Question

Let 'a' be a positive constant number. Consider two curves $C_1:y=e^x,C_2:y=e^{a-x}$. Let S be the area of the part surrounded by $C_1$,$C_2$ and the y-axis, then

• A. $\underset{a\to \infty }{lim}S=1$
• B. $\underset{a\to 0}{lim}\frac{S}{a^2}=\frac{1}{4}$
• C. Range of S is $(0,\infty )$
• D. S(a) is neither odd nor even

Answer 1

The correct option is B $\underset{a\to 0}{lim}\frac{S}{a^2}=\frac{1}{4}$

$C_1:y=e^x,C_2:y=e^{a-x}$

Point of intersection

$e^x=e^{(a-x)}$

$x=\frac{a}{2},y=e^{\frac{a}{2}}$

$S={\int }_0^{\frac{a}{2}}{C}_{2}dx-{\int }_0^{\frac{a}{2}}{C}_{1}dx$

$S={\int }_0^{\frac{a}{2}}{e}^{a-x}dx-{\int }_0^{\frac{a}{2}}{e}^{x}dx={\left[e^{a-x}\right]}_0^{\frac{a}{2}}-{\left[e^x\right]}_0^{\frac{a}{2}}$

$S=-(e^{\frac{a}{2}}-e^a)-(e^{\frac{a}{2}}-1)=(e^a{-2e}^{\frac{a}{2}}+1)={\left(\right(e^{\frac{a}{2}}-1\left)\right)}^2$

${lim}_{a\to 0}\frac{s}{a^2}=\left[\frac{{(e^{\frac{a}{2}}-1)}^2}{\frac{a}{2}}\right]\ast \frac{a^2}{4a^2}=\frac{1}{4}$