Let alpha, beta, gamma and delta be the coeff | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Let $y = \sqrt{x^3-1}$. The expression is $(x+y)^5 + (x-y)^5$. Using the binomial expansion,$(x+y)^5 + (x-y)^5 = 2[\binom{5}{0}x^5 + \binom{5}{2}x

Vedclass · Accepted Answer

$5$

Vedclass · Answer

(A) Let $y = \sqrt{x^3-1}$. The expression is $(x+y)^5 + (x-y)^5$. Using the binomial expansion,$(x+y)^5 + (x-y)^5 = 2[\binom{5}{0}x^5 + \binom{5}{2}x^3y^2 + \binom{5}{4}xy^4]$. Substituting $y^2 = x^3-1$ and $y^4 = (x^3-1)^2 = x^6 - 2x^3 + 1$: $= 2[1 \cdot x^5 + 10 \cdot x^3(x^3-1) + 5 \cdot x(x^6 - 2x^3 + 1)]$. $= 2[x^5 + 10x^6 - 10x^3 + 5x^7 - 10x^4 + 5x]$. $= 10x^7 + 2x^5 - 20x^4 - 20x^3 + 10x$. Comparing coefficients: $\alpha = 10, \beta = 2, \gamma = -20, \delta = 10$. Given equations: $10u + 2v = 18$ and $-20u + 10v = 20$. Divide the first by $2$: $5u + v = 9$. Divide the second by $10$: $-2u + v = 2$. Subtracting the equations: $(5u + v) - (-2u + v) = 9 - 2$ $\Rightarrow 7u = 7$ $\Rightarrow u = 1$. Substituting $u=1$ into $5u+v=9$ gives $v=4$. Therefore,$u+v = 1+4 = 5$.

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 the equations $\alpha u+\beta v=18$ and $\gamma u+\delta v=20$,then $u+v$ equals:

Similar Questions

In the expansion of the expression $1 + (1 + x) + (1 + x)^2 + \dots + (1 + x)^n$,the coefficient of $x^k$ $(0 \le k \le n)$ is

If the sum of odd terms is $A$ and the sum of even terms is $B$ in the expansion of $(x + a)^n$,then:

Let $a$ be the largest real root and $b$ be the smallest real root of the polynomial equation $x^6-6x^5+15x^4-20x^3+15x^2-6x+1=0$. Then $\frac{a^2+b^2}{a+b+1}$ is

If $f(x) = x^n$,then the value of $f(1) - \frac{f'(1)}{1!} + \frac{f''(1)}{2!} - \frac{f'''(1)}{3!} + ...... + \frac{(-1)^n f^n(1)}{n!}$ is

If $f(x) = 1 + nx + \frac{n(n-1)}{2} x^{2} + \frac{n(n-1)(n-2)}{6} x^{3} + \ldots + x^{n}$,then $f^{\prime \prime}(1)$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module