Q34 / 75 · JEE Main 2026 (22 January Shift 2) · Mathematics
If $y=y(x)$ satisfies the differential equation $16(\sqrt{x+9 \sqrt{x}})(4+\sqrt{9+\sqrt{x}}) \cos y \mathrm{~d} y=(1+2 \sin y) \mathrm{d} x, x>0$ and $y(256)=\frac{\pi}{2}, y(49)=\alpha$, then $2 \sin \alpha$ is equal to :
- A$2(\sqrt{2}-1)$
- B$3(\sqrt{2}-1)$
- Correct answerC$2 \sqrt{2}-1$
- D$\sqrt{2}-1$
Solution
$\int \frac{\cos y}{1+2 \sin y} d y=\int \frac{d x}{16(\sqrt{9 \sqrt{x}+x})(4+\sqrt{9+\sqrt{x}})}$
$4+\sqrt{9+\sqrt{x}}=t$
$\frac{1}{2 \sqrt{9+\sqrt{x}}} \times \frac{d x}{2 \sqrt{x}}=1 d x$
$\frac{1}{2} \ell n|1+2 \sin y|=\int \frac{4 d t}{16 t}+C$
$\left.\frac{1}{2} \ell n|1+2 \sin y|=\frac{1}{4} \ell n \right\rvert\, 4+\sqrt{9+\sqrt{x}}+C$
$\frac{1}{2} \ell n(2 \sin y+1)=\frac{1}{4} \ell n|4+\sqrt{9+\sqrt{x}}|+C$
Substituting $\left(256, \frac{\pi}{2}\right)$
$\frac{1}{2} \ell n 3=\frac{1}{2} \ell n 3+C \quad C=0$
Substituting $(49, \alpha)$
$\frac{1}{2} \ell \mathrm{n}(2 \sin \alpha+1)=\frac{1}{4} \ell \mathrm{n} 8$
$\ell \mathrm{n}(2 \sin \alpha+1)=\frac{1}{2} \ell \mathrm{n} 8$
$\ell \mathrm{n}(2 \sin \alpha+1)=\ell \mathrm{n} 2 \sqrt{2}$
$2 \sin \alpha+1=2 \sqrt{2}$
$2 \sin \alpha=2 \sqrt{2}-1$
$4+\sqrt{9+\sqrt{x}}=t$
$\frac{1}{2 \sqrt{9+\sqrt{x}}} \times \frac{d x}{2 \sqrt{x}}=1 d x$
$\frac{1}{2} \ell n|1+2 \sin y|=\int \frac{4 d t}{16 t}+C$
$\left.\frac{1}{2} \ell n|1+2 \sin y|=\frac{1}{4} \ell n \right\rvert\, 4+\sqrt{9+\sqrt{x}}+C$
$\frac{1}{2} \ell n(2 \sin y+1)=\frac{1}{4} \ell n|4+\sqrt{9+\sqrt{x}}|+C$
Substituting $\left(256, \frac{\pi}{2}\right)$
$\frac{1}{2} \ell n 3=\frac{1}{2} \ell n 3+C \quad C=0$
Substituting $(49, \alpha)$
$\frac{1}{2} \ell \mathrm{n}(2 \sin \alpha+1)=\frac{1}{4} \ell \mathrm{n} 8$
$\ell \mathrm{n}(2 \sin \alpha+1)=\frac{1}{2} \ell \mathrm{n} 8$
$\ell \mathrm{n}(2 \sin \alpha+1)=\ell \mathrm{n} 2 \sqrt{2}$
$2 \sin \alpha+1=2 \sqrt{2}$
$2 \sin \alpha=2 \sqrt{2}-1$
Hence, the correct answer is option: 'C'.