1
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

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

2
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

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

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

A

$-\frac{\pi}{4} \leq \theta \leq \frac{\pi}{4}$

B

$0 \leq \theta \leq \frac{\pi}{4}$

C

$\frac{\pi}{4} \leq \theta \leq \frac{3 \pi}{4}$

D

$0 \leq \theta \leq \frac{3 \pi}{4}$

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

In the given figure, the equation of the large circle is $x^2+y^2+4 y-5=0$ and the distance between centre is 4 . Then, the equation of smaller circle is

VITEEE 2025 Mathematics - Circle Question 2 English
A

$(x-\sqrt{7})^2+(y-1)^2=1$

B

$(x+\sqrt{7})^2+(y-1)^2=1$

C

$x^2+y^2=2 \sqrt{7} x+2 y$

D

None of the above

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