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let f(x)=min{1-tan^n x,1-sin^n x,1-x^n } xbelongs to(-pi/2,pi/2) where n belong to N.Then the left hand derivative of the f(x)atx=pi/4 is : A.-2n b.-2n+1 c.-n(pi/4)^n-1 D.n(pi/4)^ n-1

let f(x)=min{1-tan^n x,1-sin^n x,1-x^n } xbelongs to(-pi/2,pi/2) where n belong to N.Then the left hand derivative of the f(x)atx=pi/4 is :
A.-2n
b.-2n+1
c.-n(pi/4)^n-1
D.n(pi/4)^ n-1

Grade:12

1 Answers

Jitender Singh IIT Delhi
askIITians Faculty 158 Points
6 years ago
Ans:
f(x) = min(1-sin^{n}x ,1 -x^{n},1-tan^{n}x)
sinx < x < tanx
sin^{n}x < x^{n} < tan^{n}x
-sin^{n}x > -x^{n} > -tan^{n}x
1-sin^{n}x >1 -x^{n} > 1-tan^{n}x
LHD = \lim_{h\rightarrow 0}\frac{((1-tan^{n}(\frac{\pi }{4}-h))-(1-1))}{-h}
LHD = \lim_{h\rightarrow 0}\frac{(1-tan^{n}(\frac{\pi }{4}-h)}{-h}
It is zero by zero indeterminent form. Apply L’Hospital Rule,
LHD = \lim_{h\rightarrow 0}\frac{(ntan^{n-1}(\frac{\pi }{4}-h).sec^{2}(\frac{\pi }{4}-h)}{-1} = -2n
Thanks & Regards
Jitender Singh
IIT Delhi
askIITians Faculty

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