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Quantitative Aptitude

Functions

If | b | ≥ 1 ...

Question

$I f | b | \geq 1$ and x = - |a| b, then which one of the following is necessarily true?

A

a - x b < 0

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B

a - x b \geq 0

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C

a - x b > 0

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D

a - x b \leq 0

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Solution

The correct option is B $a - x b \geq 0$
$| b | \geq 1 \Rightarrow b \geq 1 o r b \leq - 1$
x = - |a| b (given)
Consider a - xb = a - (-|a|b)b
$= a + | a | b^{2} \geq 0$ since, $b^{2} \geq 1$

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Similar questions

∣ ∣ ∣ ∣ \begin{matrix} a + x & b + x & c + x b + x & c + x & a + x c + x & a + x & b + x \end{matrix} ∣ ∣ ∣ ∣ = 0

x =

\frac{y + z}{a} = \frac{z + x}{b} = \frac{x + y}{c}

\frac{x}{b + c - a} = \frac{y}{c + a - b} = \frac{z}{a + b - c}

\frac{x}{b + c - a} = \frac{y}{c + a - b} = \frac{z}{a + b - c} .

then show that x(b

-

-

-

Q. Prove that:

(i)

\frac{1}{1 + x^{a - b}} + \frac{1}{1 + x^{b - a}} = 1

\frac{1}{1 + x^{b - a} + x^{c - a}} + \frac{1}{1 + x^{a - b} + x^{c - b}} + \frac{1}{1 + x^{b - c} + x^{a - c}} = 1

y = \frac{1}{1 + x^{a - b} +^{c - b}} + \frac{1}{1 + x^{b - c} + x^{a - c}} + \frac{1}{1 + x^{b - a} + x^{c - a}}

\frac{d y}{d x}

is equal to
(a) 1
(b)

{(a + b + c)}^{x^{a + b + c - 1}}

(c) 0
(d) none of these

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