Question

The expression ($\sqrt{2x^{2} + 1}$+$\sqrt{2x^{2} - 1}$)⁶ + $\left( \frac{2}{\sqrt{2x^{2} + 1} + \sqrt{2x^{2} - 1}} \right)^{6}$is a polynomial of degree-

• A. 5
• B. 6
• C. 7
• D. None of these

Answer 1

Answer: (B)

6

Answer 2

($\sqrt{2x^{2} + 1}$+$\sqrt{2x^{2} - 1}$)⁶+($\sqrt{2x^{2} + 1}$–$\sqrt{2x^{2} - 1}$)⁶

= 2 (T₁ + T₃ + T₅)

Degree of T₁ ŗ ($\sqrt{2x^{2} + 1}$)⁶ ŗ 6

Degree of T₃ ŗ ($\sqrt{2x^{2} + 1}$)⁴ ($\sqrt{2x^{2} - 1}$)²

= (2x² + 1)² (2x² –1) ŗ 6

Degree of T₅ ŗ ($\sqrt{2x^{2} + 1}$)² ($\sqrt{2x^{2} - 1}$)⁴

= (2x² + 1) (2x² –1)² ŗ 6