Thank you for registering.

One of our academic counsellors will contact you within 1 working day.

Please check your email for login details.
MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping

Define a function h : [0, 1] ? [-1, 1] as h(x) = sin p( 2x 2-1 ). Using Inverse function Theorem, find v -1/ 2. Hint: Show that h(x) is a bijective function. d -1 dx h at x = 0 , x =

Define a function h : [0, 1] ? [-1, 1] as h(x) = sin p( 2x 2-1 ). Using Inverse function Theorem, find
v
-1/ 2.
Hint: Show that h(x) is a bijective function.
d -1
dx h
at x = 0 , x =

Grade:12

1 Answers

bharat bajaj IIT Delhi
askIITians Faculty 122 Points
7 years ago
goes from [-1,1].sin p ({2x^{2}-1}) As x belongs to [0,1],sin p ({2x^{2}-1})=sin p ({2y^{2}-1}) Take f(x) = f(y) for some x,y in domain, At x=0, Answer = 0 \frac{d}{dx} {h^{-1}(x)} = \frac{1}{\frac{d}{dx} h(x))} = \frac{p(4x)}{cos p(2x^{2}-1})} THerefore, x = y. Also, for all b in range, there is some a in the domain. Hence, h(x) is a bijective function. p ({2x^{2}-1})= p ({2y^{2}-1}) It implies that
Thanks
Bharat bajaj
IIT Delhi
askiitians faculty

Think You Can Provide A Better Answer ?

Provide a better Answer & Earn Cool Goodies See our forum point policy

ASK QUESTION

Get your questions answered by the expert for free