Question

For some parameter R > 0, the curves $y_1=x^2$ and $y_2=Rlnx$ touch each other and have no other point of intersection. Let A be the area bounded between the two curves and y-axis. Then

• A. $R=\sqrt{e}$
• B. $A=-\frac{8}{3}e\sqrt{e}$
• C. $R=2e$
• D. $A=\frac{\sqrt{e}}{6}(2e-3)+1$

Answer 1

Answer: (B), (C)

The correct options are
B $A=-\frac{8}{3}e\sqrt{e}$
C $R=2e$

The two curves have the same gradient at their point of contact.
At the common point of tangency $y_1=y_2$
$\Rightarrow x^2=Rlnx$ .....(i)
Also, $y_1^{{}^{\prime }}=y_2^{{}^{\prime }}\Rightarrow 2x=\frac{R}{x}\Rightarrow x^2=\frac{R}{2}$ ..... (ii)
Hence, $Rlnx=\frac{R}{2}\Rightarrow lnx=\frac{1}{2}\Rightarrow x=\sqrt{e}$
Hence, we get \(R = 2x^2=2e\) and point of contact i.e. $(\sqrt{e},e)$

Required area is the integral of $(y_1-y_2)$ with limits of x from 0 to $\sqrt{e}$ (touching point)
Hence \(A = \int_{0}^{\sqrt{e}}(y_1-y_2)dx=A=\int_{0}^{\sqrt{e}}(y_1-y_2)dx\)
$=\underset{h\to 0^{+}}{lim}(x^2-2elnx)dx$
$\underset{h\to 0^{+}}{lim}{[\frac{x^3}{3}-2e(xlnx-x\left)\right]}_h^{\sqrt{e}}$
$=-\frac{8}{3}e\sqrt{e}$