Let $f : [0,2] \rightarrow R$ be a function which is continuous on
$[0,2]$ and is differentiable on $(0,2)$ with $f(0) = 1$.
Let $F(x)= \int\limits_0^{x^2} \, f(\sqrt t)dt, \,$ for $\, x \in \, [0,2], if F'(x) \, = f'(x) , \forall $
$x \in \, (0,2) ,\,$ then $\, F(2)$ equals
- $e^2 -1$
- $e^4 -1$
e-1
- $e^4$
The Correct Option is B
Solution and Explanation
\(F (0)=0\)
\(F ^{\prime}( x )=2 x f ( x )= f ( x )\)
\(f ( x )= e ^{ x ^{2}+ c }\)
\(f ( x )= e ^{ x ^{2}}(\because f (0)=1)\)
\(F ( x )=\int\limits_{0}^{ x ^{2}} e ^{ x } d x\)
\(F ( x )= e ^{ x ^{2}}-1(\because F (0)=0)\)
\(\Rightarrow F (2)= e ^{4}-1\)
Therefore, the correct option is (B): \(e^4 -1\)
Top Questions on Continuity
- If \[ f(x) = \begin{cases} \frac{\sin^2(ax)}{x^2}, & x \neq 0 \\ 1, & x = 0 \end{cases} \] is continuous at \( x = 0 \), then the value of \( a \) is:
Let $ f(x) = \begin{cases} (1+ax)^{1/x} & , x<0 \\1+b & , x = 0 \\\frac{(x+4)^{1/2} - 2}{(x+c)^{1/3} - 2} & , x>0 \end{cases} $ be continuous at x = 0. Then $ e^a bc $ is equal to
- JEE Main - 2025
- Mathematics
- Continuity
- Assertion (A): $f(x) = \begin{cases} x \sin \frac{1}{x}, & x \neq 0 \\ 0, & x = 0 \end{cases}$ is continuous at $x = 0$.
Reason (R): When $x \to 0$, $\sin \frac{1}{x}$ is a finite value between -1 and 1. - If $f(x) = \begin{cases} ax + b, & x \leq 3 \\ 7, & 5<x \end{cases}$ is continuous in $\mathbb{R}$, then the values of $a$ and $b$ are:
- If $f(x) = \left\{ \begin{array}{ll} \frac{1 - \sin^3 x}{3 \cos^2 x} & \text{for} \, x \neq \frac{\pi}{2}, \\ k & \text{for} \, x = \frac{\pi}{2}, \end{array} \right. $ is continuous at $x = \frac{\pi}{2}$, then the value of $k$ is:
Questions Asked in JEE Advanced exam
- Consider the vectors $$ \vec{x} = \hat{i} + 2\hat{j} + 3\hat{k},\quad \vec{y} = 2\hat{i} + 3\hat{j} + \hat{k},\quad \vec{z} = 3\hat{i} + \hat{j} + 2\hat{k}. $$ For two distinct positive real numbers $ \alpha $ and $ \beta $, define $$ \vec{X} = \alpha \vec{x} + \beta \vec{y} - \vec{z},\quad \vec{Y} = \alpha \vec{y} + \beta \vec{z} - \vec{x},\quad \vec{Z} = \alpha \vec{z} + \beta \vec{x} - \vec{y}. $$ If the vectors $ \vec{X}, \vec{Y}, \vec{Z} $ lie in a plane, then the value of $ \alpha + \beta - 3 $ is ________.
Let $ a_0, a_1, ..., a_{23} $ be real numbers such that $$ \left(1 + \frac{2}{5}x \right)^{23} = \sum_{i=0}^{23} a_i x^i $$ for every real number $ x $. Let $ a_r $ be the largest among the numbers $ a_j $ for $ 0 \leq j \leq 23 $. Then the value of $ r $ is ________.
- JEE Advanced - 2025
- binomial expansion formula
- The total number of real solutions of the equation $$ \theta = \tan^{-1}(2 \tan \theta) - \frac{1}{2} \sin^{-1} \left( \frac{6 \tan \theta}{9 + \tan^2 \theta} \right) $$ is
(Here, the inverse trigonometric functions $ \sin^{-1} x $ and $ \tan^{-1} x $ assume values in $[-\frac{\pi}{2}, \frac{\pi}{2}]$ and $(-\frac{\pi}{2}, \frac{\pi}{2})$, respectively.)- JEE Advanced - 2025
- Inverse Trigonometric Functions
A temperature difference can generate e.m.f. in some materials. Let $ S $ be the e.m.f. produced per unit temperature difference between the ends of a wire, $ \sigma $ the electrical conductivity and $ \kappa $ the thermal conductivity of the material of the wire. Taking $ M, L, T, I $ and $ K $ as dimensions of mass, length, time, current and temperature, respectively, the dimensional formula of the quantity $ Z = \frac{S^2 \sigma}{\kappa} $ is:
- JEE Advanced - 2025
- Dimensional Analysis
- Let $$ \alpha = \frac{1}{\sin 60^\circ \sin 61^\circ} + \frac{1}{\sin 62^\circ \sin 63^\circ} + \cdots + \frac{1}{\sin 118^\circ \sin 119^\circ}. $$ Then the value of $$ \left( \frac{\csc 1^\circ}{\alpha} \right)^2 $$ is \rule{1cm}{0.15mm}.
- JEE Advanced - 2025
- Some Applications of Trigonometry
Concepts Used:
Continuity
A function is said to be continuous at a point x = a, if
limx→a
f(x) Exists, and
limx→a
f(x) = f(a)
It implies that if the left hand limit (L.H.L), right hand limit (R.H.L) and the value of the function at x=a exists and these parameters are equal to each other, then the function f is said to be continuous at x=a.
If the function is undefined or does not exist, then we say that the function is discontinuous.
Conditions for continuity of a function: For any function to be continuous, it must meet the following conditions:
- The function f(x) specified at x = a, is continuous only if f(a) belongs to real number.
- The limit of the function as x approaches a, exists.
- The limit of the function as x approaches a, must be equal to the function value at x = a.