Let dx = f(k) - f(1), then one of the possible value of k is.

JEE MainMathematicsToughDefinite Integration

Let
$\frac{d}{dx}(f(x)) = \frac{e^{sinx}}{x},x > 0,\ If\ \int_{1}^{4}\mspace{2mu}\frac{2e^{sinx^{2}}}{x}$
dx = f(k) - f(1), then one of the possible value of k is.

Show answer and solution

Given,

\frac{d}{dx}(f(x)) = \frac{e^{sinx}}{x},x > 0

Now,

I = \int_{1}^{4}\mspace{2mu}\frac{2e^{sinx^{2}}}{x}dx

put z = x^2, dz =2 dx

\therefore

I = \int_{1}^{4}\mspace{2mu}\frac{2xe^{sinx^{2}}}{x^{2}}dx = \int_{1}^{16}\mspace{2mu}\frac{e^{sinz}}{z}dz

= \lbrack f(z)\rbrack_{1}^{16} = f(16) - f(1)

\therefore

f (K) = f(16)

\therefore

One possible value of k = 16

Source: Grass-G / M-13 Mathematics Paper 4 Nov.docx

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