The value of the integral Ι = ∫ e^ x (sin x + cos x)dx is equal to e^ x ⋅ f(x) + C,C being the constant of integration. Then the maximum value of y = f(x^ 2 ), x ∈ ℝ is equal to&

Correct answer: C. 1 As we know, ∫ e^ x (f(x) + f^ ' (x))dx= e^ x ⋅ f (x) + C Thus, ∫ e^ x (sin x + cos x)dx = e^ x ⋅ sin x + C i.e. f(x) = sin x Hence, f(x^ 2 ) = (x^ 2 ): which has the maximum value of ‘1’

NTA Abhyas 2020

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

NTA Abhyas 2020 · Mathematics · Indefinite Integration

The value of the integral Ι = ∫ e^ x (sin x + cos x)dx is equal to...

The value of the integral

Ι = ∫ e^{x}

\left(sin x + cos x\right)dx

is equal to

e^{x} ⋅ f\left(x\right) + C,C

being the constant of integration. Then the maximum value of

y = f\left(x^{2}\right), \forall x \in \mathbb{R} is equal to&

Options

A $0$
B $- 1$
C $1$
D $\dfrac{1}{2}$

Correct answer

C. $1$

Step-by-step solution

As we know,

$∫ e^{x}\left(f\left(x\right) + f^{'}$ \left(x\right)\right)dx= e^{x} ⋅ f $\left(x\right) + C$
Thus, $∫ e^{x}\left(sin x + cos x\right)dx = e^{x} ⋅ sin x + C$
i.e. $f\left(x\right) = sin x$
Hence, $f\left(x^{2}\right) = \sin \left(x^{2}\right):$ which has the maximum value of $‘1’$

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