The Expansion - x 2- x 6-11-2-5- x 2- x 6-11-2-5 is a polynomial of degree-A- 8B- 10C- 13D- 14

The correct option is D 14(x^2+(x^6 1)^1/2)^5+(x^2 (x^6 1)^1/2)^5 =2 (^5C_0(x^2)^5+^5C_2(x^2)^3(x^6 1)+^5C_4 x^2(x^6 1)^2) Here last term is of 14 degree. ...

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Byju's Answer

Standard XII

Mathematics

Order of a Differential Equation

The Expansion...

Question

The Expansion $[x^{2} + (x^{6} - 1)^{\frac{1}{2}}]^{5} + [x^{2} - (x^{6} - 1)^{\frac{1}{2}}]^{5}$ is a polynomial of degree.

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Solution

The correct option is D 14
$(x^{2} + (x^{6} - 1)^{\frac{1}{2}})^{5} + (x^{2} - (x^{6} - 1)^{\frac{1}{2}})^{5} = 2 (^{5} C_{0} (x^{2})^{5} +^{5} C_{2} (x^{2})^{3} (x^{6} - 1) +^{5} C_{4} x^{2} (x^{6} - 1)^{2})$
Here last term is of 14 degree.

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Similar questions

The Expansion ${[x^{2} + {(x^{6} - 1)}^{1 / 2}]}^{5} + {[x^{2} - {(x^{6} - 1)}^{1 / 2}]}^{5}$ is a polynomial of degree.

Q. The expansion

[x + (x^{3} - 1)^{1 / 2}]^{5} + [x - (x^{3} - 1)^{1 / 2}]^{5}

is a polynomial of degree

\lim_{x \to 0} \frac{{(1 + x)}^{6} - 1}{{(1 + x)}^{2} - 1}

Q. Evaluate :

{lim}_{x \to 0} \frac{(1 + x)^{6} - 1}{(1 + x)^{2} - 1}

Q.
(a) Prove that

sin [{t a n}^{- 1} \frac{1 - x^{2}}{2 x} + {cos}^{- 1} \frac{1 - x^{2}}{1 + x^{2}}] = 1

(b) If

{sin}^{- 1} (x - \frac{x^{2}}{2} + \frac{x^{3}}{4} - . . . . .) + {cos}^{- 1} (x^{2} - \frac{x^{4}}{2} + \frac{x^{6}}{4} - . . . . . . .) = \frac{π}{2}

for

0 < | x | < \sqrt{2}

, then x equals

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The Expansion [x2+(x6−1)12]5+[x2−(x6−1)12]5 is a polynomial of degree.

The correct option is D 14(x2+(x6−1)12)5+(x2−(x6−1)12)5=2(5C0(x2)5+5C2(x2)3(x6−1)+5C4x2(x6−1)2) Here last term is of 14 degree.

The Expansion $[x^{2} + (x^{6} - 1)^{\frac{1}{2}}]^{5} + [x^{2} - (x^{6} - 1)^{\frac{1}{2}}]^{5}$ is a polynomial of degree.

The correct option is D 14
$(x^{2} + (x^{6} - 1)^{\frac{1}{2}})^{5} + (x^{2} - (x^{6} - 1)^{\frac{1}{2}})^{5} = 2 (^{5} C_{0} (x^{2})^{5} +^{5} C_{2} (x^{2})^{3} (x^{6} - 1) +^{5} C_{4} x^{2} (x^{6} - 1)^{2})$
Here last term is of 14 degree.