Question

Assertion :

Consider the two curves $y=x-\lambda x^2$ and $y=\frac{x^2}{\lambda },(\lambda >0)$

The area bounded between the curves is maximum when $\lambda =1$ Reason: The area bounded between the curves is $\frac{{\lambda }^2}{{(1+{\lambda }^2)}^2}$ square units

• A. Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
• B. Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
• C. Assertion is correct but Reason is incorrect
• D. Both Assertion and Reason are incorrect

Answer 1

The correct option is C Assertion is correct but Reason is incorrect

We obtain point of intersection of two curves by equating them
$\therefore x-\lambda x^2=\frac{x^2}{\lambda }$
$⟹\lambda x-{\lambda }^2{x}^2=x^2$
$⟹x^2(1+{\lambda }^2)-\lambda x=0$
$⟹x\left[x\right(1+{\lambda }^2)-\lambda ]=0$
$⟹x=0$ or $x=\frac{\lambda }{1+{\lambda }^2}$
Let $\frac{\lambda }{1+{\lambda }^2}=k$
Hence, the required area will be
${\int }_0^k\frac{x^2}{\lambda }-x+\left(\lambda \right)x^{2}dx$
$=\frac{1}{\lambda }{\int }_0^k{x}^2(1+{\lambda }^2)-\lambda xdx$
$=\frac{1}{\lambda }{\left[\frac{x^3}{3}\right(1+{\lambda }^2)-\lambda \frac{x^2}{2}]}_0^k$

$=\frac{1}{\lambda }\left[\frac{k^3}{3}\right(1+{\lambda }^2)-\lambda \frac{k^2}{2}]$
$=\frac{1}{\lambda }[\frac{{\lambda }^3}{3(1+{\lambda }^2{)}^2}-\frac{{\lambda }^3}{2(1+{\lambda }^2{)}^2}]$
$=-\frac{{\lambda }^2}{6(1+{\lambda }^2{)}^2}$
Hence, area $=\frac{1}{6}{\left(\frac{\lambda }{1+{\lambda }^2}\right)}^2$
Hence, area is maximum at $\lambda =1$.