Question

If $co{s}^{4}x+si{n}^{4}x-sin2x+\frac{3}{4}si{n}^{2}2x=y$, then choose the correct option.

• A. $y=1$ if $x=\frac{15\pi }{2}$
• B. $y\ne 0$ for any value of x
• C. $y-0$ if $x=15\pi$
• D. $y=1$ if $sin2x=0$

Answer 1

Answer: (A), (B), (D)

The correct options are
A $y=1$ if $x=\frac{15\pi }{2}$
B $y\ne 0$ for any value of x
D $y=1$ if $sin2x=0$

$co{s}^{4}x+si{n}^{4}x-sin2x+\frac{3}{4}si{n}^{2}2x=y$
$(co{s}^{2}x+si{n}^{2}x{)}^2-2co{s}^{2}x.si{n}^{2}x-sin2x+\frac{3}{4}si{n}^{2}2x=y$
$1-\frac{2si{n}^{2}2x}{4}+\frac{3si{n}^{2}2x}{4}-sin2x=y$
$1+\frac{si{n}^{2}2x}{4}-sin2x=y$
Hence
$y=1$ if $sin2x=0$
$x=\frac{n\pi }{2}$ where $nϵN$

and $y\ne 0$ for any values of x.

So options A, B, D are true.