1.

If n be a natural number define polynomial f_(n)(x) of n^(th) degree as follows f_(n)(costheta)cosntheta i.e. f_(2)(x)=2x^(2)-1 f_(3)(x)=4x^(3)-3x_(1) Then f_(6)(x) is equal to

Answer»

`36x^(6)-48X^(4)+18X^(2)-5`
`32x^(6)-48x^(4)+18x^(2)-1`
`36x^(6)-45X^(4)+18x^(2)-8`
`36x^(6)-48x^(4)+18x^(2)-7`

SOLUTION :As, `f(x)=x` & `cos(n+1)theta+cos(n-1)theta=2cos(ntheta).COSTHETA`
`f_((n+1))(x)+f_((n-1))=2x.f_(n)(x)`
`f_(n)(x)=1/(2x)[f_(n+1)(x)+f_(n-1)(x)]`
Now, Put `x=costheta, impliessqrt(x^(2)-1)=isintheta`
`(x+sqrt(x^(2)-1))^(10)+(x-sqrt(x^(2)-1))^(10)=(costheta+isintheta)^(10)+(costheta-isintheta)^(10)`
`=2cos(10theta)=2f_(10)(costheta)=2f_(10)(x)`


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