If $x$ is so small that $x^3$ and higher powers of $x$ may be neglected, then $\frac{(1+x)^{3 / 2}-\left(1+\frac{1}{2} x\right)^3}{(1-x)^{1 / 2}}$ may be approximate as
$1-\frac{3}{8} x^2$
$3 x+\frac{3}{8} x^2$
$-\frac{3}{8} x^2$
$\frac{x}{2}-\frac{3}{8} x^2$
The focal chord of $y^2=16 x$ is a tangent to $(x-6)^2+y^2=2$, then the possible values of the slope of this chord are
$1,-1$
$-\frac{1}{2}, 2$
$-2, \frac{1}{2}$
$\frac{1}{2}, 2$
14
15
16
None of these
If $\hat{\mathbf{a}} \cdot \hat{\mathbf{b}}=0$, where $\hat{\mathbf{a}}$ and $\hat{\mathbf{b}}$ are unit vectors and the unit vector $\hat{\complement}$ is inclined at an angle $\theta$ to both $\hat{\mathbf{a}}$ and $\hat{\mathbf{b}}$. If $\hat{\mathbf{c}}=m \hat{\mathbf{a}}+n \hat{\mathbf{b}}+p(\hat{\mathbf{a}} \times \hat{\mathbf{b}})$, where, $m, n, p \in R$, then
$-\frac{\pi}{4} \leq \theta \leq \frac{\pi}{4}$
$0 \leq \theta \leq \frac{\pi}{4}$
$\frac{\pi}{4} \leq \theta \leq \frac{3 \pi}{4}$
$0 \leq \theta \leq \frac{3 \pi}{4}$
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