TS EAMCET · Maths · Application of Derivatives
If the slope of the tangent drawn to the curve \(y=e^{a+b x^2}\) at the point \(\mathrm{P}(1,1)\) is -2 , then the value of \(2 \mathrm{a}-3 \mathrm{~b}\) is
- A \(5\)
- B \(6\)
- C \(7\)
- D \(8\)
Answer & Solution
Correct Answer
(A) \(5\)
Step-by-step Solution
Detailed explanation
\begin{aligned} & \frac{d y}{d x}=\frac{d}{d x} e^{a+b x^2} \\ & \quad=2 b x \cdot e^{a+b x^2}=2 b x \cdot y \\ & \left.\frac{d y}{d x}\right|_{(1,1)}=2 b(1) \cdot(1)=-2 \\ & \Rightarrow \quad 2 b=-2 \Rightarrow b=-1 \\ & \because \quad(1,1) \text { lie on } y=e^{a+b x^2} \\ &…
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