Q236 / 270 · NDA 2025 (Phase 2) · Mathematics
For the following two (02) items:
Let $u = \int e^x \cos x\, dx$ and $v = \int e^x \sin x\, dx$.
Consider the following:
I. $\dfrac{du}{dx} = -v$
II. $\dfrac{dv}{dx} = -u$
Which of the above is/are correct?
Let $u = \int e^x \cos x\, dx$ and $v = \int e^x \sin x\, dx$.
Consider the following:
I. $\dfrac{du}{dx} = -v$
II. $\dfrac{dv}{dx} = -u$
Which of the above is/are correct?
- AI only
- BII only
- CBoth I and II
- Correct answerDNeither I nor II
Solution
Given $u = \int e^x \cos x\, dx$ and $v = \int e^x \sin x\, dx$.
Differentiating both sides with respect to $x$, we get:
$\dfrac{du}{dx} = e^x \cos x$
$\dfrac{dv}{dx} = e^x \sin x$
Evaluating the integrals, we have:
$u = \dfrac{e^x}{2}(\cos x + \sin x) + C_1$
$v = \dfrac{e^x}{2}(\sin x - \cos x) + C_2$
Comparing the derivatives with $u$ and $v$, it is clear that:
$\dfrac{du}{dx} \neq -v$
$\dfrac{dv}{dx} \neq -u$
Therefore, neither statement I nor statement II is correct.
Differentiating both sides with respect to $x$, we get:
$\dfrac{du}{dx} = e^x \cos x$
$\dfrac{dv}{dx} = e^x \sin x$
Evaluating the integrals, we have:
$u = \dfrac{e^x}{2}(\cos x + \sin x) + C_1$
$v = \dfrac{e^x}{2}(\sin x - \cos x) + C_2$
Comparing the derivatives with $u$ and $v$, it is clear that:
$\dfrac{du}{dx} \neq -v$
$\dfrac{dv}{dx} \neq -u$
Therefore, neither statement I nor statement II is correct.
Hence, the correct answer is option: 'D'.