Let x in mathbb{R} be so small that the power | vedclass

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

Question

(A) Given the expression $(1-x)^3(2+x)^6$. Since $x$ is very small,we neglect terms with $x^3$ and higher powers. \ $(1-x)^3 = 1 - 3x + 3x^2$. \ $(2

Vedclass · Accepted Answer

-$80$

Vedclass · Answer

(A) Given the expression $(1-x)^3(2+x)^6$. Since $x$ is very small,we neglect terms with $x^3$ and higher powers. \ $(1-x)^3 = 1 - 3x + 3x^2$. \ $(2+x)^6 = 2^6 + 6 	imes 2^5 	imes x + \frac{6 	imes 5}{2} 	imes 2^4 	imes x^2 = 64 + 192x + 240x^2$. \ Now,multiply the two expansions: \ $(1-3x+3x^2)(64+192x+240x^2) = 64 + 192x + 240x^2 - 192x - 576x^2 + 192x^2$ (neglecting $x^3$ and higher). \ $= 64 + (192-192)x + (240-576+192)x^2 = 64 + 0x - 144x^2$. \ Comparing with $a+bx+cx^2$,we get $a=64, b=0, c=-144$. \ Therefore,$a+b+c = 64 + 0 - 144 = -80$.

Let $x \in \mathbb{R}$ be so small that the powers of $x$ beyond two are insignificant and negligibly small. For such $x$,if $(1-x)^3(2+x)^6$ is approximated by $a+bx+cx^2$,then $a+b+c=$

Similar Questions

If $(1 + ax)^n = 1 + 8x + 24x^2 + ....,$ then the value of $a$ and $n$ is

Find the remainder when $3 \times 7^{22} + 2 \times 10^{22} - 44$ is divided by $18$.

The coefficient of $x^{100}$ in the expansion of $\sum_{j=0}^{200} (1 + x)^j$ is

The number of non-zero terms in the expansion of $(1 + 3\sqrt{2}x)^9 + (1 - 3\sqrt{2}x)^9$ is

The number of terms in the expansion of $(x^{2}+y^{2})^{25}-(x^{2}-y^{2})^{25}$ after simplification is

Vedclass Test Series

Exam Paper Generator

Online Exam Module