5
4
3
2
Correct answer is D
\(S_{n} = \frac{n}{2}(2a + (n - 1)d \)
\(\frac{n}{2}(2a + (n - 1) d = n^{2} + 2n\)
\(n(2a + (n - 1) d = 2n^{2} + 4n\)
\(2an + n^{2}d - nd = 2n^{2} + 4n\)
Comparing the equations, d = 2 (coefficient of \(n^{2}\)).
\(f^{-1} : x \to \frac{2x + 3}{x - 1}, x \neq 1\)
\(f^{-1} : x \to \frac{x + 3}{x + 2}, x \neq -2\)
\(f^{-1} : x \to \frac{x - 1}{2x + 3}, x \neq -\frac{3}{2}\)
\(f^{-1}: x \to \frac{x - 2}{x + 3}, x \neq -3\)
Correct answer is A
\(f(x) = \frac{x + 3}{x - 2}\)
\(f(y) = \frac{y + 3}{y - 2}\)
Let f(y) = x,
\(x = \frac{y + 3}{y - 2}\)
\(x(y - 2) = y + 3\)
\(xy - y = 2x + 3 \implies y(x - 1) = 2x + 3\)
\(y = \frac{2x + 3}{x - 1}\)
Find the range of values of x for which \(2x^{2} + 7x - 15 > 0\).
\(x < -\frac{3}{2}\) or \(x > 5\)
\(x < -5\) or \(x > \frac{3}{2}\)
\(-\frac{3}{2} < x < 5\)
\(-5 < x < \frac{3}{2}\)
Correct answer is B
\(2x^{2} + 7x - 15 > 0 \implies 2x^{2} - 3x + 10x - 15 > 0\)
\(x(2x - 3) + 5(2x - 3) > 0\)
\((x + 5)(2x - 3) > 0\)
For their product to be positive, they are either both +ve or -ve.
\(x + 5 > 0 \implies x > -5\)
\(2x - 3 > 0 \implies 2x > 3\)
\(x > \frac{3}{2}\)
Check:
\(x > -5: x = -3\)
\(2(-3)^{2} + 7(-3) - 15 = 18 - 21 - 15 = -18 < 0\) (Not satisfied)
\(\therefore x < -5\)
\(x > \frac{3}{2}: x = 2\)
\(2(2^{2}) + 7(2) - 15 = 8 + 14 - 15 = 7 > 0\) (Satisfied)
Simplify \(\frac{\sqrt{3} + \sqrt{48}}{\sqrt{6}}\)
\(3\sqrt{2}\)
\(5\sqrt{2}\)
\(\frac{5\sqrt{2}}{2}\)
\(\frac{3\sqrt{2}}{2}\)
Correct answer is C
\(\frac{\sqrt{3} + \sqrt{48}}{\sqrt{6}} = \frac{\sqrt{3} + 4\sqrt{3}}{\sqrt{6}}\)
\(\frac{5\sqrt{3}}{\sqrt{6} = \frac{5\sqrt{3} \times \sqrt{6}}{\sqrt{6} \times \sqrt{6}}\)
\(\frac{5\sqrt{18}{6} = \frac{15\sqrt{2}}{6}\)
= \(\frac{5\sqrt{2}}{2}\)
\(\frac{\sqrt{3} \tan x - 1}{\sqrt{3} + \tan x}\)
\(\frac{\sqrt{3} \tan x}{\sqrt{3} + \tan x}\)
\(\frac{\sqrt{3} \tan x + 1}{\sqrt{3} - \tan x}\)
\(\frac{3 \tan x - 1}{\sqrt{3} - \tan x}\)
Correct answer is A
No explanation has been provided for this answer.
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