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): \(e^4 -1\)
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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.