1
TS EAMCET 2022 (Online) 19th July Evening Shift
MCQ (Single Correct Answer)
+1
-0

$\int \sqrt{4 \cos ^2 x-5 \sin ^2 x} \cos x d x=$

A

$\frac{1}{2} \sin x \sqrt{4-9 \sin ^2 x}+\frac{2}{3} \sin ^{-1}\left(\frac{3 \sin x}{2}\right)+c$

B

$\frac{1}{2} \cos x \sqrt{4-9 \cos ^2 x}+\frac{2}{3} \sin ^{-1}\left(\frac{3 \cos x}{2}\right)+c$

C

$\frac{1}{2} \sin x \sqrt{4-9 \sin ^2 x}+\frac{2}{3} \cos ^{-1}\left(\frac{3 \cos x}{2}\right)+c$

D

$\frac{1}{2} \cos x \sqrt{4-9 \sin ^2 x}+\frac{2}{3} \sin ^{-1}\left(\frac{3 \sin x}{2}\right)+c$

2
TS EAMCET 2022 (Online) 19th July Morning Shift
MCQ (Single Correct Answer)
+1
-0

If $\frac{42-13 x}{x^2+x-6}=\frac{A}{l x+m}+\frac{B}{p x+q}$, where $l m>0$ and $p q<0$, then $\frac{A l p}{B m q}=$

A

$\frac{27}{32}$

B

$\frac{27}{8}$

C

$\frac{8}{243}$

D

$\frac{243}{32}$

3
TS EAMCET 2022 (Online) 19th July Morning Shift
MCQ (Single Correct Answer)
+1
-0

Given that $\frac{d}{d x}\left(\tan ^{-1} x\right)=\frac{1}{1+x^2}$ and $\frac{d}{d x}\left(\sin h^{-1} x\right)=\frac{1}{\sqrt{1+x^2}}$. Then, $\int \frac{3 x^6-2 x^4+x^2-2}{x^2+1} d x=$

A

$\frac{3}{7} x^7-\frac{2}{5} x^5+\frac{1}{3} x^3-2 x+c$

B

$\frac{\frac{3}{7} x^7-\frac{2}{5} x^5+\frac{1}{3} x^3-2 x}{\frac{x^3}{3}+x}+c$

C

$\frac{3}{5} x^5-\frac{5}{3} x^3+6 x-8 \tan ^{-1} x+c$

D

$\frac{3}{5} x^5-\frac{5}{3} x^3+6 x-8 \sin h^{-1} x+c$

4
TS EAMCET 2022 (Online) 19th July Morning Shift
MCQ (Single Correct Answer)
+1
-0

$$ \int \frac{\sin x \cdot \sec ^2 x-\tan x \cdot \sin x+\cos x}{(1-\cos 2 x)} d x= $$

A

$\frac{1}{2}\left[\sec x-\operatorname{cosec} x-\log \left|\tan \left(\frac{x}{2}\right) \tan \left(\frac{\pi}{4}+\frac{x}{2}\right)\right|\right]+C$

B

$\sec x-\operatorname{cosec} x+\log \left|\frac{\tan \left(\frac{\pi}{2}\right)}{\tan \left(\frac{\pi}{4}+\frac{x}{2}\right)}\right|+C$

C

$\frac{1}{2}\left[\sec x-\operatorname{cosec} x-\log \left|\frac{\tan \left(\frac{\pi}{4}+\frac{x}{2}\right)}{\tan \left(\frac{x}{2}\right)}\right|\right]+C$

D

$\sec x+\operatorname{cosec} x-\log \left|\tan \left(\frac{x}{2}\right)\right|+C$

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