- Ax5 + 25x4 + 90x3 - 270x2 + 405x - 243
- Bx5 - 15x4 + 90x3 - 270x2 - 405x - 243
- Cx5 - 15x4 + 80x3 - 270x2 + 405x - 243
- Dx5 - 15x4 + 90x3 - 270x2 + 405x - 243
Solution:
(x - 3)5 = 5C0x5 + 5C1x4 (-3)1 + 5C2x3 (-3)2 + 5C3x2 (-3)3 + 5C4x (-3)4 + 5C5 (-3)5
= x5 -15x4 + 90x3 - 270x2 + 405x - 243
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
If in the expansion of $\Big(\text{x}^{4}-\frac{1}{\text{x}^{3}}\Big)^{15},\text{x}^{-17}$ occurs in rth term, then
Choose the correct answer.
If the middle term of $\Big(\frac{1}{\text{x}}+\text{x}\sin\text{x}\Big)^{10}$ is equal to $7\frac{7}{8},$ then value of x is:
$2\text{n}\pi+\frac{\pi}{6}.$
$\text{n}\pi+\frac{\pi}{6}.$
$\text{n}\pi+(-1)^\text{n}\frac{\pi}{6}.$
$\text{n}\pi+(-1)^\text{n}\frac{\pi}{3}.$
Hint: $\text{T}_6=\ ^{10}\text{C}_5\frac{1}{\text{x}^5}.\text{x}^5\ \sin^5\text{x}=\frac{63}{8}\Rightarrow\sin^5\text{x}=\frac{1}{2^5}\sin\frac{1}{2}$
$\Rightarrow\text{x}=\text{n}\pi+(-1)^\text{n}\frac{\pi}{6}$