A figure is bounded by the curves $y=x^2+1, y=0, x=0$ and $x=1$. The point at which a tangent should be drawn to the curve $y=x^2+1$ for it to cut off trapezium of the greatest area from the figure is
$(1,2)$
$(-1,2)$
$\left(\frac{1}{2}, \frac{5}{4}\right)$
$(0,1)$
The ends $A$, $B$ of a straight line segment of constant length $c$ slide upon the fixed rectangular axes $O X, O Y$ respectively. If the rectangle $O A P B$ completed, then the locus of the foot of perpendicular drawn from $P$ to $A B$ is
$x^2+y^2=c^2$
$\mathrm{x}^{2 / 3}+\mathrm{y}^{2 / 3}=\mathrm{c}^{2 / 3}$
$\sqrt{x}+\sqrt{y}=\sqrt{c}$
$x y=c^2$
Let 1 lies between the roots of the equation $y^2-m y+1=0$ and $[x]$ denotes the greatest integer function. Then the value of $\left[\left(\frac{4|x|}{x^2+16}\right)^m\right]$ is
5
4
0
1
Let $f(x)$ be a twice differentiable function in $[1,3]$ and $f(1)=f(3)$. Further if $\left|f^{\prime \prime}(x)\right| \leq 2$, then for all $x$ in $[1,3]$
$\left|\mathrm{f}^{\prime}(\mathrm{x})\right| \geq 4$
$\left|\mathrm{f}^{\prime}(\mathrm{x})\right| \leq-1$
$\left|\mathrm{f}^{\prime}(\mathrm{x})\right|>2$
$\left|\mathrm{f}^{\prime}(x)\right|<4$
WB JEE Papers
All year-wise previous year question papers