NEW
New Website Launch
Experience the best way to solve previous year questions with mock tests (very detailed analysis), bookmark your favourite questions, practice etc...
1

JEE Advanced 2015 Paper 2 Offline

MCQ (More than One Correct Answer)
If $$\alpha $$ $$ = 3{\sin ^{ - 1}}\left( {{6 \over {11}}} \right)$$ and $$\beta = 3{\cos ^{ - 1}}\left( {{4 \over 9}} \right),$$ where the inverse trigonimetric functions take only the principal values, then the correct options(s) is (are)
A
$$cos\beta > 0$$
B
$$\sin \beta < 0$$
C
$$\cos \left( {\alpha + \beta } \right) > 0$$
D
$$\cos \alpha < 0$$

Explanation

Here, $$\alpha = 3{\sin ^{ - 1}}\left( {{6 \over {11}}} \right)$$

and $$\beta = 3{\cos ^{ - 1}}\left( {{4 \over 9}} \right)$$ as $${6 \over {11}} > {1 \over 2}$$

$$ \Rightarrow {\sin ^{ - 1}}\left( {{6 \over {11}}} \right) > {\sin ^{ - 1}}\left( {{1 \over 2}} \right) = {\pi \over 6}$$

$$\therefore$$ $$\alpha = 3{\sin ^{ - 1}}\left( {{6 \over {11}}} \right) > {\pi \over 2} \Rightarrow \cos \alpha < 0$$

Now, $$\beta = 3{\cos ^{ - 1}}\left( {{4 \over 9}} \right)$$

As $${4 \over 9} < {1 \over 2} \Rightarrow {\cos ^{ - 1}}\left( {{4 \over 9}} \right) > {\cos ^{ - 1}}\left( {{1 \over 2}} \right) = {\pi \over 3}$$

$$\therefore$$ $$\beta = 3{\cos ^{ - 1}}\left( {{4 \over 9}} \right) > \pi $$

$$\therefore$$ $$\cos \beta < 0$$ and $$\sin \beta < 0$$

Now, $$\alpha + \beta $$ is slightly greater than $${{3\pi } \over 2}$$.

$$\therefore$$ $$\cos (\alpha + \beta ) > 0$$

Joint Entrance Examination

JEE Main JEE Advanced WB JEE

Graduate Aptitude Test in Engineering

GATE CSE GATE ECE GATE EE GATE ME GATE CE GATE PI GATE IN

Medical

NEET

CBSE

Class 12