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  • Let , <img alt="\mathrm{f(x)=\lim _{x \rightarrow 0} \frac{[\cos (x+4 \theta)-4 \cos (x+3 \theta)+6 \cos (x+2 \theta)-4 \cos (x+\theta)+\cos x]}{\theta^4}}" src="https://entrancecorner.oncodecogs.c

Let , \mathrm{f(x)=\lim _{x \rightarrow 0} \frac{[\cos (x+4 \theta)-4 \cos (x+3 \theta)+6 \cos (x+2 \theta)-4 \cos (x+\theta)+\cos x]}{\theta^4}}then absolute value of difference between maximum value of \mathrm{f(x)} is

Option: 1

2


Option: 2

0


Option: 3

1


Option: 4

5


Answers (1)

best_answer

\mathrm{f(x)=\lim _{\theta \rightarrow 0} \frac{(\cos (x+4 \theta)+\cos x)-4(\cos (x+3 \theta)+\cos (x+\theta))+6 \cos (x+2 \theta)}{ \theta^4} }
\mathrm{ =\lim _{\theta \rightarrow 0} \frac{2 \cos (x+2 \theta) \cos 2 \theta-8 \cos (x+2 \theta) \cos \theta+6 \cos (x+2 \theta)}{\theta^4} }

\mathrm{=\lim _{\theta \rightarrow 0} \frac{2 \cos (x+2 \theta)\left(2 \cos ^2 \theta-4 \cos \theta+2\right)}{\theta^4} }

\mathrm{ =\lim _{\theta \rightarrow 0} 4 \cos (x+2 \theta) \frac{(1-\cos \theta)^2}{\theta^4}=\cos x }

So, maximum value of \mathrm{ f(x)=1 } and minimum value of \mathrm{ f(x)=-1 }

\mathrm{\therefore}  Difference between maximum and minimum value \mathrm{ =2}

 

Posted by

Shailly goel

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