A vector given by $\vec{P}=f(t) \hat{i}+g(t)+\hat{k}$ moves in such a way that it is always parallel to the vector $\vec{Q}=-f^{\prime \prime}(t) \hat{i}+f^{\prime}(t) \hat{j}+\hat{k}$.
a linear function of time
a quadratic function of time
a cubic function of time
constant
The expression $\sum_{k=1}^{32}(3 K+2)\left\{\sum_{r=1}^{10}\left(\sin \frac{2 r \pi}{11}-i \cos \frac{2 r \pi}{11}\right)\right\}^k$ represents
$48(1+\mathrm{i})$
$-48(1-\mathrm{i})$
$-\frac{48}{11}(1-\mathrm{i})$
$48(1-\mathrm{i})$
$\theta$ elimination from the equation $x^2+y^2=\frac{x \cos 3 \theta+y \sin 3 \theta}{\cos ^3 \theta}=\frac{y \cos 3 \theta-x \sin 3 \theta}{\sin ^3 \theta}$ will be
$4\left(x^4+y^4\right)=3 x+4 y$
$\left(x^2+y^2+2 x\right)\left(x^2+y^2-x\right)=2 y^2$
$\left(x^2+y^2-2 x\right)\left(x^2+y^2+x\right)=9 y$
$x^{2 / 3}+y^{2 / 3}=1$
If $t_n$ denotes the $n^{\text {th }}$ term of an A.P. and $t_p=\frac{1}{q}, t_q=\frac{1}{p}$, then which one of the following options is a root of the equation $(p+2 q-3 r) x^2+(q+2 x-3 p) x+(r+2 p-3 q)=0 ?$
$t_{p q}$
$t_q$
$t_p$
$t_{p+q}$
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