jee-main 2026 Q10

jee-main · India · session1_21jan_shift1 Differential equations First-Order Linear DE: General Solution
If $y = y(x)$ and $\left(1 + x^{2}\right)dy + \left(1 - \tan^{-1}x\right)dx = 0$ and $y(0) = 1$ then $\mathbf{y}(1)$ is
(A) $\frac{\pi^{2}}{32} + \frac{\pi}{4} + 1$ (B) $\frac{\pi^{2}}{32} - \frac{\pi}{2} + 1$ (C) $\frac{\pi^{2}}{32} + \frac{\pi}{2} - 1$ (D) $\frac{\pi^{2}}{32} - \frac{\pi}{4} + 1$
If $y = y(x)$ and $\left(1 + x^{2}\right)dy + \left(1 - \tan^{-1}x\right)dx = 0$ and $y(0) = 1$ then $\mathbf{y}(1)$ is

(A) $\frac{\pi^{2}}{32} + \frac{\pi}{4} + 1$
(B) $\frac{\pi^{2}}{32} - \frac{\pi}{2} + 1$
(C) $\frac{\pi^{2}}{32} + \frac{\pi}{2} - 1$
(D) $\frac{\pi^{2}}{32} - \frac{\pi}{4} + 1$