The integral I = ∫ e ^ x ( e ^ x ) x dx = f(x) + c (where, c is the constant of integration) and f ( ((π)/(4))) ^ 2 = 2 . Then, the number of solutions of f(x) = 2e( x ∈ R - \ 0\ ) is equal to

Correct answer: 0 Given integral is I = ∫ e ^ x ( e ^ x ) x dx Let e ^ x = t ⇒ e ^ x 2 x dx = dt ∴ I = 2∫ cos tdt = 2sin t + c = 2 ( e ^ x ) + c ∴ f(x) = 2 ( e ^ x ) which has range = [- 2,2] Hence, f(x) = 2e has no solution

NTA Abhyas 2020

Exam: NTA Abhyas
Year: 2020
Subject: Mathematics
Chapter: Indefinite Integration

NTA Abhyas 2020 · Mathematics · Indefinite Integration

The integral I = ∫ e ^ x ( e ^ x ) x dx = f(x) + c (where, c is the...

The integral

I = ∫ \dfrac{{e}^{\sqrt{x}}\cos \left({e}^{\sqrt{x}}\right)}{\sqrt{x}}dx

= f\left(x\right) + c

(where,

c

is the constant of integration)

and f{\left(\ln \left(\dfrac{\pi}{4}\right)\right)}^{2} = \sqrt{2} .

Then, the number of solutions of

f\left(x\right) = 2e\left(\forall x \in R - \left\{0\right\}\right)

is equal to

Correct answer

Step-by-step solution

Given integral is

I = ∫ \dfrac{{e}^{\sqrt{x}}\cos \left({e}^{\sqrt{x}}\right)}{\sqrt{x}}dx

Let

{e}^{\sqrt{x}} = t

⇒ \dfrac{{e}^{\sqrt{x}}}{2\sqrt{x}}dx = dt

∴ I = 2∫ cos tdt

= 2sin t + c

= 2\sin \left({e}^{\sqrt{x}}\right) + c

∴ f\left(x\right) = 2\sin \left({e}^{\sqrt{x}}\right)

which has range

= \left[- 2,2\right]

Hence,

f\left(x\right) = 2e

has no solution

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