If fx-sin 2 x-sin 2x-3-cos x cosx-3 and g5-4-1- then gofx-

The correct option is D 1f'(x) = 2 sin x cos + 2 sin (x+ π/3) cos (x+ π/3) sin x cos (x + π/3) cos x sin(x + π/3) = sin 2x + sin (2x + 2 π/3) sin (x+x+ π ...

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Byju's Answer

Standard IX

Mathematics

Composite Function

If fx=sin 2 x...

Question

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)=

-2

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Solution

The correct option is D 1
$f^{'} (x) = 2 s i n x c o s + 2 s i n (x + \frac{π}{3}) c o s (x + \frac{π}{3}) - s i n x c o s (x + \frac{π}{3}) - c o s x s i n (x + \frac{π}{3})$
= $s i n 2 x + s i n (2 x + \frac{2 π}{3}) - s i n (x + x + \frac{π}{3})$
= $2 s i n (2 x + \frac{π}{3}) c o s (\frac{π}{3}) - s i n (2 x + \frac{π}{3}) = 0$
Since $f^{^{'}} (x) = 0$ $\Rightarrow f (x) = k$ Where k is a constant.
And $f (0) = s i n^{2} 0 + s i n^{2} (\frac{π}{3}) + c o s 0 c o s (\frac{π}{3}) = \frac{5}{4}$
Thus $f (x) = \frac{5}{4}, \forall x ϵ R .$
Therefore, $(g o f) (x) = g [f (x)] = g (\frac{5}{4}) = 1$ .

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Similar questions

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) =

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)

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. If

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

and

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

then

g o f (x)

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Q. Let

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

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If f(x)=sin2x+sin2(x+π3)+cosx cos(x+π3)and g(54)=1, then (gof)(x)=

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)=