JEE Mains · Maths · STD 12 - 5. continuity and differentiation
Let \(\int_0^x \sqrt{1-\left(y^{\prime}(t)\right)^2} d t=\int_0^x y(t) d t, 0 \leq x \leq 3, y \geq 0\), \(\mathrm{y}(0)=0\). Then at \(\mathrm{x}=2, \mathrm{y}^{\prime \prime}+\mathrm{y}+1\) is equal to :
- A \(1\)
- B \(2\)
- C \(\sqrt{2}\)
- D \(1 / 2\)
Answer & Solution
Correct Answer
(A) \(1\)
Step-by-step Solution
Detailed explanation
\(\sqrt{1-\left(y^{\prime}(x)\right)^2}=y(x) \) \( 1-\left(\frac{d y}{d x}\right)^2=y^2 \) \( \left(\frac{d y}{d x}\right)^2=1-y^2\) \( \frac{d y}{\sqrt{1-y^2}}=d x \text { OR } \frac{d y}{\sqrt{1-y^2}}=-d x \) \( \Rightarrow \sin ^{-1} y=x+c, \sin ^{-1} y=-x+c \)…
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