Please solve q 33?

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Answer 1 · 2018-04-11T17:33:10

The answer is #"option (b)"#

Let

#x/(b+c-a)=k#

#y/(c+a-b)=k#

#z/(a+b-c)=k#

Therefore,

#x=(b+c-a)k#

#y=(c+a-b)k#

#z=(a+b-c)k#

So,

#x(b-c)+y(c-a)+z(a-b)#

#=k(b+c-a)(b-c)+k(c+a-b)(c-a)+k(a+b-c)(a-b)#

#=k(b^2-c^2-a(b-c)+c^2-a^2-b(c-a)+a^2-b^2-c(a-b))#

#=0#

The answer is #"option (b)"#

#"Question (31)"#

Let

#a/(b+c)=k#, #=>#, #a=(b+c)k#

#b/(c+a)=k#, #=>#, #b=(a+c)k#

#c/(a+b)=k#, #=>#, #c=(a+b)k#

Therefore,

#(a+b+c)=(b+c)k+(a+c)k+(a+b)k#

#(a+b+c)=(bk+ck)+(ak+ck)+(ak+bk)#

#(a+b+c)=2k(a+b+c))#

#k=1/2#

Similarly if #k'=1/k#

#k'=2#

The answer is #"option (d)"#