If f x -sin x -cos x - g x - x 2-1- then g f x is invertible in the domainA- -4- -4-B- -2- -2- -0- -D- -0- -2-

By definition of composition of function, g(f (x)) = (sin x + cos x)^2 1, is invertible (i.e. bijective) ⇒ gf(x) = sin 2x is bijective. We know, sin x is bijec ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XI

Mathematics

Continuity in an Interval

If f x =sin x...

Question

If $f (x) = s i n x + c o s x, g (x) = x^{2} - 1$ , then g{f(x)} is invertible in the domain

$[0, \frac{π}{2}]$

No worries! We‘ve got your back. Try BYJU‘S free classes today!

[- \frac{π}{4}, \frac{π}{4}]

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

[- \frac{π}{2}, \frac{π}{2}]

No worries! We‘ve got your back. Try BYJU‘S free classes today!

[0, π]

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $[- \frac{π}{4}, \frac{π}{4}]$
By definition of composition of function,
g(f (x)) $= (s i n x + c o s x)^{2} - 1$ , is invertible (i.e. bijective)
$\Rightarrow$ g{f(x)} = sin 2x is bijective.
We know, sin x is bijective, only when $x ϵ [- \frac{π}{2}, \frac{π}{2}]$
Thus, g{f(x)} is bijective, if - $\frac{π}{2} \leq 2 x \leq \frac{2 x}{2}$
$\Rightarrow - \frac{π}{4} \leq x \leq \frac{π}{4}$

Suggest Corrections

Similar questions

Q. If

f (x) = s i n x + c o s x, g (x) = x^{2} - 1 t h e n g (f (x))

in invertible in the Domain

Q. Let f(x) = sin x + cos x,

g (x) = x^{2} - 1

. Thus g(f(x)) is invertible for x

\in

Let $f (x) = s i n x + c o s x$ , $g (x) = x^{2} - 1$ . Thus $g (f (x))$ is invertible for $x ϵ$

{x \in R : | cos x | \geq sin x} \cap [0, \frac{3 π}{2}] =

Q. Verify Rolle's theorem for each of the following functions on the indicated intervals
(i) f(x) = cos 2 (x − π/4) on [0, π/2]

(ii) f(x) = sin 2x on [0, π/2]

(iii) f(x) = cos 2x on [−π/4, π/4]

(iv) f(x) = e^x sin x on [0, π]

(v) f(x) = e^xcos x on [−π/2, π/2]

(vi) f(x) = cos 2x on [0, π]

(vii) f(x) =

\frac{\sin x}{e^{x}}

on 0 ≤ x ≤ π

(viii) f(x) = sin 3x on [0, π]

(ix) f(x) =

{e^{1 - x}}^{2}

on [−1, 1]

(x) f(x) = log (x² + 2) − log 3 on [−1, 1]

(xi) f(x) = sin x + cos x on [0, π/2]

(xii) f(x) = 2 sin x + sin 2x on [0, π]

(xiii)

f (x) = \frac{x}{2} - \sin \frac{π x}{6} on [- 1, 0]

(xiv)

f (x) = \frac{6 x}{π} - 4 \sin^{2} x on [0, π / 6]

(xv) f(x) = 4^sin^x on [0, π]

(xvi) f(x) = x² − 5x + 4 on [1, 4]

(xvii) f(x) = sin⁴ x + cos⁴ x on

[0, \frac{π}{2}]

(xviii) f(x) = sin x − sin 2x on [0, π]

Join BYJU'S Learning Program

If f(x)=sinx+cosx,g(x)=x2−1, then g{f(x)} is invertible in the domain

The correct option is B [−π4,π4]By definition of composition of function, g(f (x)) =(sinx+cosx)2−1, is invertible (i.e. bijective) ⇒ g{f(x)} = sin 2x is bijective. We know, sin x is bijective, only when xϵ[−π2, π2] Thus, g{f(x)} is bijective, if - π2≤2x≤2x2 ⇒ −π4≤x≤π4

If $f (x) = s i n x + c o s x, g (x) = x^{2} - 1$ , then g{f(x)} is invertible in the domain