Calculate the lowest form of the expression- b2-11-b2-bb3-

The correct option is D 1 b^2 Given, b^2 11 b^2×bb^3 →b^2 11 b^2×bb^3=b^2 1 (b^2 1)×1b^2 = 1 1×1b^2 = 1 b^2 ...

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Standard XII

Mathematics

Algebra of Complex Numbers

Calculate the...

Question

Calculate the lowest form of the expression: $\frac{b^{2} - 1}{1 - b^{2}} \times \frac{b}{b^{3}}$ .

\frac{1}{b^{2}}

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\frac{1}{- b}

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\frac{b - 1}{- b}

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\frac{1}{- b^{2}}

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Solution

The correct option is D $\frac{1}{- b^{2}}$
Given, $\frac{b^{2} - 1}{1 - b^{2}} \times \frac{b}{b^{3}}$
$\to \frac{b^{2} - 1}{1 - b^{2}} \times \frac{b}{b^{3}} = \frac{b^{2} - 1}{- (b^{2} - 1)} \times \frac{1}{b^{2}}$
$=$ $\frac{1}{- 1} \times \frac{1}{b^{2}}$
$=$ $\frac{1}{- b^{2}}$

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(a) (2a + b − 1) (4a² + b² + 1 − 3ab − 2a)

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Q. If

x = 1 + a + a^{2} + a^{3} + . . . .

\infty (| a | < 1)

and

y = 1 + b + b^{2} + b^{3} + . . .

\infty (| b | < 1)

then

1 + a b + a^{2} b^{2} + a^{3} b^{3} + . . .

\infty = \frac{x y}{x + y - 1}

If x = 1 + a + $a^{2}$ .......... to $\infty$ (|a|<1), y = 1 + b + $b^{2}$ ......... to $\infty$ (|b| < 1), then Z = 1 + ab + $a^{2}$ $b^{2}$ + $a^{3}$ $b^{3}$ ..... to ∞ is

The lowest form of $\frac{(a - b)}{(a^{2} - b^{2})}$ is $\frac{1}{(a + b)}$ .

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Calculate the lowest form of the expression: b2−11−b2×bb3.

The correct option is D 1−b2Given, b2−11−b2×bb3→b2−11−b2×bb3=b2−1−(b2−1)×1b2= 1−1×1b2= 1−b2

Calculate the lowest form of the expression: $\frac{b^{2} - 1}{1 - b^{2}} \times \frac{b}{b^{3}}$ .

The correct option is D $\frac{1}{- b^{2}}$
Given, $\frac{b^{2} - 1}{1 - b^{2}} \times \frac{b}{b^{3}}$
$\to \frac{b^{2} - 1}{1 - b^{2}} \times \frac{b}{b^{3}} = \frac{b^{2} - 1}{- (b^{2} - 1)} \times \frac{1}{b^{2}}$
$=$ $\frac{1}{- 1} \times \frac{1}{b^{2}}$
$=$ $\frac{1}{- b^{2}}$