Applying L'Hospital's Rule, for numerator we have to apply Newton Leibnitz integral rule
x→1lim((x−1)cos(x−1)+sin(x−1)2(x−1)×(x−1)2cos(x−1)4−0)(00)
Divide (x−1)
=x→1limcos(x−1)+(x−1)sin(x−1)2(x−1)2cos(x−1)4
=1+10
=0
JEE Main 2020 — Mathematics Calculus
x→1lim((x−1)sin(x−1)∫0(x−1)2tcost2dt)
Held on 6 Sept 2020 · Verified 6 Jul 2026.
is equal to 21.
is equal to 1.
is equal to −21.
is equal to 0.
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.