Question

Assertion (A) : If $sin^{-1} x + sin^{-1}y + sin^{-1}z = \frac{3\pi}{2}$,then
$\frac{3 \displaystyle\sum_{r = 1}^{2008}(x^{2008r} + y^{2008 r})}{2008 \sum x^{2008} y^{2008}}$ is $2$
Reason (R) : $sin^{-1} x + sin^{-1}y + sin^{-1}z = \frac{3\pi}{2}$, is possible only if $x = y = z = 1$.

• A. Both (A) & (R) are individually true & (R) is correct explanation of (A).
• B. Both (A) & (R) are individually true but (R) is not the correct (proper) explanation of (A).
• C. (A) is true but (R) is false.
• D. (A) is false but (R) is true.

Answer 1

Answer: (A)

$ x = y = z = 1$
$\therefore x^{2008} + y^{2008} = 2$& $x^{2008} y^{2008} = 1$
$\therefore \frac{3 \displaystyle\sum_{r = 1}^{2008}(x^{2008r} + y^{2008r})}{2008 \sum x^{2008} \cdot y^{2008}} $
$= \frac{3 \times 2008 \times 2}{2008 \times 3} = 2$