Question

$f(x) = \sin^{2}x + \sin^{2}{}\left( x + \frac{\pi}{3} \right) + \cos x\cos\left( x + \frac{\pi}{3} \right)\text{and }g\left( \frac{5}{4} \right) = 1,\text{then }(gof)(x)$ is equal to

• A. 1
• B. –1
• C. 2
• D. – 2

Answer 1

Answer: (A)

1

Answer 2

$\lbrack x\rbrack - x$

= $\frac{1 - \cos 2x}{2} + \frac{1 - \cos(2x + 2\pi ⥂ / ⥂ 3)}{2} + \frac{1}{2}\{ 2\cos x\cos(x + \pi/3)\}$

= $\frac{1}{2}\lbrack 1 - \cos 2x + 1 - \cos(2x + 2\pi ⥂ / ⥂ 3) + \cos(2x + \pi/3) + \cos\pi/3\rbrack$

= $\frac{1}{2}\left\lbrack \frac{5}{2} - \{\cos 2x + \cos\left( 2x + \frac{2\pi}{3} \right)\} + \cos\left( 2x + \frac{\pi}{3} \right) \right\rbrack$

= $\frac{1}{2}\left\lbrack \frac{5}{2} - 2\cos\left( 2x + \frac{\pi}{3} \right)\cos\frac{\pi}{3} + \cos\left( 2x + \frac{\pi}{3} \right) \right\rbrack = 5/4$ for all x.

∴ $gof(x) = g(f(x)) = g(5/4) = 1$ [$\because$g(5/4) =1 (given)]

Hence, $\frac{1}{x - \lbrack x\rbrack}$ for all x.