MHT CET · Maths · Application of Derivatives
The equation of the tangent to curve \(y=4 \mathrm{xe}^{\mathrm{x}}\) at \(\left(-1, \frac{-4}{\mathrm{e}}\right)\) is
- A \(6 x-\frac{e}{4} y=-5\)
- B \(\mathrm{x}-\frac{\mathrm{e}}{4} \mathrm{y}=0\)
- C \(x=-1\)
- D \(y=\frac{-4}{e}\)
Answer & Solution
Correct Answer
(D) \(y=\frac{-4}{e}\)
Step-by-step Solution
Detailed explanation
\(y=4 \mathrm{x}^{\mathrm{x}} \)
\( \therefore\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)_{\left(-1, \frac{-4}{\mathrm{e}}\right)}=4(-1) \mathrm{e}^{-1}+4 \mathrm{e}^{-1}=\frac{-4}{\mathrm{e}}\) \(+\frac{4}{\mathrm{e}}=0\)
Thus tangent is parallel to \(\mathrm{X}\) axis.
Hence required equation of tangent is \(y=\frac{-4}{e}\)
\( \therefore\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)_{\left(-1, \frac{-4}{\mathrm{e}}\right)}=4(-1) \mathrm{e}^{-1}+4 \mathrm{e}^{-1}=\frac{-4}{\mathrm{e}}\) \(+\frac{4}{\mathrm{e}}=0\)
Thus tangent is parallel to \(\mathrm{X}\) axis.
Hence required equation of tangent is \(y=\frac{-4}{e}\)
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