1
TS EAMCET 2020 (Online) 14th September Morning Shift
MCQ (Single Correct Answer)
+1
-0

A line with direction cosines proportional to $2,1,2$ meets the line $L_1$ passing through $(0,-1,0)$ with direction ratios $1,1,1$ at $A(x, y, z)$ and another line $L_2$ at $B(1,1,1)$ then $x+y+z=$

A

7

B

8

C

9

D

10

2
TS EAMCET 2020 (Online) 14th September Morning Shift
MCQ (Single Correct Answer)
+1
-0

If a plane $\pi$ passes through the point $(-1,6,2)$ is perpendicular to the planes $x+2 y+2 z-5=0$ and $3 x+3 y+2 z-8=0$, then, the perpendicular distance from the point $(1,-1,1)$ to the plane $\pi$ is

A

$\frac{20}{\sqrt{29}}$

B

$\frac{21}{\sqrt{29}}$

C

$\frac{27}{\sqrt{29}}$

D

$\sqrt{29}$

3
TS EAMCET 2020 (Online) 14th September Morning Shift
MCQ (Single Correct Answer)
+1
-0

If $f: \mathbf{R} \rightarrow \mathbf{R}$ and $g: \mathbf{R} \rightarrow \mathbf{R}$ be defined by $f(x)=\left\{\begin{array}{cc}x+2, & x>0 \\ 2-x, & x \leq 0\end{array}\right.$ and $g(x)=\left\{\begin{array}{cc}x^2-2 x-2, & 1 \leq x<2 \\ x-7 & x \geq 2 \\ x+5, & x<1\end{array}\right.$ then $\lim _{x \rightarrow 0} g \circ f(x)$

A

is equal to -7

B

is equal to -5

C

is equal to 2

D

does not exist

4
TS EAMCET 2020 (Online) 14th September Morning Shift
MCQ (Single Correct Answer)
+1
-0

Define $f: R \rightarrow R$ by $f(x)= \begin{cases}(x-a) \frac{e^{\frac{1}{(x-a)}}-1}{\frac{1}{(x-a)}}+1 & \text { for } x \neq a \\ 0, \quad \text { at } x=a\end{cases}$

Then which one of the following is true?

A

Left and right limits of $f$ at $x=a$ are equal and they are not equal to $f(a)$

B

Both left and right limits of $f$ at $x=a$ exist and are not equal

C

The function $f(x)$ is continuous at $x=a$

D

The function $f(x)$ has a simple discontinuity at a point other than $a$

TS EAMCET Papers

All year-wise previous year question papers