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

If $f(x)=\frac{x\left(a^x-1\right)}{1-\cos x}$ and $g(x)=\frac{x\left(1-a^x\right)}{a^x\left(\sqrt{1-x^2}-\sqrt{1+x^2}\right)}$, then $\lim _{x \rightarrow 0}(f(x)-g(x))=$

A

$3 \log a$

B

$e^a$

C

$2 \log a$

D

$\log a$

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

If $f(x)=\left\{\begin{array}{cc}\frac{a \sin x-b x+c x^2+x^3}{2 \log (1+x)-2 x^3+x^4} & , x \neq 0 \\ 0 & , x=0\end{array}\right.$

is continuous at $x=0$, then

A

$a=2 b$

B

$a=b$

C

$a=b=c$

D

$b=c$

3
TG EAPCET 2025 (Online) 2nd May Evening Shift
MCQ (Single Correct Answer)
+1
-0

If the function $g(x)=\left\{\begin{array}{cl}K \sqrt{x+1} & , 0 \leq x \leq 3 \\ m x+2 & , 3 < x \leq 5\end{array}\right.$ is differentiable, then $K+m=$

A

4

B

2

C

6

D

0

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

Consider the following statements

Assertion (A) For $x \in R-\{1\}$;

$$ \frac{d}{d x}\left(\tan ^{-1}\left(\frac{1+x}{1-x}\right)\right)=\frac{d}{d x}\left(\tan ^{-1} x\right) $$

Reason (R) For $x<1, \tan ^{-1}\left(\frac{1+x}{1-x}\right)=\frac{\pi}{4}+\tan ^{-1} x$, for

$$ x>1, \tan ^{-1}\left(\frac{1+x}{1-x}\right)=-\frac{3 \pi}{4}+\tan ^{-1} x $$

The correct answer is

A

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

B

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

C

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

D

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

TS EAMCET Papers

All year-wise previous year question papers