Question

Assertion :The area bounded by the curves $y=x^2+2x-3$ and the line $y=\lambda x+1$ is least, if $\lambda =2$ Reason: The area bounded by the curve $y=x^2+2x-3$ and $y=\lambda x+1$ is $=\frac{1}{6}{\{{(\lambda -2)}^2+16\}}^{\frac{3}{2}}$.sq.unit.

• 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. Assertion is false and Reason are correct

Answer 1

The correct option is A Both Assertion and Reason are correct and Reason is the correct explanation for Assertion

The given curves are
$y=x^2+2x-3$ ......... $\left(1\right)$
and $y=\lambda x+1$ ......... $\left(2\right)$
Solving eqns$\left(1\right)$ and $\left(2\right)$ we get
$x^2+(2-\lambda )x-4=0$
$\alpha ,\beta$ are the roots of the quadratic , then
$\alpha +\beta =\lambda -2,\alpha \beta =-4$
Hence, Required area
$S\left(\lambda \right)=\left|{\int }_{\alpha }^{\beta }(\lambda x+1)-(x^2+2x-3)dx\right|$
$=\left|{[4x+(\lambda -2)\frac{x^2}{2}-\frac{x^3}{3}]}_{\alpha }^{\beta }\right|$
$=|4(\beta -\alpha )+\frac{(\lambda -2)}{2}({\beta }^2-{\alpha }^2)-\frac{1}{3}({\beta }^3-{\alpha }^3)|$
$=\sqrt{{(\beta +\alpha )}^2-4\beta \alpha }\left|\{4+\frac{(\lambda -2)}{2}(\beta +\alpha )-\frac{1}{3}\{{(\alpha +\beta )}^2-\alpha \beta \}\}\right|$
$=\frac{1}{6}{\{{(\lambda -2)}^2+16\}}^{\frac{3}{2}}$
For least value of $S\left(\lambda \right),\lambda -2=0$
$\therefore \lambda =2$

Hence, the assertion and reason are correct and Reason is the correct explanation of assertion.