Find the fourth term of the binomial expansion of (x−k)5 in descending powers of x.<
10x3k2
5x3k2
−5x2k3
−10x2k3
Correct answer is D
(x−k)5=5C0x5(−k)0+5C1x4(−k)1+...
The fourth term in the expansion = ^{5}C_{4 - 1}(x)^{5 - 3}(-k)^{3 = 10 \times x^{2} \times -k^{3}
= −10x2k3
2xx−3,x≠3
2xx+3,x≠−3
3xx−3,x≠3
3xx+3,x≠−3
Correct answer is A
From the rules of binary operation, x∗e=x
⟹x∗e=3x+3e−xe=x
3e−xe=x−3x=−2x
e=2xx−3,x≠3
Simplify \frac{\tan 80° - \tan 20°}{1 + \tan 80° \tan 20°}
3\sqrt{2}
2\sqrt{3}
\sqrt{3}
\sqrt{2}
Correct answer is C
\tan (x - y) = \frac{\tan x - \tan y}{1 + \tan x \tan y}
\implies \frac{\tan 80 - \tan 20}{1 + \tan 80 \tan 20} = \tan (80 - 20) = \tan 60°
\tan 60 = \frac{\sin 60}{\cos 60} = \frac{\sqrt{3}}{2} ÷ \frac{1}{2}
= \sqrt{3}
Simplify \frac{\sqrt{3}}{\sqrt{3} - 1} + \frac{\sqrt{3}}{\sqrt{3} +1}
\frac{1}{2}
\frac{1}{2}\sqrt{3}
3
2\sqrt{3}
Correct answer is C
\frac{\sqrt{3}}{\sqrt{3} - 1} + \frac{\sqrt{3}}{\sqrt{3} + 1}
= \frac{\sqrt{3}(\sqrt{3} + 1) + \sqrt{3}(\sqrt{3} - 1)}{(\sqrt{3} - 1)(\sqrt{3} + 1)}
= \frac{6}{3 - 1}
= 3
-4
0
4
12
Correct answer is D
Gradient = \frac{y_{1} - y_{2}}{x_{1} - x_{2}}
\frac{1}{2} = \frac{5 - 9}{4 - x}
\frac{1}{2} = \frac{-4}{4 - x} \implies -8 = 4 - x
x = 4 + 8 = 12