1
TS EAMCET 2023 (Online) 14th May Morning Shift
MCQ (Single Correct Answer)
+1
-0

If a line having slope 2 is a tangent to the curve $y=x^4-6 x^3+13 x^2-12 x+5$ at points $P\left(x_1, y_1\right)$ and $Q\left(x_2, y_2\right), x_1, x_2 \in N$, then $x_1 x_2-y_1 y_2=$

A

17

B

3

C

-17

D

-13

2
TS EAMCET 2023 (Online) 14th May Morning Shift
MCQ (Single Correct Answer)
+1
-0

Let $m$ be the slope of the normal $L$ drawn at $(1,2)$ to the curve $x=t^2-7 t+7, y=t^2-4 t-10$ and $a x+b y+c=0$ be the equation of the normal $L$. If GCD of $(a, b, c)$ is 1 , then $m(a+b+c)=$

A

8

B

$-64 / 5$

C

-8

D

5

3
TS EAMCET 2023 (Online) 14th May Morning Shift
MCQ (Single Correct Answer)
+1
-0

If the function $f(x)=x e^{-x}, x \in R$ attains its maximum value $\beta$ at $x=\alpha$, then $(\alpha, \beta)=$

A

$\left(2, \frac{1}{e}\right)$

B

$\left(1, \frac{1}{e}\right)$

C

$\left(1, \frac{-1}{e}\right)$

D

$\left(\frac{1}{e}, 1\right)$

4
TS EAMCET 2023 (Online) 14th May Morning Shift
MCQ (Single Correct Answer)
+1
-0

If $\int \frac{x^{49} \tan ^{-1}\left(x^{50}\right)}{\left(1+x^{100}\right)} d x=k\left(\tan ^{-1}\left(x^{50}\right)\right)^2+C$, then $k=$

A

$\frac{-1}{100}$

B

$\frac{1}{50}$

C

$\frac{-1}{50}$

D

$\frac{1}{100}$

TS EAMCET Papers

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