1
TG EAPCET 2025 (Online) 4th May Evening Shift
MCQ (Single Correct Answer)
+1
-0

In a quadrilateral $A B C D, \mathbf{A}=\frac{2 \pi}{3}$ and $A C$ is the bisector of angle $\mathbf{A}$. If $15|\mathbf{A C}|=5|\mathbf{A D}|=3|\mathbf{A B}|$, then angle between $\mathbf{A B}$ and $\mathbf{B C}$ is

A

$\cos ^{-1}\left(\frac{\sqrt{3}}{\sqrt{7}}\right)$

B

$\cos ^{-1}\left(\frac{3 \sqrt{3}}{2 \sqrt{7}}\right)$

C

$\cos ^{-1}\left(\frac{4 \sqrt{3}}{5 \sqrt{7}}\right)$

D

$\cos ^{-1}\left(\frac{3 \sqrt{3}}{4 \sqrt{7}}\right)$

2
TG EAPCET 2025 (Online) 4th May Evening Shift
MCQ (Single Correct Answer)
+1
-0

$\mathbf{a}, \mathbf{b}, \mathbf{c}$ are three non- coplanar and mutually perpendicular vectors of same magnitude $K . r$ is any vectors satisfying $\mathbf{a} \times((\mathbf{r}-\mathbf{b}) \times \mathbf{a})+\mathbf{b} \times((\mathbf{r}-\mathbf{c}) \times \mathbf{b})+\mathbf{c} \times((\mathbf{r}-\mathbf{a}) \times \mathbf{c})=\mathbf{0}$, then $\mathbf{r}=$

A

$\frac{K^2(\mathbf{a}+\mathbf{b}+\mathbf{c})}{3 K^2-1}$

B

$\frac{\mathbf{a}+\mathbf{b}+\mathbf{c}}{2}$

C

$\frac{K(\mathbf{a}+\mathbf{b}+\mathbf{c})}{K+1}$

D

$\frac{\mathbf{a}+\mathbf{b}+\mathbf{c}}{K^2+1}$

3
TG EAPCET 2025 (Online) 4th May Evening Shift
MCQ (Single Correct Answer)
+1
-0

Consider the following

Assertion (A) The two lines $\mathbf{r}=\mathbf{a}+t(\mathbf{b})$ and $\mathbf{r}=\mathbf{b}+s(\mathbf{a})$ intersect each other.

Reason (R) The shortest distance between the lines $\mathbf{r}=\mathbf{p}+t(\mathbf{q})$ and $\mathbf{r}=\mathbf{c}+s(\mathbf{d})$ is equal to the length of projection of the vector ( $\mathbf{p}-\mathbf{c}$ ) on ( $\mathbf{q} \times \mathbf{d}$ )

The correct answer is

A

Both $(A)$ and $(R)$ are true and $(R)$ is the correct explanation of (A).

B

Both $(A)$ and $(R)$ are true and $(R)$ is not the correct explanation of (A).

C

(A) is true, but (R) is false.

D

(A) is false, but (R) is true.

4
TG EAPCET 2025 (Online) 4th May Evening Shift
MCQ (Single Correct Answer)
+1
-0

The mean deviation about median of the numbers $3 x, 6 x, 9 x, \ldots .81 x$ is 91 , then $|x|=$

A

4

B

$\frac{5}{2}$

C

$\frac{9}{2}$

D

8

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