CameraIcon

SearchIcon

MyQuestionIcon

1

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

Inverse of a Function

If f(x)=sinx+...

Question

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

A

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

B

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

C

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

D

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

Open in App

Solution

The correct option is B
$g (f (x)) = (s i n x + c o s x)^{2} - 1 = s i n 2 x$ which is invertible in $[- \frac{π}{4}, \frac{π}{4}]$ because only in this domain the function is one one and onto. When we go out of this domain the function starts repeating itself, and hence more than one poins in co domain will have same pre-image in the domain

Suggest Corrections

thumbs-up

0

Similar questions

g (f (x)) = |\sin x| and f (g (x)) = {(\sin \sqrt{x})}^{2}, then

f (x) = \sin^{2} x, g (x) = \sqrt{x}

f (x) = \sin x, g (x) = | x |

f (x) = x^{2}, g (x) = \sin \sqrt{x}

(d) f and g cannot be determined.

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

g (f (x))

in invertible in the Domain

f (x) = sin x + cos x

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

g (f (x))

is invertible in the domain.

f (x) = sin x + cos x

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

g {f (x)}

is invertible in the domain

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

, then g{f(x)} is invertible in the domain

View More

Join BYJU'S Learning Program

similar_icon

Related Videos

lock

Inverse of a Function

MATHEMATICS

Watch in App

explore_more_icon

Explore more

Inverse of a Function

Standard IX Mathematics

Join BYJU'S Learning Program

CrossIcon