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Grade: 11
        
I didn't understand how to solve this question...                                                       .                          
3 months ago

Answers : (2)

Aditya Gupta
825 Points
							
this ques merely looks difficult but is easy, and is a direct application of LAGRANGES MEAN VALUR THEOREM.
let mu=u and lambda=m
note that ∫xf”(x)dx= xf’(x) – f(x) [integration by parts]
so ∫xf”(x)dx from a to b= bf’(b) – f(b) – [af’(a) – f(a)] = bf’(b) – af’(a)+f(a) – f(b) =LHS.......(1)
Now consider the following functions:
g(x)= xf’(x) and h(x)= f(x) [note that g’(x)= f’(x)+xf”(x) and h’(x)= f’(x)]
applying LMVT on both we have
[g(b)–g(a)]/(b-a)=g’(u) for some u in (a,b) and [h(b)–h(a)]/(b-a)=h’(m) for some m in (a,b)
bf’(b)–af’(a)=(b-a)[f’(u)+uf”(u)].....(2)
and f(b)-f(a)=(b-a)f’(m).....(3)
subtracting 3 from 2
bf’(b)–af’(a) – f(b)+f(a)= (b-a)[f’(u)+uf”(u) – f’(m)]
from 1,
LHS=(b-a)[f’(u)+uf”(u) – f’(m)]
note that ∫f”(x)dx from m to u= f’(u) – f’(m)
so that
LHS=(b-a)[∫f”(x)dx from m to u+uf”(u)]
3 months ago
Aditya Gupta
825 Points
							
sorry i answered the wrong ques.
we use formula tan^2a=(1-cos2a)/(1+cos2a)
tan^(pi/16)+tan^2(7pi/16)+tan^(2pi/16)+tan^2(6pi/16)+tan^2(3pi/16)+tan^2(5pi/16)+tan^(4pi/16)
tan^(pi/16)+tan^2(pi/2-pi/16)+tan^2(2pi/16)+tan^2(pi/2-2pi/16)+tan^2(3pi/16)+tan^2(pi/2-3pi/16)+tan^2(pi/4)
tan^2(pi/16)+cot^2(pi/16)+tan^2(2pi/16)+cot^2(2pi/16)+tan^2(3pi/16)+cot^2(3pi/16)+1
we use formula
tan^2a+cot^2a=(1-cos2a)/(1+cos2a)+(1+cos2a)/(1-cos2a)
=2(1+cos^2(2a))/sin^2(2a))
2(1+cos^2(pi/8))/sin^2(pi/8) +2(1+cos^2(pi/4))/sin^2(pi/4) +2(1+cos^2(3pi/8))/sin^2(3pi/8) +1
2(cosec^2(pi/8)+cot^2(pi/8)+cosec^2(3pi/8)+cot^2(3pi/8)+2(1+1/2)/(1/2) +1
2(1+2cot^2(pi/8)+1+2cot^2(3pi/8))+6+1
4+6+1+4(cot^2(pi/8)+cot^2(3pi/8))
11+4{(1+cospi/4)/(1-cos(pi/4)) + (1+cos(3pi/4))/(1-cos(3pi/4)}
11+4{(1+1/sqrt2)/(1-1/sqrt2) +(1-1/sqrt2)/(1+1/sqrt2)}
11+4{2*3/2)/(1/2)
11+4(6)
11+24
35 Ans.
3 months ago
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