Q96 / 150 · MHT CET 2026 (17 April Shift 2) · Mathematics
The function $f(x) = \int_0^x\dfrac{dt}{1 + \cos t}$ satisfies which of the following differential equations?
- Correct answerA$2\dfrac{df}{dx} = 1 + [f(x)]^2$
- B$\dfrac{df}{dx} = 1 + [f(x)]^2$
- C$2\dfrac{df}{dx} = 1 - [f(x)]^2$
- D$\dfrac{df}{dx} = 1 - [f(x)]^2$
Solution
Given $f(x) = \int_0^x \dfrac{dt}{1 + \cos t}$
Using the half-angle formula $1 + \cos t = 2 \cos^2(t/2)$, we get:
$f(x) = \int_0^x \dfrac{dt}{2 \cos^2(t/2)} = \dfrac{1}{2} \int_0^x \sec^2(t/2) dt$
Integrating, we obtain:
$f(x) = \left[ \tan(t/2) \right]_0^x = \tan(x/2)$
Differentiating $f(x)$ with respect to $x$:
$\dfrac{df}{dx} = \dfrac{1}{2} \sec^2(x/2)$
Using the identity $\sec^2(x/2) = 1 + \tan^2(x/2)$:
$\dfrac{df}{dx} = \dfrac{1}{2} \left( 1 + \tan^2(x/2) \right)$
Substituting $f(x) = \tan(x/2)$:
$\dfrac{df}{dx} = \dfrac{1}{2} \left( 1 + [f(x)]^2 \right)$
$2\dfrac{df}{dx} = 1 + [f(x)]^2$
Using the half-angle formula $1 + \cos t = 2 \cos^2(t/2)$, we get:
$f(x) = \int_0^x \dfrac{dt}{2 \cos^2(t/2)} = \dfrac{1}{2} \int_0^x \sec^2(t/2) dt$
Integrating, we obtain:
$f(x) = \left[ \tan(t/2) \right]_0^x = \tan(x/2)$
Differentiating $f(x)$ with respect to $x$:
$\dfrac{df}{dx} = \dfrac{1}{2} \sec^2(x/2)$
Using the identity $\sec^2(x/2) = 1 + \tan^2(x/2)$:
$\dfrac{df}{dx} = \dfrac{1}{2} \left( 1 + \tan^2(x/2) \right)$
Substituting $f(x) = \tan(x/2)$:
$\dfrac{df}{dx} = \dfrac{1}{2} \left( 1 + [f(x)]^2 \right)$
$2\dfrac{df}{dx} = 1 + [f(x)]^2$
Hence, the correct answer is option: 'A'.