Differentiation and its Application: If the tangent to the curve, y=f(x)=xlog_ex , ( x>0 ) at a point… | JEE Main 2020

JEE Main 2020 (Online) 6th September Evening ShiftQuestion 1

If the tangent to the curve, $y=f(x)=x\log_ex$, ($x>0$) at a point $(c,f(c))$ is parallel to the line-segment joining the points $(1,0)$ and $(e,e)$, then $c$ is equal to :

Differentiation and its Application

$e^{\left(\frac1{e-1}\right)}$✓ Correct answer$e^{\left(\frac1{1-e}\right)}$$\frac{e-1}e$$\frac1{e-1}$

Solution

Concept. Set $f'(c)$ equal to the slope of the segment. Formulas used. (1) $\dfrac{d}{dx}(x\ln x)=\ln x+1$Step 1 - Slope of the segment. $\frac{e-0}{e-1}=\frac{e}{e-1}$. Step 2 - Solve. By (1), $\ln c+1=\frac{e}{e-1}$, so $\ln c=\frac{1}{e-1}$ and $c=e^{1/(e-1)}$. Answer: (A)

Differentiation and its Application - JEE Mathematics question (JEE Main 2020 (Online) 6th September Evening Shift)

More from Differentiation and its Application