1
AP EAPCET 2025 - 26th May Evening Shift
MCQ (Single Correct Answer)
+1
-0

If $P(\alpha, \beta)$ is a point on the curve $9 x^2+4 y^2=144$ in the first quadrant and the minimum area of the triangle formed by the tangent of the curve at $P$ with the coordinate axis is $S$, then

A

$S=\sqrt{\alpha \beta}$

B

$S=\alpha \beta$

C

$S=2 \sqrt{\alpha \beta}$

D

$S=2 \alpha \beta$

2
AP EAPCET 2025 - 26th May Evening Shift
MCQ (Single Correct Answer)
+1
-0

$$ \int(\log 2 x)^3 d x= $$

A

$x\left((\log 2 x)^3-3(\log 2 x)^2+6(\log 2 x)-6\right]+C$

B

$\frac{x}{4}\left[4(\log 2 x)^3-6(\log 2 x)^2+6(\log 2 x)-3\right]+C$

C

$\frac{x}{2}\left[(\log 2 x)^3-3(\log 2 x)^2+3(\log 2 x)-6\right]+C$

D

$x\left[(\log 2 x)^3-6(\log 2 x)^2+18(\log 2 x)-54\right]+C$

3
AP EAPCET 2025 - 26th May Evening Shift
MCQ (Single Correct Answer)
+1
-0

$$ \int \frac{x+1}{(x-2) \sqrt{1-x}} d x= $$

A

$\log (x+1)-\log (x-2) \sqrt{1-x}+C$

B

$\log (x-2) \sqrt{1-x}+C$

C

$6 \tan ^{-1} \sqrt{1-x}-2 \sqrt{1-x}+C$

D

$4 \tan ^{-1} \sqrt{1-x}-2 \sqrt{1-x}+C$

4
AP EAPCET 2025 - 26th May Evening Shift
MCQ (Single Correct Answer)
+1
-0

$$ \int \frac{1}{1+x+x^2} d x= $$

A

$\frac{2}{\sqrt{3}} \log \left(\frac{2 x+1+\sqrt{3}}{2 x-1-\sqrt{3}}\right)+C$

B

$\frac{1}{\sqrt{3}} \log \left(\frac{2 x+1-\sqrt{3}}{2 x+1+\sqrt{3}}\right)+C$

C

$\frac{2}{\sqrt{3}} \tan ^{-1}\left(\frac{2 x+1}{\sqrt{3}}\right)+C$

D

$\frac{2}{\sqrt{5}} \tan ^{-1}\left(\frac{2 x+1}{\sqrt{5}}\right)+C$