JEE Mains · Maths · STD 12 - 6. Application of derivatives
The tangent to the curve, \(y = xe^{x^2}\) passing, through the point \((1, e)\) also passes through the point
- A \((2, 3e)\)
- B \(\left( {\frac{4}{3},2e} \right)\)
- C \(\left( {\frac{5}{3},2e} \right)\)
- D \((3, 6e)\)
Answer & Solution
Correct Answer
(B) \(\left( {\frac{4}{3},2e} \right)\)
Step-by-step Solution
Detailed explanation
\(y = x{e^{{x^2}}}\) \((1,e)\) lies on this Now \(\frac{{dy}}{{dx}} = x{e^{{x^2}}}.2x + {e^{{x^2}}}.1\) Put \(x=1\) \(m=2e+e=3e\) Equation of tangent at \((1,e)\) \(y-e=3e(x-1)\) \(y-e=3ex-3e\) \(y=3ex-2e\) \(\left( {\frac{4}{3},2e} \right)\) satisfies it \(\therefore \) Answer…
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