Review exercise

On the following page we have given the solution of Review exercise of Mathematics 9 (Science) published by Caravan Book House, Lahore.

Chose the correct answers.

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  <question title="(i). H.C.F. of $p^3q-pq^3$ and is $p^5q^2-p^2q^5$ is ..." type="radio"> A. $p q(p^2-q^2)$ | B. $p q(p-q)$| C. $p^2q^2((p-q)$ | D. $pq(p^3-q^3)$ </question>
      <question title="(ii) H.C.F. of $x^2 y^2$ and is $20 x^3 y^3$ is ..." type="radio"> A. $5 x^2 y^2$ | B. $20 x^3 y^3$| C. $100 x^5 y^5$ | D. $5 xy$</question>

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(iii) H.C.F. of $x-2$ and is $x^2+x-6$ is —

(a) $x^2+x-6$ (b) $x+3$

© $x-2$ (d) $x+2$

Answer:

$c$

(iv) H.C.F. of $a^3+b^3$ and is $a^2-ab+b^2$ is —

(a) $a+b$ (b) $a^2-ab+b^2$

© $(a-b)^2$ (d) $a^2+b^2$

Answer:

$c$

(v) H.C.F. of $x^2-5x+6$ and is $x^2-x-6$ is —

(a) $x-3$ (b) $x+2$

© $x^2-4$ (d) $x-2$

Answer:

$a$

(vi) H.C.F. of $a^2-b^2$ and is $a^3-b^3$ is —

(a) $a-b$ (b) $a+b$

© $a^2+ab+b^2$ (d) $a^2-ab+b^2$

Answer:

$a$

(vii) H.C.F. of $x^2+3x+2$ ,$x^2+4x+3$ and is $x^2+5x+4$ is —

(a) $x+1$ (b) $(x+1)(x+2)$

© $x+3$ (d) $(x+4)((x+1)$

Answer:

$a$

(viii) L.C.M. of $15x^2$ ,$45xy$ and is $30xyz$ is —

(a) $90xyz$ (b) $90x^2yz$

© $15xyz$ (d) $15x^2yz$

Answer:

$b$

(ix) L.C.M. of $a^2+b^2$ ,$45xy$ and is $a^4-b^4$ is —

(a) $a^2+b^2$ (b) $a^2-b^2$

© $a^4-b^4$ (d) $a-b$

Answer:

$c$

(x) The product of two algebraic expressions is equal to the — of their H.C.F. and L.C.M.

(a) Sum (b) Difference

© Product (d) Quotient

Answer:

$c$

(xi) Simplify $\frac{a}{9a^2-b^2}+\frac{1}{3a-b}$= —

(a) $\frac{4a}{9a^2-b^2}$ (b) $\frac{4a-b}{9a^2-b^2}$

© $\frac{4a+b}{9a^2-b^2}$ (d) $\frac{b}{9a^2-b^2}$

Answer:

$c$

(xii) Simplify $\frac{a^2+5a-14}{a^2-3a-18}\times \frac{a+3}{a-2}$= —

(a) $\frac{a+7}{a-6}$ (b) $\frac{a+7}{a-2}$

© $\frac{a+3}{a-6}$ (d) $\frac{a-2}{a+3}$

Answer:

$a$

(xiii) Simplify $\frac{a^3-b^3}{a^4-b^4}\div \frac{a^2+ab+b^2}{a^2+b^2}$= —

(a) $\frac{1}{a+b}$ (b) $\frac{1}{a-b}$

© $\frac{a-b}{a^2+b^2}$ (d) $\frac{a+b}{a^2+b^2}$

Answer:

$a$

(xiv) Simplify $\left(\frac{2x+y}{x+y}-1\right)\div \left(1-\frac{x}{x+y}\right)$= —

(a) $\frac{x}{x+y}$ (b) $\frac{y}{x+y}$

© $\frac{y}{x}$ (d) $\frac{x}{y}$

Answer:

$d$

(xv) The square root of $a^2-2a+1$ is —

(a) $\pm (a+1)$ (b) $\pm (a-b)$

© $(a-1)$ (d) $(a+1)$

Answer:

$b$

(xvi) What should be added to complete the square of $x^4+64$ ? —

(a) $8 x^2$ (b) $-8x^2$

© $16x^2$ (d) $4x^2$

Answer:

$c$

(xvii) The square root of $x^4+\frac{1}{x^4}+2$ is —

(a) $\pm \left(x+\frac{1}{x}\right)$ (b) $\left(x^2-\frac{1}{x^2}\right)$

© $\pm \left(x-\frac{1}{x}\right)$ (d) $\pm \left(x^2-\frac{1}{x^2}\right)$

Answer:

$b$

Find the H.C.F. of the following by factorization. $8x^4-128$ , $12x^3-96$

Solution:

$\begin{align}8x^4-128 &= 8(x^4-16)\\&=8[(x^2)^2-(4)^2]\\&= 2 \times 2\times 2 \times (x^2+4)(x^2-4)\\&=2 \times 2\times 2 \times (x^2+4)(x-2)(x+2)\end{align}$

$\begin{align}12 x^3-96&=12(x^3-8)\\&=2 \times 2\times 3 \times (x-2)(x^2+ 2 x+4)\end{align}$

$\begin{align} H.C.F. &= 2 \times 2 (x-2)\\&= 4(x-2)\end{align}$

Find the $L.C.M.$ of the following by factorization.

$12x^2-75, 6x^2-13x-5, 4x^2-20x+25$

Solution:

$\begin{align}12x^2-75 &= 3(4x^2-25)\\&=3[(2x)^2-(5)^2]\\&= 3(2x+5)(2x-5)\end{align}$

$\begin{align}6x^2-13x-5 &= 6x^2-15x+2x-5\\&=3x(2x-5)+1(2x-5)\\&= (2x-5)(3x+1)\end{align}$

$\begin{align}4x^2-20x+25 &= 4x^2-10x-10x+25\\&=2x(2x-5)-5(2x-5)\\&= (2x-5)(2x-5)\end{align}$

$\begin{align} L.C.M. &= 3(2x+5)(2x-5)(3x+1)(2x-5)\\&= 3(2x+5)(2x-5)^2(3x+1)\end{align}$

If H.C.F. of $x^4+3x^3+5x^2+26x+56$ and $x^4+2x^3-4x^2-x+28$ is $x^2+5x+7$, find their L.C.M.

Solution:

$\begin{align}p(x)&= x^4+3 x^3+5 x^2+26 x+56\end{align}$

$\begin{align}q(x)&= x^4+2 x^3-4 x^2-x+28\end{align}$

$\begin{align}L.C.M. &= ?\end{align}$

$\begin{align}L.C.M.& =\frac{p(x) \times q(x)}{H.C.F.}\\&=\frac{(x^4+3x^3+5x^2+26x+56)(x^4+2x^3-4x^2-x+28)}{x^2+5x+7}\\&= \frac{x^4+3x^3+5x^2+26x+56}{x^2+5x+7}\\&=x^2-2x+8\end{align}$

$\begin{align} L.C.M. &= x^2-2x+8\end{align}$

Simplify:

(i) $\frac{3}{x^3+x^2+x+1}-\frac{3}{x^3-x^2+x-1}$

(ii) $\frac{a+b}{a^2-b^2} \div \frac{a^2-ab}{a^2-2ab+b^2}$

Solution:

(i) $\frac{3}{x^3+x^2+x+1}-\frac{3}{x^3-x^2+x-1}$

$\begin{align}\frac{3}{x^3+x^2+x+1}-\frac{3}{x^3-x^2+x-1}&=\frac{3}{x^2(x+1)+1(x+1)}-\frac{3}{x^2(x-1)+1(x-1)}\\&=\frac{3}{(x+1)(x^2+1)}-\frac{3}{(x-1)(x^2+1)}\\&= \frac{3}{(x^2+1)}\left[\frac{1}{x+1}-\frac{1}{x-1}\right]\\&= \frac{3}{(x^2+1)}\left[\frac{x-1-x-1}{(x+1)(x-1)}\right]\\&= \frac{3}{(x^2+1)}\left[\frac{-2}{(x+1)(x-1)}\right]\\&=\frac{-6}{(x^2+1)(x^2-1)}\end{align}$

*Solution:
(ii) $\frac{a+b}{a^2-b^2} \div \frac{a^2-ab}{a^2-2ab+b^2}$
$\begin{align}\frac{a+b}{a^2-b^2} \div \frac{a^2-ab}{a^2-2ab+b^2}&= \frac{a+b}{(a-b)(a+b)} \div \frac{a(a-b)}{(a-b)^2}\\&=\frac{1}{(a-b)} \div \frac{a}{(a-b)}\\&= \frac{1}{(a-b)} \times \frac{(a-b)}{a}\\&= \frac{1}{a}\end{align}$
====Question 7:==== Find square root by using factorization.
$\left(x^2+\frac{1}{x^2}\right)+10\left(x+\frac{1}{x}\right)+27, (x \neq 0)$
Solution:**

$\begin{align}\left(x^2+\frac{1}{x^2}\right)+10\left(x+\frac{1}{x}\right)+27&= \left(x^2+\frac{1}{x^2}\right)+10\left(x+\frac{1}{x}\right)+27-2+2\\&=\left(x+\frac{1}{x}\right)^2+2\left(x+\frac{1}{x}(5)\right)+25\\&=\left(x+\frac{1}{x}\right)^2+2\left(x+\frac{1}{x}(5)\right)+(5)^2\\&= \left(x+\frac{1}{x}+5\right)^2\end{align}$

$\begin{align} \sqrt{ \left(x^2+\frac{1}{x^2}\right)+10\left(x+\frac{1}{x}\right)+27 }&=\pm \left(x+\frac{1}{x}+5\right)\end{align}$