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
$\frac{7}{3}$
If $A_1, A_2, A_3, \ldots, A_{1006}$ be independent events such that $P\left(A_l\right)=\frac{1}{2 i},(i=1,2, \ldots, 1006)$ and the probability that none of the events occurs be $\frac{\alpha!}{2^a(\beta!)^2}$ ,then
$\beta$ is of the form $4 k+2, k \in I$
$\alpha=2 \beta$
$\beta$ is of the form $4 k+1, k \in I$
$\beta$ is a prime number
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