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Consider f(x)=|1-x| \quad 1 \leq x \leq 2 and

g(x)=f(x)+b \sin \frac{\pi}{2} x, \quad 1 \leq x \leq 2,
then which of the following is correct?
 
 

Option: 1

Rolle's theorem is applicable to both f, g and b=\frac{3}{2}


Option: 2

LMVT is not applicable to f and Rolle's theorem is applicable to g with b=\frac{1}{2}


Option: 3

LMVT is applicable to f and Rolle's theorem is applicable to g with b=1
 


Option: 4

Rolle's theorem is not applicable to both f, g for any real b


Answers (1)

best_answer

f(x)=x-1,1 \leq x \leq 2
g(x)=x-1+b \sin \frac{\pi}{2} x, \quad 1 \leq x \leq 2f(1)=0 ; f(2)=1
\Rightarrow Rolle's theorem is not applicable to ' f ' but LMVT is applicable to f.
(\because x-1 is continuous and differentiable in [1,2] and (1,2) respectively).

Now, g(1)=b ; g(2)=1 and function x-1, \sin \frac{\pi}{2} x are both continuous in [1,2] and (1,2).
\because For Rolle's theorem to be applicable to g. We must have b=1.

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Gaurav

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