α=x→∞lim((1−ee)(e1−1+xx))x(1∞ form )∴α=eL Where L=x→∞limx((1−ee)(e1−1+xx)−1)⇒L=x→∞lim(1−ee)x(e1−1+xx−(e1−e))⇒L=1−eex→∞limx(1−1+xx)⇒L=1−eex→∞limx+1x⇒ L=1−ee⋅1⇒ L=1−ee∴α=e1−ee⇒logα=1−ee∴ Required value =1+1−ee1−ee=e
JEE Main 2025 — Mathematics Calculus
If x→∞lim((1−ee)(e1−1+xx))x=α, then the value of 1+logeαlogeα equals :
Held on 22 Jan 2025 · Verified 6 Jul 2026.
e−1
e2
e−2
e
Sign in to track your attempts and accuracy.
Sign in to keep a private note on this question. Nothing you write is ever public.
Let $[\cdot]$ denote the greatest integer function. Then the value of $\displaystyle\int_0^3 \left(\dfrac{e^x + e^{-x}}{[x]!}\right) dx$ is :
The value of $\sum_{r=1}^{20}\left(\left|\sqrt{\pi\left(\int_{0}^{r} x|\sin \pi x| d x\right)}\right|\right)$ is $\_\_\_\_$
The value of ∫₀¹ x·eˣ dx is:
If the area of the region bounded by $16x^2 - 9y^2 = 144$ and $8x - 3y = 24$ is A, then $3(A + 6 \log_e(3))$ is equal to _______.
Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be a differentiable function such that $f\left(\dfrac{x+y}{3}\right) = \dfrac{f(x) + f(y)}{3}$ for all $x, y \in \mathbb{R}$, and $f'(0) = 3$. Then the minimum value of the function $g(x) = 3 + e^x f(x)$, is:
Work through every JEE Main Calculus PYQ, year by year.