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

A five digit number (having all different digits) is formed using the digits $1,2,3,4,5$, 6,78 and 9 . The probability that the formed number either begins or ends with an odd digit is equal to

A

$\frac{5}{6}$

B

$\frac{1}{3}$

C

$\frac{1}{6}$

D

$\frac{2}{3}$

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

The function, $f(x)=(3 x-7) x^{2 / 3}, x \in R$ is increasing for all $x$ lying in

A

$(-\infty, 0) \cup\left(\frac{3}{7}, \infty\right)$

B

$(-\infty, 0) \cup\left(\frac{14}{15}, \infty\right)$

C

$\left(-\infty, \frac{14}{15}\right)$

D

$\left(-\infty, \frac{14}{15}\right) \cup(0, \infty)$

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

Let $A=\left[\begin{array}{ll}x & 1 \\ 1 & 0\end{array}\right], x \in R$ and $A^4=\left[a_{i j}\right]$.

It $a_{11}=109$, then $a_{22}$ is equal to

A

21

B

10

C

9

D

14

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

If $\alpha$ and $\beta$ are the roots of equation $x^2+p x+2=0$ and $\frac{1}{\alpha}$ and $\frac{1}{\beta}$ are roots of equation $2 x^2+2 q x+1=0$, then $\left(\alpha-\frac{1}{\alpha}\right)\left(\beta-\frac{1}{\beta}\right)\left(\alpha+\frac{1}{\beta}\right)\left(\beta+\frac{1}{\alpha}\right)$ is equal to

A

$\frac{9}{4}\left(9+p^2\right)$

B

$\frac{9}{4}\left(9-q^2\right)$

C

$\frac{9}{4}\left(9-p^2\right)$

D

$\frac{9}{4}\left(9+q^2\right)$

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