Question

The curvilinear trapezoid is bounded by the curve $y=x^2+1$ and the straight lines $x=1$ and $x=2$. The co-ordinates of the point ( on the given curve) with abscissa $x \in[1,2]$ where tangent drawn cut off from the curvilinear trapezoid an ordinary trapezium of the greatest area, is

• A. $(1,2)$
• B. $(2,5)$
• C. $\left(\frac{3}{2}, \frac{13}{4}\right)$
• D. none

Answer 1

Answer: (C)

$\left.\frac{ dy }{ dx }\right]_{ x _1 y _1}=2 x _1$
$T : y - y _1=2 x _1\left( x - x _1\right)$
For co-ordinates of the point $P$ put $x=1$ in (1)
$y =y_1+2 x_1\left(1-x_1\right) $
$ =1+x_1^2+2 x_1+2 x_1^2=1+2 x_1-x_1^2$
Hence $P\left(1,1+2 x_1-x_1^2\right)$
For $Q$ put $x =2$ in (1)
$y = y _1+2 x _1\left(2- x _1\right)=1+ x _1^2+4 x _1-2 x _1^2=1+4 x _1- x _1^2$
hence $Q=\left(2,1+4 x_1-x_1^2\right)$
$\therefore \quad A =\left[\left(1+2 x _1- x _1^2\right)+\left(1+4 x _1- x _1^2\right)\right]=1+3 x _1- x _1^2$
$\frac{ dA }{ dx _1}=3-2 x _1 \Rightarrow x _1=3 / 2$