Consider the curve $x=1-3 t^2, y=t-3 t^3$. The tangent to the curve at the point $t$ is inclined at an angle $\phi$ to OX and the tangent at $\mathrm{P}(-2,2)$ meets the curve again at Q . Then
the curve is symmetrical about $x$-axis
the curve is symmetrical about $y$-axis
$3 t=\tan \phi+\sec \phi$
tangents at P and Q are at right angle
If $f(x)=x\left(1331 x^2-3630 x+3300\right)$, then for $a=\cos ^2\left(\tan ^{-1}\left(\sin \left(\cot ^{-1} 3\right)\right)\right)$
$f(a+1)=2331$
$f^{\prime}(a)=11$
$\mathop {\lim }\limits_{x \to a}f(x)=1000$
$\int_0^a(f(x)-1000) d x=\frac{2500}{11}$
Let $\vec{r}=\sin x(\vec{a} \times \vec{b})+\cos y(\vec{b} \times \vec{c})+2(\vec{c} \times \vec{a})$ ,where $\vec{a}, \vec{b}$ and $\vec{c}$ are three non-coplanar vectors. It is given that $\vec{r}$ is perpendicular to $(\vec{a}+\vec{b}+\vec{c})$ .Then the possible value(s)of $\left(x^2+y^2\right)$ is/are
$\frac{5 \pi^2}{4}$
$\frac{35 \pi^2}{4}$
$\frac{37 \pi^2}{4}$
$\frac{\pi^2}{4}$
The parabola $y=4-x^2$ has vertex P. It intersects $x$-axis at A and B. If the parabola is translated from its initial position to a new position by moving its vertex along the line $y=x+4$, so that it intersects $x$-axis at B and C , then the abscissa of C will be
12
8
6
$\frac{7}{3}$
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