2480
1240
620
124
Correct answer is B
Given the first and last term of an A.P, the sum of the terms is given by
\(S_{n} = \frac{n}{2} [a + l]\)
where a = first term; l = last term and n = number of terms.
\(\therefore S_{20} = \frac{20}{2} [7 + 117]\)
= \(10 (124)\)
= 1240
At what value of x is the function y = x2 - 2x - 3 minimum?
1
-1
-4
4
Correct answer is A
For y = ax2 - bx + c for minimum y
\(\frac{dy}{dx}\) = 2x - 2
= \(\frac{dy}{dx}\) = 0
∴ 2x - 2 = 0
x = 1
Find the gradient of the line passing through the points (-2, 0) and (0, -4)
2
-4
-2
4
Correct answer is C
Given (-2, 0) and (0, -4)
Gradient = \(\frac{-4 - 0}{0 - (-2)}\)
= \(\frac{-4}{2}\)
= -2
Evaluate x2(x2 - 1)-\(\frac{1}{2}\) - (x2 - 1)\(\frac{1}{2}\)
(x2 - 1)-\(\frac{1}{2}\)
(x2 - 1)1
(x2 - 1)
(x2 - 1)-1
Correct answer is A
x2(x2 - 1)-\(\frac{1}{2}\) - (x2 - 1)\(\frac{1}{2}\) = \(\frac{x^2}{(x^2 - 1)^\frac{1}{2}}\) - \(\frac{(x^2 - 1)^\frac{1}{2}}{1}\)
= \(\frac{x^2 - (x^2 - 1)}{(x^2 - 1) ^\frac{1}{2}}\)
= \(\frac{x^2 - x^2 + 1}{(x^2 - 1)^\frac{1}{2}}\)
= (x2 - 1)-\(\frac{1}{2}\)
Simplify \(\frac{\sqrt{1 + x} + \sqrt{x}}{\sqrt{1 + x} - \sqrt{x}}\)
-2x - 2\(\sqrt{x (1 + x)}\)
1 + 2x + 2\(\sqrt{x (1 + x)}\)
\(\sqrt{x (1 + x)}\)
1 + 2x - 2\(\sqrt{x (1 + x)}\)
Correct answer is B
\(\frac{\sqrt{1 + x} + \sqrt{x}}{\sqrt{1 + x} - \sqrt{x}}\)
= \((\frac{\sqrt{1 + x} + \sqrt{x}}{\sqrt{1 + x} - \sqrt{x}}) (\frac{\sqrt{1 + x} + \sqrt{x}}{\sqrt{1 + x} + \sqrt{x}})\)
= \(\frac{(1 + x) + \sqrt{x(1 + x)} + \sqrt{x(1 + x)} + x}{(1 + x) - x}\)
= \(\frac{1 + 2x + 2\sqrt{x(1 + x)}}{1}\)
= \(1 + 2x + 2\sqrt{x(1 + x)}\)