Q.12. Let and , where inverse trigonometric functions take only the principal values. Given below a

Q: What is the correct answer for JEE Main Mathematics Q12?

The correct answer is A. Given $\alpha = 3 \sin^{-1}(\frac{6}{11})$ and $\beta = 3 \cos^{-1}(\frac{4}{9})$. For $\alpha = 3 \sin^{-1}(\frac{6}{11})$: Let $\theta = \sin^{-1}(\frac{6}{11})$. Since $0 \frac{1}{2}$, we have $\frac{\pi}{6} < \theta < \frac{\pi}{2}$. Multiplying

Q: Q.12. Let and , where inverse trigonometric functions take only the principal values. Given below are two statements: Statement I: Statement II: In the light of the above statements, choose the…

The correct answer is A. Given $\alpha = 3 \sin^{-1}(\frac{6}{11})$ and $\beta = 3 \cos^{-1}(\frac{4}{9})$. For $\alpha = 3 \sin^{-1}(\frac{6}{11})$: Let $\theta = \sin^{-1}(\frac{6}{11})$. Since $0 \frac{1}{2}$, we have $\frac{\pi}{6} < \theta < \frac{\pi}{2}$. Multiplying

Q: What are the options for this JEE Main Mathematics question?

A. Both Statement I and Statement II are true · B. Both Statement I and Statement II are false · C. Statement I is true but Statement II is false · D. Statement I is false but Statement II is true

Question

Q.12. Let  and , where inverse trigonometric functions take only the principal values. Given below are two statements: Statement I:  Statement II:  In the light of the above statements, choose the correct answer from the options given below:

FirstInTest · Accepted Answer

Given $\alpha = 3 \sin^{-1}(\frac{6}{11})$ and $\beta = 3 \cos^{-1}(\frac{4}{9})$. For $\alpha = 3 \sin^{-1}(\frac{6}{11})$: Let $ heta = \sin^{-1}(\frac{6}{11})$. Since $0 < \frac{6}{11} < 1$, we have $0 < heta < \frac{\pi}{2}$. Specifically, $\sin(\frac{\pi}{6}) = \frac{1}{2} = \frac{5.5}{11}$. Since $\frac{6}{11} > \frac{1}{2}$, we have $\frac{\pi}{6} < heta < \frac{\pi}{2}$. Multiplying by 3: $\frac{3\pi}{6} < 3 heta < \frac{3\pi}{2} \Rightarrow \frac{\pi}{2} < \alpha < \frac{3\pi}{2}$. For $\beta = 3 \cos^{-1}(\frac{4}{9})$: Let $\phi = \cos^{-1}(\frac{4}{9})$. Since $0 < \frac{4}{9} < 1$, we have $0 < \phi < \frac{\pi}{2}$. Specifically, $\cos(\frac{\pi}{3}) = \frac{1}{2} = \frac{4.5}{9}$. Since $\frac{4}{9} < \frac{1}{2}$, we have $\frac{\pi}{3} < \phi < \frac{\pi}{2}$. Multiplying by 3: $\frac{3\pi}{3} < 3\phi < \frac{3\pi}{2} \Rightarrow \pi < \beta < \frac{3\pi}{2}$. Now let's check the statements: Statement I: $\cos (\alpha + \beta) > 0$ We have $\frac{\pi}{2} < \alpha < \frac{3\pi}{2}$ and $\pi < \beta < \frac{3\pi}{2}$. Adding the inequalities: $\frac{\pi}{2} + \pi < \alpha + \beta < \frac{3\pi}{2} + \frac{3\pi}{2}$ $\frac{3\pi}{2} < \alpha + \beta < 3\pi$. In the interval $(\frac{3\pi}{2}, 3\pi)$, the cosine function is positive in $(\frac{3\pi}{2}, 2\pi)$ and negative in $(2\pi, \frac{5\pi}{2})$ and positive in $(\frac{5\pi}{2}, 3\pi)$. Let's refine the ranges for $\alpha$ and $\beta$. For $\alpha$: $\sin^{-1}(\frac{6}{11})$. $\frac{6}{11} \approx 0.545$. $\sin^{-1}(0.545)$ is slightly larger than $\pi/6 \approx 0.523$ radians (30 degrees). So $ heta \in (\pi/6, \pi/2)$. Thus $\alpha = 3 heta \in (\pi/2, 3\pi/2)$. For $\beta$: $\cos^{-1}(\frac{4}{9})$. $\frac{4}{9} \approx 0.444$. $\cos^{-1}(0.444)$ is larger than $\pi/3 \approx 0.523$ radians (60 degrees). So $\phi \in (\pi/3, \pi/2)$. Thus $\beta = 3\phi \in (\pi, 3\pi/2)$. Statement II: $\cos (\alpha) < 0$ Since $\frac{\pi}{2} < \alpha < \frac{3\pi}{2}$, $\alpha$ lies in the second or third quadrant. In these quadrants, $\cos(\alpha)$ is negative. So, Statement II is TRUE. Now for Statement I: $\cos (\alpha + \beta) > 0$ We have $\frac{\pi}{2} < \alpha < \frac{3\pi}{2}$ and $\pi < \beta < \frac{3\pi}{2}$. Adding them: $\frac{3\pi}{2} < \alpha + \beta < 3\pi$. Let's check the solution's steps for $\alpha$ and $\beta$ ranges: Sol. $\frac{1}{2} < \frac{6}{11} < \frac{1}{\sqrt{2}}$ is false. $\frac{1}{\sqrt{2}} \approx 0.707$. $\frac{6}{11} \approx 0.545$. So $\frac{1}{2} < \frac{6}{11} < \frac{1}{\sqrt{2}}$ is true. This means $\sin^{-1}(\frac{1}{\sqrt{2}}) < \sin^{-1}(\frac{6}{11}) < \sin^{-1}(\frac{1}{2})$ is wrong. It should be $\sin^{-1}(\frac{1}{2}) < \sin^{-1}(\frac{6}{11}) < \sin^{-1}(\frac{1}{\sqrt{2}})$. So $\frac{\pi}{6} < heta < \frac{\pi}{4}$. Then $\alpha = 3 heta \in (\frac{3\pi}{6}, \frac{3\pi}{4}) = (\frac{\pi}{2}, \frac{3\pi}{4})$. In this range, $\cos(\alpha)$ is negative. So Statement II is TRUE. For $\beta$: $\frac{1}{2} < \frac{4}{9} < 1$ is true. $\frac{4}{9} \approx 0.444$. So $\cos^{-1}(1) < \cos^{-1}(\frac{4}{9}) < \cos^{-1}(\frac{1}{2})$ is wrong. It should be $\cos^{-1}(\frac{1}{2}) < \cos^{-1}(\frac{4}{9}) < \cos^{-1}(0)$. So $\frac{\pi}{3} < \phi < \frac{\pi}{2}$. Then $\beta = 3\phi \in (\pi, \frac{3\pi}{2})$. So we have $\alpha \in (\frac{\pi}{2}, \frac{3\pi}{4})$ and $\beta \in (\pi, \frac{3\pi}{2})$. For Statement II: $\cos(\alpha) < 0$. Since $\alpha \in (\frac{\pi}{2}, \frac{3\pi}{4})$, $\alpha$ is in the second quadrant, so $\cos(\alpha) < 0$. Statement II is TRUE. For Statement I: $\cos(\alpha + \beta) > 0$. Adding the ranges: $\frac{\pi}{2} + \pi < \alpha + \beta < \frac{3\pi}{4} + \frac{3\pi}{2}$ $\frac{3\pi}{2} < \alpha + \beta < \frac{9\pi}{4}$. In the interval $(\frac{3\pi}{2}, \frac{9\pi}{4})$, $\cos(x)$ is positive for $x \in (\frac{3\pi}{2}, 2\pi)$ and positive for $x \in (2\pi, \frac{9\pi}{4})$. Since $2\pi = \frac{8\pi}{4}$, the entire interval $(\frac{3\pi}{2}, \frac{9\pi}{4})$ spans from the 4th quadrant into the 1st quadrant (after one full rotation). So $\cos(\alpha + \beta)$ will be positive in this range. Thus, Statement I is TRUE. Both Statement I and Statement II are true.

Answer

Both Statement I and Statement II are true

Answer

Both Statement I and Statement II are false

Answer

Statement I is true but Statement II is false

Answer

Statement I is false but Statement II is true

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