Let f x -sin 2 x -sin 2 x -3-cos x cos x -3- h x - f 2 x -1 and g 3-4-1 - Then g - h x isA- xB- - x 2-1C- x 2D- 1

The correct option is D 1f(x)=sin^2x+sin^2 (x+π3)+cos xcos(x+π3) =sin^2x+1 cos^2 (x+π3)+cos xcos nbsp; (x+π3) =1+sin^2x+cos (x+π3)[cos x cos nbsp; (x+π3 ...

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Byju's Answer

Standard XII

Mathematics

General Solution of Trigonometric Equation

Let f x =sin ...

Question

Let $f (x) = {sin}^{2} x + {sin}^{2} (x + \frac{π}{3}) + cos x cos (x + \frac{π}{3})$ , $h (x) = \sqrt{f^{2} (x) - 1}$ and $g (\frac{3}{4}) = 1$ . Then $(g \circ h) (x)$ is

x

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\sqrt{x^{2} - 1}

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x^{2}

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1

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Solution

The correct option is D $1$
$f (x) = {sin}^{2} x + {sin}^{2} (x + \frac{π}{3}) + cos x cos (x + \frac{π}{3}) = {sin}^{2} x + 1 - {cos}^{2} (x + \frac{π}{3}) + cos x cos (x + \frac{π}{3}) = 1 + {sin}^{2} x + cos (x + \frac{π}{3}) [cos x - cos (x + \frac{π}{3})] = 1 + {sin}^{2} x + cos (x + \frac{π}{3}) [2 sin \frac{π}{6} sin (x + \frac{π}{6})]$

$= 1 + {sin}^{2} x + \frac{1}{2} [2 cos (x + \frac{π}{3}) sin (x + \frac{π}{6})]$

$= 1 + {sin}^{2} x + \frac{1}{2} [sin (2 x + \frac{π}{2}) - sin \frac{π}{6}] = 1 + {sin}^{2} x + \frac{1}{2} [cos 2 x - \frac{1}{2}] = 1 + {sin}^{2} x + \frac{1}{2} [1 - 2 {sin}^{2} x - \frac{1}{2}]$
$\Rightarrow f (x) = \frac{5}{4}$

$h (x) = \sqrt{f^{2} (x) - 1} = \sqrt{\frac{25}{16} - 1} = \frac{3}{4}$
$∴ (g \circ h) (x) = g (h (x)) = g (\frac{3}{4}) = 1$

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Similar questions

Q. If

f (x) = s i n^{2} x + s i n^{2} (x + \frac{π}{3}) + c o s x c o s (x + \frac{π}{3}) a n d g (\frac{5}{4}) = 1

, then (gof)(x)=

Q. lf

f (x) = {sin}^{2} x + {sin}^{2} (x + \frac{π}{3}) + cos x cos (x + \frac{π}{3})

and

g (\frac{5}{4}) = 1

g (1) = 0

then

(g o f) (x) =

f (x) = {s i n}^{2} x + s i n^{2} (x + \frac{π}{3}) + c o s (x + \frac{π}{3}) c o s x

and

g (\frac{5}{4}) = 1

then

g (f (x))

Q. Assertion :

f (x) = {sin}^{2} x + {sin}^{2} (x + \frac{π}{3}) + cos x cos (x + \frac{π}{3})

then

f^{'} (x) = 0

Reason: Derivative of a constant function is always zero

Q. If

f (x) = {sin}^{2} x + {sin}^{2} (x + \frac{π}{3}) + cos x cos (x + \frac{π}{3})

and

g (x)

is a one-one function defined in

R \to R

, then

(g o f) (x)

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Let f(x)=sin2x+sin2(x+π3)+cosxcos(x+π3), h(x)=√f2(x)−1 and g(34)=1. Then (g∘h)(x) is

Let $f (x) = {sin}^{2} x + {sin}^{2} (x + \frac{π}{3}) + cos x cos (x + \frac{π}{3})$ , $h (x) = \sqrt{f^{2} (x) - 1}$ and $g (\frac{3}{4}) = 1$ . Then $(g \circ h) (x)$ is