If a differentiable function satisfies $(x-y) f(x+y)-(x+y) f(x-y)=2\left(x^2 y-y^3\right) \forall x, y \in \mathbb{R}$ and $f(I)=2$, then
$\mathrm{f}(\mathrm{x})$ must be a polynomial function
$f(3)=13$
$f(3)=12$
$f(0)=0$
Let $f(x)>0$ for all $x \in \mathbb{R}$ and $f(x)$ is bounded. If $\mathop {\lim }\limits_{n \to \infty } \sum_{r-1}^n a^{r-1} \int_{(r-1) a}^{r a} \frac{f(x) d x}{f(x)+f(2 r a-a-x)}=\frac{3}{5}$ where $0< a< 1$, then the value(s) of a is are
$\frac{5}{11}$
$\frac{7}{11}$
$\frac{1}{11}$
$\frac{6}{11}$
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 $a=\cos ^2\left(\tan ^{-1}\left(\sin \left(\cot ^{-1} 3\right)\right)\right)$
$f(a+1)=2331$
$f^{\prime}(a)=11$
$\mathop {\lim }\limits_{x \to a}f(x)=1000$
$\int_0^a(f(x)-1000) d x=\frac{2500}{11}$
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