1
VITEEE 2025
MCQ (Single Correct Answer)
+4
-1

For any four vectors $\mathbf{a , b , c , d}$ the expression $(\mathrm{b} \times \mathrm{c}) \cdot(\mathrm{a} \times \mathrm{d})+(\mathrm{c} \times \mathrm{a}) \cdot(\mathrm{b} \times \mathrm{d})+(\mathrm{a} \times \mathrm{b}) \cdot(\mathrm{c} \times \mathrm{d})$ is always equal to

A

$[\mathrm{a} \mathrm{b} \mathrm{c}]$

B

$[b \subset c]$

C

$[\mathbf{a} \mathbf{c} \mathbf{c} \mathbf{d}]$

D

None of these

2
VITEEE 2025
MCQ (Single Correct Answer)
+4
-1

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

A

$1-\frac{3}{8} x^2$

B

$3 x+\frac{3}{8} x^2$

C

$-\frac{3}{8} x^2$

D

$\frac{x}{2}-\frac{3}{8} x^2$

3
VITEEE 2025
MCQ (Single Correct Answer)
+4
-1

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

A

$1,-1$

B

$-\frac{1}{2}, 2$

C

$-2, \frac{1}{2}$

D

$\frac{1}{2}, 2$

4
VITEEE 2025
MCQ (Single Correct Answer)
+4
-1
Let $f(x)$ be a polynomial of degree 6 divisible by $x^3$ and having a point of extremum at $x=2$. If $f^{\prime}(x)$ is divisible by $1+x^2$, then find the value of $\frac{3 f(2)}{f(1)}$
A

14

B

15

C

16

D

None of these

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