1.

If the lines : L_(1):x=1+s,y=-3-lambdas,z=1+lambdas & L_(2):x=(t)/(2),y=1+t,z=2-t with the parameters s & t respectively are coplanar then lambda=

Answer»

0
`-1`
`-(1)/(2)`
`-2`

Solution :`f(f(X))`
`(x+x^(2)+x^(4)+x^(8)+........)^(1)rarr "Co-efficient of" x^(10)=0`
`+`
`(x+x^(2)+x^(4)+x^(8)+........)^(2)rarr "Co-efficient of" x^(10)=2 (x^(2).x^(8),x^(8).x^(2))`
`+`
`(x+x^(2)+x^(4)+x^(8)+........)^(4)rarr "Co-efficient of" x^(10)(2+2+2+4=10):(4!)/(3!)=4(x^(2).x^(2).x^(2).x^(4),x.x.x^(4).x^(4))`
`+ ""(1+1+4+4=10):(4!)/(2!2!)=6`
`(x+x^(2)+x^(4)+x^(8)+........)^(8)rarr "Co-efficient of" x^(10)(1+1+1+1+1+1+2+2):(8!)/(2!6!)=28`
+
`{:( : ),( : ):}}"Co-efficient of" x^(10)=0`
`therefore "Co-efficient of" x^(10)=2+4+6+28=40`


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