The curvilinear trapezoid is bounded by the curve $y = x^2 + 1$ and the straight lines $x=1$ and $x=2$. The coordinates of the point $(x_1, y_1)$ on the given curve with abscissa $x_1 \in [1, 2]$,where the tangent drawn cuts off an ordinary trapezium of the greatest area from the curvilinear trapezoid,are

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

Explore More

Similar Questions

Let $R$ denote the set of all real numbers. Let $f: R \rightarrow R$ be defined by $f(x)=\begin{cases} \frac{6x+\sin x}{2x+\sin x} & \text{if } x \neq 0 \\ \frac{7}{3} & \text{if } x=0 \end{cases}$. Then which of the following statements is (are) True?
$(A)$ The point $x=0$ is a point of local maxima of $f$
$(B)$ The point $x=0$ is a point of local minima of $f$
$(C)$ Number of points of local maxima of $f$ in the interval $[\pi, 6\pi]$ is $3$
$(D)$ Number of points of local minima of $f$ in the interval $[2\pi, 4\pi]$ is $1$

The curve $y(x) = ax^{3} + bx^{2} + cx + 5$ touches the $x$-axis at the point $P(-2, 0)$ and cuts the $y$-axis at the point $Q$,where the derivative $y'(0) = 3$. Find the local maximum value of $y(x)$.

If the function $f(x) = x^3 - 3(a - 2)x^2 + 3ax + 7$,for some $a \in R$,is increasing in $(0, 1]$ and decreasing in $[1, 5)$,then a root of the equation $\frac{f(x) - 14}{(x - 1)^2} = 0$ $(x \neq 1)$ is

If $x+y=6, x \geqslant 0, y \geqslant 0$,then the maximum value of $x^2 y$ is

The minimum value of the slope of the tangent to the curve $y=x^3-3x^2+2x+93$ is

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo