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Grade: 12
        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 =
6 years ago

Answers : (1)

bharat bajaj
IIT Delhi
askIITians Faculty
122 Points
							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
6 years ago
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