$\theta$ 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 $(p+2 q-3 r) x^2+(q+2 x-3 p) x+(r+2 p-3 q)=0 ?$
$t_{p q}$
$t_q$
$t_p$
$t_{p+q}$
Consider the sequence of numbers $(1,2,3, \ldots \ldots, 13)$. A person choose three numbers at random from the sequence. The probability that the chosen three number form an A.P. is
$\frac{21}{157}$
$\frac{18}{143}$
$\frac{29}{180}$
$\frac{24}{163}$
If $f(x)=\frac{1+x}{1-x}$ and $A$ is a matrix such that $A^3=0$, then $f(A)=$
$1+2 \mathrm{~A}+2 \mathrm{~A}^2$
$1+2 A+A^2$
$1-2 \mathrm{~A}+\mathrm{A}^2$
$1+A+A^2$
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