Question

The coefficient of $x^{8}$ in the expansion of $(x^{2}+\sqrt{1-x^{2}})^{5}+(x^{2}-\sqrt{1-x^{2}})^{5}$ is

• A. -20
• B. -10
• C. -30
• D. -40
• E. 20

Hint:

Math Tip: When expanding $(A+B)^n + (A-B)^n$, all terms with odd powers of $B$ will cancel out perfectly, saving you from calculating half of the binomial theorem expansion.

Answer 1

The Correct Option is A

Concept:
Algebra - Binomial Expansion Identity.
$(a+b)^n + (a-b)^n = 2 [^nC_0 a^n + ^nC_2 a^{n-2}b^2 + ^nC_4 a^{n-4}b^4 + \dots]$
Step 1: Identify the components of the binomial expansion.
Here, let $a = x^2$, $b = \sqrt{1-x^2}$, and $n = 5$.
Step 2: Apply the identity formula.
Since we are adding conjugate binomials to the 5th power, the odd power terms cancel out: $$ 2 [^5C_0 (x^2)^5 (\sqrt{1-x^2})^0 + ^5C_2 (x^2)^3 (\sqrt{1-x^2})^2 + ^5C_4 (x^2)^1 (\sqrt{1-x^2})^4] $$
Step 3: Simplify the individual terms inside the bracket.
Evaluate combinations and powers:

• Term 1: $1 \cdot x^{10} \cdot 1 = x^{10}$
• Term 2: $10 \cdot x^6 \cdot (1-x^2) = 10x^6 - 10x^8$
• Term 3: $5 \cdot x^2 \cdot (1-x^2)^2 = 5x^2(1 - 2x^2 + x^4) = 5x^2 - 10x^4 + 5x^6$

Step 4: Extract the coefficient of $x^8$.
Examine the expanded terms for $x^8$:

• Term 1 has $x^{10}$ (No $x^8$).
• Term 2 has $-10x^8$. The coefficient here is $-10$.
• Term 3 has max power $x^6$ (No $x^8$).

Step 5: Calculate the final coefficient.
Do not forget the $2$ outside the main bracket from the identity in Step 2: $$ \text{Total Coefficient} = 2 \times (-10) = -20 $$