace{f'(h(x))}_{\geq\, 0} \cdot h'(x)$, and the sign of $g’$ is purely controlled by $h’$. Since $h$ decreases then increases on $\left(0, \frac{\pi}{2}\right)$, so does $g$ — the opposite of what both statements claim.
Alternately
(1) Given $f^{\prime \prime}(x)\gt 0$ for all $x \in \mathbb{R}$, which means $f^{\prime}(x)$ is a strictly increasing function.We are given $f^{\prime}(a-1)=0$. Since $f^{\prime}(x)$ is strictly increasing, $f^{\prime}(x)\lt 0$ for $x \lt a-1$ and $f^{\prime}(x)\gt 0$ for $x \gt a-1$. The function $g(x)$ is defined as $g(x)=f\left(\tan ^2 x-2 \tan x+a\right)$ for $x \in(0, \pi / 2)$. Let $u(x)=\tan ^2 x-2 \tan x+a=(\tan x-1)^2+a-1$. Differentiating $g(x)$ with respect to $x$ : $ \begin{aligned} & g^{\prime}(x)=f^{\prime}(u(x)) \cdot u^{\prime}(x)=f^{\prime}\left((\tan x-1)^2+a-1\right) \cdot\left(2 \tan x \sec ^2 x-2 \sec ^2 x\right) \\ & g^{\prime}(x)=f^{\prime}\left((\tan x-1)^2+a-1\right) \cdot 2 \sec ^2 x(\tan x-1) \end{aligned} $Case 1: $x \in(0, \pi / 4)$. In this interval, $0<\tan x \lt1$, so $(\tan x-1)\lt 0$. Also, $(\tan x-1)^2\gt 0$, so $u(x)=(\tan x-1)^2+a-1>a-1$. Since $f^{\prime}(x)$ is increasing and $f^{\prime}(a-1)=0$, for $u(x)>a-1$, we have $f^{\prime}(u(x))\gt 0$. Thus, $g^{\prime}(x)=(+)(+)(-)\lt 0$. So $g(x)$ is decreasing in $(0, \pi / 4)$. Statement (I) is False.Case 2: $x \in(\pi / 4, \pi / 2)$. In this interval, $\tan x \gt1$, so $(\tan x-1)\gt 0$. Also, $(\tan x-1)^2\gt 0$, so $u(x)=(\tan x-1)^2+a-1>a-1$. Thus, $f^{\prime}(u(x))\gt 0$. Then $g^{\prime}(x)=(+)(+)(+)\gt 0$. So $g(x)$ is increasing in $(\pi / 4, \pi / 2)$. Statement (II) is False.Therefore, neither (I) nor (II) is true.
Layer 1 — Understanding $f$:** Since $f”(x) > 0$ everywhere, $f’$ is strictly increasing. And since $f'(a-1) = 0$, the function $f’$ crosses zero exactly at $x = a-1$. This means $f$ has a minimum at $x = a-1$.**Layer 2 — The inner function $h(x) = \tan^2 x – 2\tan x + a$:** Let $t = \tan x$. Then $h = t^2 – 2t + a = (t-1)^2 + (a-1)$. This is a parabola in $t$ with minimum value $a-1$ at $t = 1$.**Key observation:** Since $\tan x$ increases from $0$ to $\infty$ as $x$ goes from $0$ to $\frac{\pi}{2}$, the inner function $h(x)$ **decreases** from $0$ to $\frac{\pi}{4}$ (as $\tan x$ moves from $0$ toward $1$) and **increases** from $\frac{\pi}{4}$ to $\frac{\pi}{2}$ (as $\tan x$ moves from $1$ to $\infty$). The minimum of $h(x)$ is $a-1$, achieved at $x = \frac{\pi}{4}$.
Layer 3 — The crucial sign check for $g'(x)$:
Since $g(x) = f(h(x))$, by the chain rule:
$g'(x) = f'(h(x)) \cdot h'(x)$The minimum value of $h(x)$ is $a – 1$, so $h(x) \geq a – 1$ for all $x$. But we know $f'(a-1) = 0$ and $f’$ is increasing. Therefore:$h(x) \geq a-1 \implies f'(h(x)) \geq f'(a-1) = 0$So $f'(h(x)) \geq 0$ always. The sign of $g'(x)$ is entirely determined by $h'(x)$!Now let’s see the complete picture of $g(x)$:Verdict on the two statements:
– Statement (I): “$g$ is increasing on $\left(0, \frac{\pi}{4}\right)$” → **FALSE**. $g$ is actually *decreasing* there.
– Statement (II): “$g$ is decreasing on $\left(\frac{\pi}{4}, \frac{\pi}{2}\right)$” → **FALSE**. $g$ is actually *increasing* there.
Answer: Option (1) — Neither (I) nor (II) is True.
—The 3-step mental model to remember:
1. $f” > 0$ means $f’$ is a rising curve that hits zero at $x = a-1$, so $f'(x) \geq 0$ whenever $x \geq a-1$. 2. $h(x) = (\tan x – 1)^2 + (a-1)$ has its minimum value $a-1$ at $x = \frac{\pi}{4}$, so $h(x) \geq a-1$ always — meaning $f'(h(x)) \geq 0$ always. 3. Therefore $g'(x) = \underbrace{f'(h(x))}_{\geq\, 0} \cdot h'(x)$, and the sign of $g’$ is purely controlled by $h’$. Since $h$ decreases then increases on $\left(0, \frac{\pi}{2}\right)$, so does $g$ — the opposite of what both statements claim.
Alternately
(1) Given $f^{\prime \prime}(x)\gt 0$ for all $x \in \mathbb{R}$, which means $f^{\prime}(x)$ is a strictly increasing function.We are given $f^{\prime}(a-1)=0$. Since $f^{\prime}(x)$ is strictly increasing, $f^{\prime}(x)\lt 0$ for $x \lt a-1$ and $f^{\prime}(x)\gt 0$ for $x \gt a-1$. The function $g(x)$ is defined as $g(x)=f\left(\tan ^2 x-2 \tan x+a\right)$ for $x \in(0, \pi / 2)$. Let $u(x)=\tan ^2 x-2 \tan x+a=(\tan x-1)^2+a-1$. Differentiating $g(x)$ with respect to $x$ : $ \begin{aligned} & g^{\prime}(x)=f^{\prime}(u(x)) \cdot u^{\prime}(x)=f^{\prime}\left((\tan x-1)^2+a-1\right) \cdot\left(2 \tan x \sec ^2 x-2 \sec ^2 x\right) \\ & g^{\prime}(x)=f^{\prime}\left((\tan x-1)^2+a-1\right) \cdot 2 \sec ^2 x(\tan x-1) \end{aligned} $Case 1: $x \in(0, \pi / 4)$. In this interval, $0<\tan x \lt1$, so $(\tan x-1)\lt 0$. Also, $(\tan x-1)^2\gt 0$, so $u(x)=(\tan x-1)^2+a-1>a-1$. Since $f^{\prime}(x)$ is increasing and $f^{\prime}(a-1)=0$, for $u(x)>a-1$, we have $f^{\prime}(u(x))\gt 0$. Thus, $g^{\prime}(x)=(+)(+)(-)\lt 0$. So $g(x)$ is decreasing in $(0, \pi / 4)$. Statement (I) is False.Case 2: $x \in(\pi / 4, \pi / 2)$. In this interval, $\tan x \gt1$, so $(\tan x-1)\gt 0$. Also, $(\tan x-1)^2\gt 0$, so $u(x)=(\tan x-1)^2+a-1>a-1$. Thus, $f^{\prime}(u(x))\gt 0$. Then $g^{\prime}(x)=(+)(+)(+)\gt 0$. So $g(x)$ is increasing in $(\pi / 4, \pi / 2)$. Statement (II) is False.Therefore, neither (I) nor (II) is true.
Q2. Let $f$ be a differentiable function satisfying $f(x)=1-2 x+\int_0^x \mathrm{e}^{(x-t)} f(t) \mathrm{dt}, x \in \mathbf{R}$ and let $\mathrm{g}(x)=\int_0^x(f(\mathrm{t})+2)^{15}(\mathrm{t}-4)^6(\mathrm{t}+12)^{17} \mathrm{dt}, x \in \mathbf{R}$. If p and q are respectively the points of local minima and local maxima of g , then the value of $|\mathrm{p}+\mathrm{q}|$ is equal to
Solution: (9) Given $f(x)=1-2 x+\int_0^x e^{(x-t)} f(t) d t$.Differentiating and simplifying:
$
e^{-x} f^{\prime}(x)-e^{-x} f(x)=-2 e^{-x}+(1-2 x) e^{-x}(-1)+e^{-x} f(x)
$This reduces to:
$
f^{\prime}(x)-2 f(x)=2 x-3
$Solving the linear ODE $\frac{d y}{d x}-2 y=2 x-3$ :
$
y \cdot e^{-2 x}=\int e^{-2 x}(2 x-3) d x
$On solving, we get $f(x)=1-x$.
Now, $g(x)=\int_0^x(3-t)^{15}(t-4)^6(t+12)^{17} d t$$
\begin{aligned}
& g^{\prime}(x)=(3-x)^{15}(x-4)^6(x+12)^{17} \\
& =-(x-3)^{15}(x-4)^6(x+12)^{17}
\end{aligned}
$Sign analysis of $g^{\prime}(x)$ :
$g^{\prime}(x)$ changes from $+\rightarrow-$ at $x=3 \rightarrow$ local maxima $(q=3)$
$g^{\prime}(x)$ changes from $-\rightarrow+$ at $x=-12 \rightarrow$ local minima $(p=-12)$
$
|p+q|=|-12+3|=9
$
Q3. Consider the following three statements for the function $f:(0, \infty) \rightarrow \mathbb{R}$ defined by $f(x)=\left|\log _{\mathrm{e}} x\right|-|x-1|$ :
(I) $f$ is differentiable at all $x\gt 0$.
(II) $f$ is increasing in $(0,1)$.
(III) $f$ is decreasing in $(1, \infty)$.Then.
(1) All (I), (II) and (III) are TRUE.
(2) Only (II) and (III) are TRUE.
(3) Only (I) is TRUE.
(4) Only (I) and (III) are TRUE.
Solution: (4) For $0 \lt x \lt 1$ : $f(x)=-\ln x+x-1$, $f^{\prime}(x)=-\frac{1}{x}+1=\frac{x-1}{x}\lt 0$ (decreasing).For $x \gt1$ : $f(x)=\ln x-x+1, f^{\prime}(x)=\frac{1}{x}-1=\frac{1-x}{x}\lt 0$ (decreasing).
At $x=1$ : LHD $=\frac{1-1}{1}=0, \mathrm{RHD}=\frac{1-1}{1}=0$. So $f$ is differentiable at $x=1$.
(I) TRUE: $f$ is differentiable for all $x\gt 0$.
(II) FALSE: $f$ is decreasing in $(0,1)$.
(III) TRUE: $f$ is decreasing in $(1, \infty)$.Only (I) and (III) are TRUE.
Q4. Let $f(x)=x^{2025}-x^{2000}, x \in[0,1]$ and the minimum value of the function $f(x)$ in the interval $[0,1]$ be $(80)^{80}(n)^{-81}$. Then $n$ is equal to
(1) -40
(2) -41
(3) -80
(4) -81
Solution: (4) $f(x)=x^{2025}-x^{2000}, f^{\prime}(x)=x^{1999}\left(2025 x^{25}-2000\right)=0$.Critical point: $x^{25}=\frac{2000}{2025}=\frac{80}{81}$, i.e., $x=\left(\frac{80}{81}\right)^{1 / 25}$.
$f(0)=f(1)=0$. At the critical point:
$f=x^{2000}\left(x^{25}-1\right)=\left(\frac{80}{81}\right)^{80}\left(\frac{80}{81}-1\right)=-\frac{80^{80}}{81^{81}}$.
This equals $(80)^{80}(n)^{-81}$, so $n^{-81}=-81^{-81} \Rightarrow n=-81$.
Q5. Let $f: \mathbf{R} \rightarrow \mathbf{R}$ be a twice differentiable function such that the quadratic equation $f(x) \mathrm{m}^2-2 f^{\prime}(x) \mathrm{m}+f^{\prime \prime}(x)=0$ in m , has two equal roots for every $x \in \mathbf{R}$. If $f(0)=1, f^{\prime}(0)=2$, and $(\alpha, \beta)$ is the largest interval in which the function $f\left(\log _{\mathrm{e}} x-x\right)$ is increasing, then $\alpha+\beta$ is equal to
Solution: (1) For equal roots in $f(x) m^2-2 f^{\prime}(x) m+f^{\prime \prime}(x)=0$, discriminant $=0$ :
$\left[f^{\prime}(x)\right]^2=f(x) f^{\prime \prime}(x)$
This gives $\frac{f^{\prime \prime}(x)}{f^{\prime}(x)}=\frac{f^{\prime}(x)}{f(x)}$
Integrating: $f^{\prime}(x)=k f(x)$, so $f(x)=A e^{k x}$
Using $f(0)=1$ and $f^{\prime}(0)=2$ : $A=1, k=2$
$f(x)=e^{2 x}$
For $g(x)=f(\ln x-x)$ :
$g^{\prime}(x)=2 e^{2(\ln x-x)} \cdot\left(\frac{1}{x}-1\right)$
$g^{\prime}(x)\gt 0$ when $\frac{1-x}{x}\gt 0$, i.e., $0 \lt x \lt1$
$(\alpha, \beta)=(0,1)$
$\alpha+\beta=1$
Q6. Let $(2 \alpha, \alpha)$ be the largest interval in which the function $f(t)=\frac{|t+1|}{t^2}, t\lt 0$, is strictly decreasing. Then the local maximum value of the function $g(x)=2 \log _{\mathrm{e}}(x-2)+\alpha x^2+4 x-\alpha, x \gt2$, is $\_\_\_\_$
Solution: (4) For $t\lt 0$, analyze $f(t)=\frac{|t+1|}{t^2}$ in two regions.For $-1 \lt t \lt 0$ :
$f(t)=\frac{t+1}{t^2}$, so $f^{\prime}(t)=\frac{-(t+2)}{t^3}\gt 0$ (increasing).
For $t \lt -1$ :
$f(t)=\frac{-t-1}{t^2}$, so $f^{\prime}(t)=\frac{t+2}{t^3}$.For $-2 \lt t \lt -1$ :
$f^{\prime}(t)\lt 0$ (decreasing).
For $t \lt -2$ :
$f^{\prime}(t)\gt 0$ (increasing).
Largest decreasing interval is $(-2,-1)=(2 \alpha, \alpha)$, so $\alpha=-1$.
For $g(x)=2 \ln (x-2)-x^2+4 x+1$ :
$g^{\prime}(x)=\frac{2}{x-2}-2 x+4=0$ gives $(x-2)^2=1$, so $x=3$.
Since $g^{\prime \prime}(x)=-\frac{2}{(x-2)^2}-2\lt 0, x=3$ is maximum.
$g(3)=2 \ln (1)-9+12+1=4$
