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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 `(x+sqrt(x^(2)-1))^(10)+(x-sqrt(x^(2)-1s))^(10)` is equal toA. `f_(10)(x)`B. `f_(11)(x)+f_(9)(x)`C. `f_(11)(x)`D. `2f_(10)(x)` |
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Answer» Correct Answer - D 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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