Question

If $A_1$ denotes area of the region bounded by the curves $C_1:y=(x-1)e^x$, tangent to $C_1$ at $(1,0)$ and y-axis and $A_2$ denotes the area of the region bounded by $C_1$ and co-ordinate axes in fourth quadrant, then

• A. $A_1>A_2$
  
• B. $A_1<A_2$
  
• C. $2A_1+A_2=2$
  
• D. $A_1+2A_2=4$

Answer 1

Answer: (B), (C)

The correct options are
B $A_1<A_2$

C $2A_1+A_2=2$


$y=(x-1)e^x;\frac{dy}{dx}=x{e}^x$ and $y\left(0\right)=-1$



Equation of tangent is $T:y-0=e(x-1)\Rightarrow y=e(x-1)$

$A_2=\left|{\int }_0^1\right(x-1\left)e^{x}dx\right|$

$=\left|\right[(x-1)e^x-e^x{]}_0^1|=e-2$

$A_1=\frac{1}{2}\times 1\times e-A_2$
$=\frac{1}{2}\times 1\times e-(e-2)=2-\frac{e}{2}$

$\therefore 2A_1+A_2=4-e+e-2=2$ and $A_1<A_2$