Question 13 Marks
If $a + b = 11$ and $a^2+ b^2= 65;$ find $a^3+ b^3.$
Answer$a + b = 11$ and $a^2 + b^2 = 65$
Now, $(a+b)^2 = a^2 + b^2 + 2ab$
$\Rightarrow (11)^2 = 65 + 2ab$
$\Rightarrow 121 = 65 + 2ab$
$\Rightarrow 2ab = 56$
$\Rightarrow ab = 28a^3 + b^3 = ( a + b )( a^2 - ab + b2)$
$= (11)(65 - 28)$
$= 11 \times 37$
$= 407$
View full question & answer→Question 23 Marks
Evaluate$:\frac{1.2 \times 1.2+1.2 \times 0.3+0.3 \times 0.3}{1.2 \times 1.2 \times 1.2-0.3 \times 0.3 \times 0.3}$
Answer$\frac{1.2 \times 1.2+1.2 \times 0.3+0.3 \times 0.3}{1.2 \times 1.2 \times 1.2-0.3 \times 0.3 \times 0.3}$
Let $1.2=\mathrm{a}$ and $0.3=\mathrm{b}$
Then, the given expression becomes
$\frac{a \times a+a+b+b \times b}{a \times a \times a-b \times b \times b}$
$=\frac{a^2+a b+b^2}{a^3-b^3}$
$=\frac{a^2+a b+b^2}{(a-b)\left(a^2+a b+b^2\right)}$
$ =\frac{1}{a-b}$
$=\frac{1}{1.2-0.3}$
$=\frac{1}{0.9}$
$=\frac{10}{9}$
$=1 \frac{1}{9}$
View full question & answer→Question 33 Marks
Evaluate $:\frac{0.8 \times 0.8 \times 0.8+0.5 \times 0.5 \times 0.5}{0.8 \times 0.8-0.8 \times 0.5+0.5 \times .5}$
Answer$\frac{0.8 \times 0.8 \times 0.8+0.5 \times 0.5 \times 0.5}{0.8 \times 0.8-0.8 \times 0.5+0.5 \times .5}$
Let $0.8=a$ and $0.5=b$
Then, the given expression becomes
$\frac{a \times a \times a+b \times b \times b}{a \times a-a \times b+b \times b}$
$ =\frac{a^3+b^3}{a^2-a b+b^2}$
$ =\frac{(a+b)\left(a^2-a b+b^2\right)}{a^2-a b+b^2}$
$ =a+b$
$ =0.8+0.5$
$ =1.3$
View full question & answer→Question 43 Marks
Simplify : $( x + 6 )( x - 4 )( x - 2 )$
AnswerUsing identity :
$(x + a)(x + b)(x + c) =$
$x^3+ (a + b + c)x^2+ (ab+bc+ ca)x +abc$
$(x + 6)(x - 4)(x - 2)=$
$ x^3+ (6 - 4 - 2)x^2+ [6 \times (-4) + (-4) \times (-2) + (-2) \times 6]x + 6 \times (-4) \times (-2)$
$= x^3- 0x^2+ (-24 + 8 - 12)x + 48$
$= x^3- 28x + 48$
View full question & answer→Question 53 Marks
Find : $(a - b)(a - b)(a - b)$
Answer$(a + b)(a + b)(a + b)$
$= (a \times a + a \times b + b \times a + b \times b)(a + b)$
$= (a^2+ab+ab+ b^2)(a + b)$
$= (a^2+ b^2+ 2ab)(a + b)$
$= a^2\times a + a^2\times b + b^2\times a + b^2\times b + 2ab \times a + 2ab \times b$
$= a^3+ a^2b + ab^2+ b^3+ 2a^2b + 2ab^2$
$= a^3+ b^3+ 3a^2b + 3ab^2$
replacing b by $-b$, we get
$= a^3+ (-b)^3+ 3a^2(-b) + 3a(-b)^2$
$= a^3- b^3- 3a^2b + 3ab^2$
View full question & answer→Question 63 Marks
Find $: (a + b)(a + b)(a + b)$
Answer$(a + b)(a + b)(a + b)$
$= (a \times a + a \times b + b \times a + b \times b)(a + b)$
$= (a^2+ab+ab+ b^2)(a + b)$
$= (a^2+ b^2+ 2ab)(a + b)$
$= a^2\times a + a^2\times b + b^2\times a + b^2\times b + 2ab \times a + 2ab \times b$
$= a^3+ a^2b + ab^2+ b^3+ 2a^2b + 2ab^2$
$= a^3+ b^3+ 3a^2b + 3ab^2$
View full question & answer→Question 73 Marks
Prove that : $x^2+ y^2 + z^2 - xy - yz - zx$ is always positive.
Answer$x^2+ y^2+ z^2-xy-yz-zx$
$= 2(x^2+ y^2+ z^2-xy-yz-zx)$
$= 2x^2+ 2y^2+ 2z^2- 2xy - 2yz - 2zx$
$= x^2+ x^2+ y^2+ y^2+ z^2+ z^2- 2xy - 2yz - 2zx$
$= (x^2+ y^2- 2xy) + (z^2+ x^2- 2zx) + (y^2+ z^2- 2yz)$
$= (x - y)^2+ (z - x)^2+ (y - z)^2$
Since square of any number is positive, the given equation is always positive.
View full question & answer→Question 83 Marks
If $3 x-\frac{4}{x}=4 ;$ and $x \neq 0$ find $: 27 x^3-\frac{64}{x^3}$
Answer$3 x-\frac{4}{x}=4$
We need to find $27 x^3-\frac{64}{x^3}$
Let us now consider the expansion of $\left(3 x-\frac{4}{x}\right)^3$ :
$\left(3 x-\frac{4}{x}\right)^3=27 x^3-\frac{64}{x^3}-3 \times 3 x \times \frac{4}{x}\left(3 x-\frac{4}{x}\right)$
$ \Rightarrow(4)^3=27 x^3-\frac{64}{x^3}-144 \quad\left[ Given:3 x-\frac{4}{x}=4\right]$
$ \Rightarrow 64+144=27 x^3-\frac{64}{x^3}$
$ \Rightarrow 27 \times 3-\frac{64}{x^3}=208$
View full question & answer→Question 93 Marks
If $a+\frac{1}{a}=m$ and $a \neq 0;$ find in terms of $' m \ ';$ the value of $ : a-\frac{1}{a}$
AnswerGiven that $\mathrm{a}+\frac{1}{a}=\mathrm{m}$
Now consider the expansion of $\left(a+\frac{1}{a}\right)^2$ :
$\left(a+\frac{1}{a}\right)^2=a^2+\frac{1}{a^2}+2$
$ \Rightarrow \mathrm{m}^2=\mathrm{a}^2+\frac{1}{a^2}+2$
$ \Rightarrow \mathrm{a}^2+\frac{1}{a^2}=\mathrm{m}^2-2$
Now consider the expansion of $\left(a-\frac{1}{a}\right)^2$ :
$\left(a-\frac{1}{a}\right)^2=a^2+\frac{1}{a^2}-2$
$ \Rightarrow\left(a-\frac{1}{a}\right)^2=m^2-2-2$
$ \Rightarrow\left(a-\frac{1}{a}\right)^2=m^2-4$
$ \Rightarrow\left(a-\frac{1}{a}\right)^2$
$= \pm \sqrt{m^2-4}$
View full question & answer→Question 103 Marks
If $x+2 y+3 z=0$ and $x^3+4 y^3+9 z^3=18 x y z;$ evaluate $:\frac{(x+2 y)^2}{x y}+\frac{(2 y+3 z)^2}{y z}+\frac{(3 z+x)^2}{z x}$
AnswerGiven that $x^3+4 y^3+9 z^3=18 x y z$ and $x+2 y+3 z=0$
$x+2 y=-3 z, 2 y+3 z=-x$ and $3 z+x=-2 y$
Now
$\frac{(x+2 y)^2}{x y}+\frac{(2 y+3 z)^2}{y z}+\frac{(3 x+x)^2}{z x}$
$ =\frac{(-3 z)^2}{x y}+\frac{(-x)^2}{y z}+\frac{(-2 y)^2}{z x}$A
$ =\frac{9 z^2}{x y}+\frac{x^2}{y z}+\frac{4 y^2}{z x}$
$ =\frac{x^3+4 y^3+9 z^3}{x y z}$
Given that $x^3+4 y^3+9 z^3=18 x y z$
$\therefore \frac{(x+2 y)^2}{x y}+\frac{(2 y+3 z)^2}{y z}+\frac{(3 z+x)^2}{z x}$
$=\frac{18 x y z}{x y z}$
$=18$
View full question & answer→Question 113 Marks
The difference between two positive numbers is $4$ and the difference between their cubes is $316.$Find $:$ Their product
AnswerGiven difference between two positive numbers is $4$ and difference between their cubes is $316$.
Let the positive numbers be $a$ and $b$
$a - b = 4$
$a^3- b^3= 316$
Cubing both sides,
$(a - b)^3= 64$
$a^3- b^3- 3ab(a - b) = 64$ Given $a^3- b^3= 316$
So $316 - 64 = 3ab(4)$
$252 = 12ab$
So $ab= 21;$ product of numbers is $21$
View full question & answer→Question 123 Marks
If $\frac{x^2+1}{x}=3 \frac{1}{3}$ and $\mathrm{x}>1 ;$ Find If $x-\frac{1}{x}$
AnswerGiven $\frac{x^2+1}{x}=3 \frac{1}{3}$
$\frac{x^2+1}{x}=\frac{10}{3}$
$ {\left[x+\frac{1}{x}\right]=\frac{10}{3}}$
Squaring on both sides, we get
$x^2+\frac{1}{x^2}+2=\frac{100}{9}$
$ x^2+\frac{1}{x^2}=\frac{100-18}{9}=\frac{82}{9}$
$ x-\frac{1}{x}=\sqrt{\left(x+\frac{1}{x}\right)^2-4}=\sqrt{\frac{100}{9}-4}=\sqrt{\frac{64}{9}}=\frac{8}{3}$
$ \therefore x-\frac{1}{x}=\frac{8}{3}$
View full question & answer→Question 133 Marks
If $x=\frac{1}{5-x}$ and $x \neq 5$ find $x^3+\frac{1}{x^3}$
AnswerGiven $\mathrm{x}=\frac{1}{5-x}$;
By cross multiplication
$\Rightarrow x(5-x)=1$
$ \Rightarrow x^2-5 x=-1$
$ \Rightarrow x^2+1=5 x$
$\frac{x^2+1}{x}=5$
$\Rightarrow\left[x+\frac{1}{x}\right]=5 ....(1)$
We know that
$\left(x^3+\frac{1}{x^3}\right)=\left(x+\frac{1}{x}\right)^3-3\left(x+\frac{1}{x}\right)$
$ =(5)^3-3(5) \quad[$ from equation $(1) ] $
$ \Rightarrow{ }^{\wedge} 3+1 / x^{\wedge} 3$
$=125-15$
$=110$
View full question & answer→Question 143 Marks
If $a^2+ b^2+ c^2= 35$ and $ab + bc + ca = 23;$ find $a+ b+ c.$
AnswerWe know that
$(a+b+c)^2=a^2+b^2+c^2+2(a b+b c+c a)\ldots(1)$
Given that, $a^2+b^2+c^2=35$ and $a b+b c+c a=23$
We need to find $a+b+c$ :
Substitute the values of $\left(a^2+b^2+c^2\right)$ and $(a b+b c+c a)$
in the identity $(1),$ we have
$(a+b+c)^2=35+2(23)$
$ \Rightarrow(a+b+c)^2=81$
$ \Rightarrow a+b+c= \pm \sqrt{81}$
$ \Rightarrow a+b+c$
$= \pm 9$
View full question & answer→Question 153 Marks
If $a + b + c = 12$ and $a^2+ b^2+ c^2= 50;$ find $ab + bc + ca.$
AnswerWe know that
$(a+b+c)^2=a^2+b^2+c^2+2(a b+b c+c a) ........(1)$
Given that, $a^2+b^2+c^2=50$ and $a+b+c=12$.
We need to find $a b+b c+c a :$
Substitute the values of $\left(a^2+b^2+c^2\right)$ and $(a+b+c)$
in the identity $(1),$ we have
$(12)^2=50+2(a b+b c+c a)$
$ \Rightarrow 144=50+2(a b+b c+c a)$
$ \Rightarrow 94=2(a b+b c+c a)$
$ \Rightarrow a b+b c+c a=\frac{94}{2}$
$ \Rightarrow a b+b c+c a$
$=47$
View full question & answer→Question 163 Marks
If $a \neq 0$ and $a-\frac{1}{a}=3;$ Find$:a^3-\frac{1}{a^3}$
Answer$a-\frac{1}{a}=3 .$
Taking a cube on both sides,
$\left(a-\frac{1}{a}\right)^3=3^3$
$ a^3-\frac{1}{a^3}-3\left(a-\frac{1}{a}\right)=27 \ldots \ldots \ldots \ldots . . .\left[(a-b)^3=a^3-b^3\right. -3 a b(a-b)]$
$ a^3-\frac{1}{a^3}-3 \times 3=27 \ldots \ldots \ldots \ldots .\left[a-\frac{1}{a}=3\right]$
$ a^3-\frac{1}{a^3}-9=27$
$ a^3-\frac{1}{a^3}=27+9$
$ a^3-\frac{1}{a^3}=36 .$
View full question & answer→Question 173 Marks
If $a + 2b + c = 0;$ then show that :$a^3+ 8b^3+ c^3= 6abc.$
AnswerGiven that $a + 2b + c = 0$ We know that
$\therefore a + 2b = - c ...(1)$
Cubing both sides of the above equation,
$a^3 + b^3 + c^3 = (a + b+ c) (a^2 + b^2 + c^2- ab - bc - ca) + 3abc$
If $a + b + c = 0,$ then,
$a^3 + b^3 + c^3= 3abc$
Here, $a= a, b = 2b$ and $c = c,$
Thus,
$a^3 + (2b)^3+ c^3 = 3(a) (2b) (c) = 6abc$
$\text{L.H.S} = \text{R.H.S}$
View full question & answer→Question 183 Marks
If $\left(a+\frac{1}{a}\right)^2=3$ and $a \neq 0$; then show $: a^3+a^{\frac{1}{3}}=0$
AnswerGiven that $\left(a+\frac{1}{a}\right)^2=3$
$\Rightarrow a+\frac{1}{a}= \pm \sqrt{3} ......(1)$
We need to find $a^3+\frac{1}{a^3} :$
Consider the identity,
$\left(a+\frac{1}{a}\right)^3=a^3+\frac{1}{a^3}+3\left(a+\frac{1}{a}\right)$
$ \Rightarrow a^3+\frac{1}{a^3}=( \pm \sqrt{3})^3-3( \pm \sqrt{3}) \quad[$ From $(1)] $
$ \Rightarrow a^3+\frac{1}{a^3}= \pm 3 \sqrt{3}-3( \pm \sqrt{3})$
$ \Rightarrow a^3+\frac{1}{a^3}$
$=0$
View full question & answer→Question 193 Marks
If $a^2+\frac{1}{a^2}=18 ; a \neq 0$ find$:(i) a-\frac{1}{a}\ (ii) a^3-\frac{1}{a^3}$
Answer$ a^2+\frac{1}{a^2}=18$
$ \left(a-\frac{1}{a}\right)^2=a^2+\frac{1}{a^2}-2$
$ \Rightarrow\left(a-\frac{1}{a}\right)^2=18-2$
$ \Rightarrow\left(a-\frac{1}{a}\right)^2=16$
$ \Rightarrow a-\frac{1}{a}= \pm \sqrt{16}$
$\Rightarrow a-\frac{1}{a}= \pm 4$ ...(1)
$ \text { (ii) }\left(a-\frac{1}{a}\right)^3=a^3-\frac{1}{a^3}-3\left(a-\frac{1}{a}\right)$
$ \Rightarrow a^3-\frac{1}{a^3}=\left(a-\frac{1}{a}\right)^3+3\left(a-\frac{1}{a}\right)$
$ \Rightarrow a^3-\frac{1}{a^3}=( \pm 4)^3+3( \pm 4)[$ From $(1)]$
$ \Rightarrow a^3-\frac{1}{a^3}= \pm 76$
View full question & answer→Question 203 Marks
Two positive numbers $x$ and $y$ are such that $x > y$. If the difference of these numbers is $5$ and their product is $24,$ find$:(i)$ Sum of these numbers$(ii)$ Difference of their cubes$(iii)$ Sum of their cubes.
AnswerGiven $x - y = 5$ and $xy= 24 (x>y)$
$(x + y)^2= (x - y)^2+ 4xy = 25 + 96 = 121$
So, $x + y = 11;$ sum of these numbers is $11.$
Cubing on both sides gives
$(x - y)^3= 5^3$
$x^3- y^3- 3xy(x - y) = 125$
$x^3- y^3- 72(5) = 125$
$x^3- y^3= 125 + 360 = 485$
So, difference of their cubes is $485.$
Cubing both sides, we get
$(x + y)^3= 11^3$
$x^3+ y^3+ 3xy(x + y) = 1331$
$x^3+ y^3= 1331 - 72(11) = 1331 - 792 = 539$
So, sum of their cubes is $539.$
View full question & answer→Question 213 Marks
The sum of two numbers is $9$ and their product is $20$. Find the sum of their $(i)$ Squares $(ii)$ Cubes
AnswerGiven sum of two numbers is $9$ and their product is $20.$
Let the numbers be $a$ and $b.$
$a + b = 9$
$ab= 20$
Squaring on both sides gives
$(a+b)^2= 9^2$
$a^2+ b^2+ 2ab = 81$
$a^2+ b^2+ 40 = 81$
So sum of squares is $81 - 40 = 41$
Cubing on both sides gives
$(a + b)^3= 9^3$
$a^3+ b^3+ 3ab(a + b) = 729$
$a^3+ b^3+ 60(9) = 729$
$a^3+ b^3= 729 - 540 = 189$
So the sum of cubes is $189.$
View full question & answer→Question 223 Marks
If $a \neq 0$ and $a-\frac{1}{a}=4 ;$ find $:\left(a^4+\frac{1}{a^4}\right)$
Answer$\left(a-\frac{1}{a}\right)^2=a^2+\frac{1}{a^2}-2$
$ \Rightarrow a^2+\frac{1}{a^2}=\left(a-\frac{1}{a}\right)^2+2$
$ \Rightarrow a^2+\frac{1}{a^2}=(4)^2+2 \quad\left[\because a-\frac{1}{a}=4\right]$
$ \Rightarrow a^2+\frac{1}{a^2}=18 \ldots(1)$
We know that,
$a^4+\frac{1}{a^4}= \left(a^2+\frac{1}{a^2}\right)^2-2$
$ ={ }^{\prime}(18)^{\wedge} 2-2 \quad[$ From$(1)]$
$ =324-2$
$\Rightarrow a^4+\frac{1}{a^4}$
$=322$
View full question & answer→Question 233 Marks
If $ a - b = 7$ and $ab = 18;$ find $a + b.$
AnswerWe know that,
$(a-b)^2=a^2-2 a b+b^2$
and
$ (a+b)^2=a^2+2 a b+b^2$
Rewrite the above equation, we have
$(a+b)^2=a^2+b^2-2 a b+4 a b$
$=(a+b)^2+4 a b\ldots(1)$
Given that $a-b=7 ; a b=18$
Substitute the values of ( $a-b)$ and $(ab)$ in equation $(1),$ we have
$(a+b)^2 =(7)^2+4(18)$
$ =49+72=121$
$\Rightarrow a+b = \pm \sqrt{121}$
$\Rightarrow a+b$
$=\pm 11$
View full question & answer→Question 243 Marks
If $a + b = 7$ and $ab = 10;$ find $a - b.$
AnswerWe know that,
$(a+b)^2=a^2+2 a b+b^2$
and
$ (a-b)^2=a^2-2 a b+b^2$
Rewrite the above equation, we have
$(a-b)^2=a^2+b^2-2 a b+4 a b$
$=(a+b)^2-4 a b$
$\ldots(1)$
Given that $a+b=7 ; a b=10$
Substitute the values of $(a+b)$ and $(ab)$ in equation $(1),$ we have
$(a-b)^2 =(7)^2-4(10)$
$ =49-40=9$
$\Rightarrow a-b = \pm \sqrt{9}$
$\Rightarrow a-b = \pm 3$
View full question & answer→Question 253 Marks
If $3x + 4y = 16$ and $xy = 4;$ find the value of $9x^2+ 16y^2.$
AnswerGiven that $( 3x + 4y ) = 16$ and $xy = 4$
We need to find $9x^2+ 16y^2.$
We know that
$( a + b )^2 = a^2 + b^2 + 2ab$
Consider the square of $3x + 4y :$
$\therefore ( 3x + 4y )^2 = (3x)^2 + (4y)^2 + 2 \times 3x \times 4y$
$\Rightarrow ( 3x + 4y )^2 = 9x^2 + 16y^2 + 24xy .....(1)$Substitute the values of $( 3x + 4y )$ and $xy$
in the above equation $(1),$ we have
$( 3x + 4y )^2 = 9x^2 + 16y^2 + 24xy$
$\Rightarrow (16)^2 = 9x^2 + 16y^2 + 24(4)$
$\Rightarrow 256 = 9x^2 + 16y^2 + 96$
$\Rightarrow 9x^2 + 16y^2$
$= 160$
View full question & answer→