JEE MAIN - Mathematics (2024 - 9th April Evening Shift - No. 4)
Let $$\int_\limits0^x \sqrt{1-\left(y^{\prime}(t)\right)^2} d t=\int_0^x y(t) d t, 0 \leq x \leq 3, y \geq 0, y(0)=0$$. Then at $$x=2, y^{\prime \prime}+y+1$$ is equal to
$$\sqrt2$$
2
1/2
1
Explanation
$$\int_\limits0^x \sqrt{1-\left(y^{\prime}(t)\right)^2} d t=\int_\limits0^x y(t) d t$$
Differentiating both side
$$\begin{aligned} & \sqrt{1-\left(y^{\prime}(x)\right)^2}=y(x) \\ & \left(\frac{d y}{d x}\right)^2+y^2=1 \\ & y^{\prime 2}+y^2=1 \\ & 2 y^{\prime} y^{\prime \prime}+2 y y^{\prime}=0 \\ & y^{\prime \prime}+y=0 \end{aligned}$$
$$\therefore$$ _9th_April_Evening_Shift_en_4_1.png)
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