1.

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 \( F(x)=\int_{0}^{x^{2}} f(\sqrt{t}) d t \) for \( x \in[0,2] \). If \( F^{\prime}(x)=f^{\prime}(x) \) for all \( x \in(0,2) \), then \( F(2) \) equals(2014)

Answer»

F(x) = 0→x²∫​​​​​f(√t)dt

F'(x) = f(x)\(d \over dx\)(x²) - f(0)\(d \over dx\)(0)

F'(x) = f(x).2x

f'(x) = f(x).2x          ....[ F'(x) =f'(x) ]

\(f'(x) \over f(x)\) = 2x

\(∫{ f'(x) \over f(x)}\)\(dx\) = \(∫ 2x dx\)

log(f(x)) = x² + C

f(x) = (\(e ^x\))².\(e^c\)

1 = \(e^0\).Q  → Q = 1

f(x) = \((e^x)^2\)

Now,

F(x) = 0 → x² \(∫e^t dt\)

F(x) = \((e^x)²\) - \(e^0\) = \((e^x)² - 1\)

F(2) = \(e^4 -1\)


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