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

$$ \begin{aligned} I_1 & =\int \frac{e^x}{e^{4 x}+e^{2 x}+1} d x, I_2 \\ & =\int \frac{e^{-x}}{e^{-4 x}+e^{-2 x}+1} d x, \text { then } I_2-I_1= \end{aligned} $$

A

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

B

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

C

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

D

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

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

If $\int \frac{1}{x} \sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}} d x=2 f(x)-2 \sin ^{-1} \sqrt{x}+c$, then $f(x)=$

A

$\operatorname{sech}^{-1} \sqrt{x}$

B

$\operatorname{cosec}^{-1} \sqrt{x}$

C

$\log \left(\frac{1+x}{\sqrt{x}}\right)$

D

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

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

$$ \begin{aligned} & \int \frac{3 x+2}{4 x^2+4 x+5} d x=A \log \\ & \left(4 x^2+4 x+5\right)+B \tan ^{-1}\left(\frac{2 x+1}{2}\right)+C, \text { then } A+B= \end{aligned} $$

A

$1 / 2$

B

$3 / 4$

C

$3 / 8$

D

$1 / 8$

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

Consider the following

Assertion

$$ \begin{aligned} & \text { (A) } \begin{array}{r} \sqrt{x-3}\left(\sin ^{-1}(\log x)+\cos ^{-1}\right. \\ (\log x) d x=\frac{\pi}{3}(x-3)^{3 / 2}+c \end{array} \end{aligned} $$

Reason $(\mathrm{R}) \sin ^{-1}(f(x))+\cos ^{-1}(f(x))=\frac{\pi}{2},|f(x)|<1$

The correct answer is

A

Both $(A)$ and $(B)$ 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.

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