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JEE Mains · Maths · STD 12 - 5. continuity and differentiation
If \(x = \int\limits_0^y {\frac{{dt}}{{\sqrt {1 + {t^2}} }}} \), then \(\frac{{{d^2}y}}{{d{x^2}}}\) is equal to
- A \(y\)
- B \(\sqrt {1 + {y^2}} \)
- C \(\frac{x}{{\sqrt {1 + {y^2}} }}\)
- D \(y^2\)
Answer & Solution
Correct Answer
(A) \(y\)
Step-by-step Solution
Detailed explanation
\(x = \int\limits_0^y {\frac{{dt}}{{\sqrt {1 + {t^2}} }}} \) \( \Rightarrow 1 = \frac{1}{{\sqrt {1 + {y^2}} }}.\frac{{dy}}{{dx}}\) [\(\because \) If \(1\left( x \right) = \int\limits_{\phi \left( x \right)}^{\psi \left( x \right)} {f\left( t \right)dt} ,\) then…
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