Binomial Theorem
Binomial Theorem
nta_pyq_2025_jan
Grade 11
Question:
Let $\alpha,\beta,\gamma$ and $\delta$ be the coefficients of $x^{7},x^{5},x^{3}$ and $x$ respectively in the expansion of $(x+\sqrt{x^{3}-1})^{5}+(x-\sqrt{x^{3}-1})^{5},\,x>1.$ If $u$ and $v$ satisfy $\alpha u+\beta v=18$ and $\gamma u+\delta v=20$, then $u+v$ equals:
Step-by-Step Solution
Key Concept: $(A+B)^{n}+(A-B)^{n}=2\bigl[\binom{n}{0}A^{n}+\binom{n}{2}A^{n-2}B^{2}+\dots\bigr]$ keeps only even powers of $B$. With $B=\sqrt{x^{3}-1}$, $B^{2}=x^{3}-1$ is polynomial, so the whole expression is a polynomial in $x$.
$(x+\sqrt{x^{3}-1})^{5}+(x-\sqrt{x^{3}-1})^{5}=2\!\left[\binom{5}{0}x^{5}+\binom{5}{2}x^{3}(x^{3}-1)+\binom{5}{4}x(x^{3}-1)^{2}\right].$
Expand:
$2\bigl[x^{5}+10x^{3}(x^{3}-1)+5x(x^{6}-2x^{3}+1)\bigr]$
$=2\bigl[x^{5}+10x^{6}-10x^{3}+5x^{7}-10x^{4}+5x\bigr]$
$=10x^{7}+20x^{6}+2x^{5}-20x^{4}-20x^{3}+10x.$
Coefficients: $\alpha=10,\,\beta=2,\,\gamma=-20,\,\delta=10.$
Solve: $10u+2v=18$ and $-20u+10v=20\Rightarrow 5u+v=9,\ -2u+v=2.$
Subtract: $7u=7\Rightarrow u=1,\,v=4.$ Hence $u+v=5.$
Correct Answer: 1