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)

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