1
TS EAMCET 2020 (Online) 11th September Evening Shift
MCQ (Single Correct Answer)
+1
-0

Assertion (A) The function $f(x)=x-\log \left(\frac{1+x}{x}\right), x>0$ has no maximum.

Reason (R) If a function $f(x)$ is strictly increasing in an interval $(a, b)$, then at any point in $(a, b) f^{\prime}(x) \neq 0$

The correct option among the following is

A

(A) is true, (R) is true and (R) is the correct explanation for $A$.

B

(A) is true, (R) is true but (R) is the not the correct explanation for A .

C

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

D

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

2
TS EAMCET 2020 (Online) 11th September Evening Shift
MCQ (Single Correct Answer)
+1
-0

$$ \int \frac{x^2}{\left(\sqrt{4-x^2}\right)^3} d x= $$

A

$\frac{x^2}{\sqrt{4-x^2}}-\sin ^{-1}\left(\frac{x}{2}\right)+C$

B

$\frac{x}{\sqrt{4-x^2}}-\tan ^{-1}\left(\frac{x}{\sqrt{4-x^2}}\right)+C$

C

$\frac{x}{\sqrt{4-x^2}}+\sin ^{-1}\left(\frac{2}{\sqrt{4-x^2}}\right)+C$

D

$\sqrt{4-x^2}-\tan ^{-1}\left(\frac{x}{2}\right)+C$

3
TS EAMCET 2020 (Online) 11th September Evening Shift
MCQ (Single Correct Answer)
+1
-0

$$ \int \frac{d x}{x \ln (x) \ln ^2(x) \ln ^3(x) \ldots \ln ^m(x)}=\frac{(\ln (x))^K}{K}+C \Rightarrow 2 K= $$

A

$(m+1)(m+2)$

B

$(2-m)(1-m)$

C

$(m+1)(2-m)$

D

$(m+2)(1-m)$

4
TS EAMCET 2020 (Online) 11th September Evening Shift
MCQ (Single Correct Answer)
+1
-0

If $I_m=\int x^m \cos n x d x=g(x)-\frac{m(m-1)}{n^2} I_{m-2}$, then $g(x)=$

A

$\frac{x^m \sin n x}{n}+\frac{m(m-1) x^{m-1} \cos n x}{n^2}$

B

$\frac{x^m \cos n x}{n}+\frac{x^{m-1} m(m-1)}{n^2} \sin n x$

C

$\frac{m}{n} \sin n x+\frac{m}{n^2} x^{m-1} \cos n x$

D

$\frac{x^m \sin n x}{n}+\frac{m}{n^2} x^{m-1} \cos n x$

TS EAMCET Papers

All year-wise previous year question papers