Question 11 Mark
If $\text{x}+\frac{1}{\text{x}}=3,$ then $\text{x}^6+\frac{1}{\text{x}^6}=$
- 927
- 414
- 364
- 322
Answer
- 322
Solution:
$\Big(\text{x}+\frac{1}{\text{x}}\Big)^2=\text{x}^2+\frac{1}{\text{x}^2}+2$
$\text{x}+\frac{1}{\text{x}}=3$ (given)
$\Rightarrow\text{x}^2+\frac{1}{\text{x}^2}=(3)^2-2$
$\Rightarrow\text{x}^2+\frac{1}{\text{x}^2}=7\ ...(1)$
Cubing both side of equation (1). we have
$\Big(\text{x}^2+\frac{1}{\text{x}^2}\Big)^3=(7)^3$
$\Rightarrow(\text{x}^2)^3+\Big(\frac{1}{\text{x}^2}\Big)^3+3(\text{x}^2)\frac{1}{\text{x}^2}\Big(\text{x}^2+\frac{1}{\text{x}^2}\Big)=7^3$
$\Rightarrow\text{x}^6+\frac{1}{\text{x}^6}+3(7)=7^3$
$\Rightarrow\text{x}^6+\frac{1}{\text{x}^6}=343-21$
$\Rightarrow\text{x}^6+\frac{1}{\text{x}^6}=322$
Hence, correct option is (d). View full question & answer→MCQ 21 Mark
If $49\text{a}^2-{\text{b}}=\Big(7\text{a}+\frac{1}{2}\Big)\Big(7\text{a}-\frac{1}{2}\Big),$ then the value of $b$ is:
- A
$0$
- ✓
$\frac{1}{4}$
- C
$\frac{1}{\sqrt2}$
- D
$\frac{1}{2}$
AnswerCorrect option: B. $\frac{1}{4}$
$\Big(7\text{a}+\frac{1}{2}\Big)\Big(7\text{a}-\frac{1}{2}\Big)$
$=(7\text{a})^2-\Big(\frac{1}{2}\Big)^2$
$[$by using identity $(a + b)(a - b) = a^2 - b^2]$
$\Rightarrow\Big(7\text{a}+\frac{1}{2}\Big)\Big(7\text{a}-\frac{1}{2}\Big)$
$=49\text{a}^2-\frac{1}{4}$
$\Rightarrow49\text{a}^2-\text{b}$
$=49\text{a}^2-\frac{1}{4}$
$\Rightarrow\text{b}=\frac{1}{4}$
View full question & answer→MCQ 31 Mark
$(a - b)^3 + (b - c)^3 + (c - a)^3 =$
AnswerCorrect option: C. $3(a - b)(b - c)(c - a)$
Let
$a - b = A$
$b - c = B$
$c - a = C$
Now $(A + B + C)^3 $
$= A^3 + B^3 + C^3 + 3(A + B)(B + C)(C + A)$
$\Rightarrow A^3 + B^3 + C^3 $
$= (A + B + C)^3- 3(A + B)(B + C)(C + A)$
Now putting values of $A, B$ and $C$. we get
$(\text{a} - \text{b})^3 + (\text{b} - \text{c})^3 + (\text{c} - \text{a})^3$
$=(\not\text{a}-\not\text{b}+\not\text{b}-\not\text{c}+\not\text{c}-\not\text{a})^3 -3(\text{a}-\not\text{b}+\not\text{b}{-\text{c})(\text{b}-\not\text{c}+\not\text{c}-\text{a})(\text{c}-\not\text{a}+\not\text{a}-\text{b})}$
$\Rightarrow (a - b)^3+ (b - c)^3 + (c - a)^3 $
$= 0 - 3 (a - c)(b - a)(c - b)$
$\Rightarrow (a - b)^3+ (b - c)^3 + (c - a)^3 $
$= 3(a - b)(b - c)(c - a)$
View full question & answer→MCQ 41 Mark
If $a + b + c = 0,$ then $\frac{\text{a}^2}{\text{bc}}+\frac{\text{b}^2}{\text{ca}}+\frac{\text{c}^2}{\text{ab}}=$
Answer$a^3 + b^3 + c^3 - 3abc $
$= (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca)$
$If a + b + c = 0,$
then, $a^3 + b^3 + c^3 - 3abc = 0$
$\Rightarrow a^3 + b^3 + c^3 = 3abc ...(1)$
Now, consider $\frac{\text{a}^2}{\text{bc}}+\frac{\text{b}^2}{\text{ca}}+\frac{\text{c}^2}{\text{ab}}$
Multiplying dividing by $a. b.$ and $c$ in $\frac{\text{a}^2}{\text{bc}}.\frac{\text{b}^2}{\text{ca}}$ and $\frac{\text{c}^2}{\text{ab}}$ respectively. we get
$\frac{\text{a}^3}{\text{abc}}+\frac{\text{b}^3}{\text{bca}}+\frac{\text{c}^3}{\text{cab}}$
$=\frac{\text{a}^3+\text{b}^3+\text{c}^3}{\text{abc}}$
$=\frac{3\text{abc}}{\text{abc}} ....[$From $(1)]$
$=3$
View full question & answer→Question 51 Mark
If $\text{x}^4+\frac{1}{\text{x}^4}=623,$ then $\text{x}+\frac{1}{\text{x}}=$
- $27$
- $25$
- $3\sqrt3$
- $-3\sqrt3$
Answer
- $3\sqrt3$
Solution:
$\Big(\text{x}+\frac{1}{\text{x}}\Big)^2=\text{x}^2+\frac{1}{\text{x}^2}+2.\text{x}.\frac{1}{\text{x}}=\text{x}^2+\frac{1}{\text{x}^2}+2$
$\Rightarrow\text{x}^2+\frac{1}{\text{x}^2}=\Big\{\Big(\text{x}+\frac{1}{\text{x}}\Big)^2-2\Big\}$
Squaring both sides.
$\Rightarrow\Big(\text{x}^2+\frac{1}{\text{x}^2}\Big)^2=\Big\{\Big(\text{x}+\frac{1}{\text{x}}\Big)^2-2\Big\}^2$
$\Rightarrow\text{x}^4+\frac{1}{\text{x}^4}+2.\text{x}^2.\frac{1}{\text{x}}=\Big\{\Big(\text{x}+\frac{1}{\text{x}}\Big)^2-2\Big\}^2$
$\Rightarrow\text{x}^4+\frac{1}{\text{x}^4}+2=\Big\{\Big(\text{x}+\frac{1}{\text{x}}\Big)^2-2\Big\}^2=(623)+2$
$\Rightarrow623+2=\Big\{\Big(\text{x}+\frac{1}{\text{x}}\Big)^2-2\Big\}^2\ \Big\{\text{x}^4+\frac{1}{\text{x}^4}=623\Big\}$
$\Rightarrow625=\Big\{\Big(\text{x}+\frac{1}{\text{x}}\Big)^2-2\Big\}^2$
$\Rightarrow\Big(\text{x}+\frac{1}{\text{x}}\Big)^2-2=\sqrt{625}=25$
$\Rightarrow\Big(\text{x}+\frac{1}{\text{x}}\Big)^2=25+2=27$
$\Rightarrow\Big(\text{x}+\frac{1}{\text{x}}\Big)^2=\sqrt{27}$
$\Rightarrow\text{x}+\frac{1}{\text{x}}=3\sqrt{3}$
Hence, correct option is (c). View full question & answer→MCQ 61 Mark
If the volume of a cuboid is $3x^2 - 27,$ then its possible dimensions are:
- A
$3, x^2, - 27x$
- ✓
$3, x - 3, x + 3$
- C
$3, x^2, 27x$
- D
$3, 3, 3$
AnswerCorrect option: B. $3, x - 3, x + 3$
Volume of a cuboid of side $a, b$ and $c = abc$
Now, Volume $= 3x^2 - 27 ($given$)$
$abc = 3(x^2 - 9)$
$abc = 3(x - 3)(x + 3)$
So, possible dimensions are $3, x - 3$ and $x + 3$
View full question & answer→MCQ 71 Mark
The product $(x^2 - 1)(x^4 + x^2 + 1)$ is equal to:
- A
$x^8 - 1$
- B
$x^8 + 1$
- ✓
$x^6 - 1$
- D
$x^6 + 1$
AnswerCorrect option: C. $x^6 - 1$
Given expression is $(x^2 - 1)(x^4 + x^2 + 1)$
Let $x^2 = A$ and $1 = B$
Then, we have
$(A - B)(A^2 + AB + B^2)$
$= A^3 - B^3$
$= (X^2)^3 - (1)^3$
$= X^6 - 1$
View full question & answer→Question 81 Mark
If $\text{x}-\frac{1}{\text{x}}=\frac{15}{4},$ then $\text{x}+\frac{1}{\text{x}}=$
- $4$
- $\frac{17}{4}$
- $\frac{13}{4}$
- $\frac{1}{4}$
Answer
- $\frac{17}{4}$
Solution:
$\Big(\text{x}-\frac{1}{\text{x}}\Big)^2=\text{x}^2+\frac{1}{\text{x}^2}-2\ ...(1)$
$\Big(\text{x}+\frac{1}{\text{x}}\Big)^2=\text{x}^2+\frac{1}{\text{x}^2}+2\ ...(2)$
Subtracting eq. (2) from eq. (1). we get
$\Big(\text{x}-\frac{1}{\text{x}}\Big)^2-\Big(\text{x}+\frac{1}{\text{x}}\Big)^2=-4$
$\Rightarrow\Big(\frac{15}{4}\Big)^2-\Big(\text{x}+\frac{1}{\text{x}}\Big)^2=-4$
$\Rightarrow\Big(\text{x}+\frac{1}{\text{x}}\Big)^2=\Big(\frac{15}{4}\Big)^2+4=\frac{225}{16}+4$
$\Rightarrow\Big(\text{x}+\frac{1}{\text{x}}\Big)^2=\frac{225+64}{16}=\frac{189}{16}$
$\Rightarrow\Big(\text{x}+\frac{1}{\text{x}}\Big)=\sqrt{\frac{289}{16}}$
$\Rightarrow\text{x}+\frac{1}{\text{x}}=\frac{17}{4}$
Hence, correct option is (b). View full question & answer→MCQ 91 Mark
$(x - y) (x + y) (x^2 + y^2) (x^4{ }+ y^4)$ is equal to:
- A
$x^{16} - y^{16}$
- ✓
$x^8 - y^8$
- C
$x^8 + y8$
- D
$x^{16} + y^{16}$
AnswerCorrect option: B. $x^8 - y^8$
$(x - y)(x + y) = x^2 - y^{2 }[$by identity $(a + b)(a - b) = a^2 - b^2]$
$(x^2 - y^2)(x^2 + y^2) = x^4 - y^4$
$(x^4 - y^4)(x^4 + y^4) = x^8 - y^8$
Now,
$(x - y)(x + y)(x^2 + y^2)(x^4 + y^4)$
$= (x^2 - y^2)(x^2 + y^2)(x^4 + y^4)$
$= (x^4 - y^4)(x^4 + y^4)$
$= x^8 - y^8$
View full question & answer→MCQ 101 Mark
If $a - b = -8$ and $ab = -12,$ then $a^3 - b^3 =$
- A
$-244$
- B
$-240$
- ✓
$-224$
- D
$-260$
AnswerCorrect option: C. $-224$
$a - b = -8$
$(a - b)^2 = 64$
$a^2 + b^2 - 2ab = 64$
$a^2 + b^2 - 2ab + 3ab = 64 + 3ab$
$a^2 + b^2 + ab = 64 + 3(-12)$
$a^2 + b^2 + ab = 64 - 36$
$a^2 + b^2 + ab = 28$
Now
$a^3 - b^3 = (a - b)(a^2 + b^2 + ab)$
$= (-8)(28)$
$= -224$
View full question & answer→MCQ 111 Mark
$\frac{(\text{a}^2-\text{b}^2)^3+(\text{b}^2-\text{c}^2)^3+(\text{c}^2-\text{a}^2)^3}{(\text{a}-\text{b})^3+(\text{b}-\text{c})^3+(\text{c}-\text{a})^3}=$
- ✓
$3(a + b)( b+ c)(c + a)$
- B
$3(a - b)(b - c)(c - a)$
- C
$(a - b)(b - c)(c - a)$
- D
AnswerCorrect option: A. $3(a + b)( b+ c)(c + a)$
If $a + b + c = 0$ then, $a^3 + b^3 + c^3 = 3abc$
Now$, (a^2 - b^2) + (b^2 - c^2) + (c^2 - a^2)$
$= a^2 - b^2 + b^2 - c^2 + c^2 - a^2 = 0$
$\Rightarrow (a^2 - b^2)^3 + (b^2 - c^2)^3 + (c^2 - a^2)^3$
$ = 3(a^2 - b^2)(b^2 - c^2)(c^2 - a^2)$
Again, $(a - b) + (b - c) + (c - a) $
$= a - b + b - c + c - a = 0$
$\Rightarrow (a - b)^3 + (b - c)^3 + (c - a)^3$
$ = 3(a - b)(b - c)(c - a)$
Thus, we have
$\frac{(\text{a}^2-\text{b}^2)^3+(\text{b}^2-\text{c}^2)^3+(\text{c}^2-\text{a}^2)^3}{(\text{a}-\text{b})^3+(\text{b}-\text{c})^3+(\text{c}-\text{a})^3}$
$=\frac{3(\text{a}^2-\text{b}^2)(\text{b}^2-\text{c}^2)(\text{c}^2-\text{a}^2)}{3(\text{a}-\text{b})(\text{b}-\text{c})(\text{c}-\text{a})}$
$=\frac{(\text{a}-\text{b})(\text{a}+\text{b})(\text{b}-\text{c})(\text{b}+\text{c})(\text{c}-\text{a})(\text{c}+\text{a})}{(\text{a}-\text{b})(\text{b}-\text{c})(\text{c}-\text{a})}$
$=(\text{a}+\text{b})(\text{b}+\text{c})(\text{c}+\text{a})$
View full question & answer→MCQ 121 Mark
If $\frac{\text{a}}{\text{b}}+\frac{\text{b}}{\text{a}}=1,$ then $a^3 + b^3 =$
- A
$1$
- B
$-1$
- C
$\frac{1}{2}$
- ✓
$0$
Answer$\frac{\text{a}}{\text{b}}+\frac{\text{b}}{\text{a}}=1$
$\Rightarrow\text{a}^2+\text{b}^2-\text{ab}=0$
Now by identity $a^3 + b^3 $
$= (a + b)(a^2 + b^2 - ab).$
if $a^2 + b^2 - ab = 0.$
then $a^3 + b^3 = 0$
View full question & answer→Question 131 Mark
If $3\text{x}+\frac{2}{\text{x}}=7,$ then $\Big(9\text{x}^2-\frac{4}{\text{x}^2}\Big)=$
-
25
-
35
-
49
-
30
Answer
- 35
Solution:
$\Big(3\text{x}+\frac{2}{\text{x}}\Big)^2=9\text{x}^2+\frac{4}{\text{x}^2}+12\ ...(1)$
$\Big(3\text{x}-\frac{2}{\text{x}}\Big)^2=9\text{x}^2+\frac{4}{\text{x}^2}-12\ ...(2)$
Subtracting eq. (1) from eq. (2). we get
$\Big(3\text{x}-\frac{2}{\text{x}}\Big)^2-\Big(3\text{x}+\frac{2}{\text{x}}\Big)^2=-24$
$\Rightarrow\Big(3\text{x}-\frac{2}{\text{x}}\Big)^2=(7)^2-24=25$
$\Rightarrow3\text{x}-\frac{2}{\text{x}}=5$
Now $\Big(3\text{x}+\frac{2}{\text{x}}\Big)-\Big(3\text{x}-\frac{2}{\text{x}}\Big)=7\times5$
$\Big(9\text{x}^2-\frac{4}{\text{x}^2}\Big)=35$
Hence, correct option is (b). View full question & answer→MCQ 141 Mark
If $a + b + c = 9$ and $ab + bc + ca =23,$ then $a^3{ }+ b^3{ }+ c^3 - 3abc =$
AnswerGiven,$ a + b + c = 9$
Hence$, (a + b + c)^2 = 81$
So$, a^2 + b^2 + c^2 + 2ab + 2bc + 2ca = 81$
$i.e. a^2 + b^2 + c^2 + 2(ab + bc + ca) = 81$
$i.e. a^2 + b^2 + c^2 + 2(23) = 81$
$i.e. a^2 + b^2 + c^2 = 81 - 46 = 35$
Now$, a^3 + b^3 + c^3 - 3abc$
$= (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca)$
$= (a + b + c)[(a^2 + b^2 + c^{2)} - (ab + bc + ca)]$
$= (9)[35 - 23]$
$= 9 \times 12$
$= 108$
View full question & answer→Question 151 Mark
If $\text{x}+\frac{1}{\text{x}}=4,$ then $\text{x}^4+\frac{1}{\text{x}^4}=$
- 196
- 194
- 192
- 190
Answer
- 194
Solution:
$\Big(\text{x}+\frac{1}{\text{x}}\Big)^2=\text{x}^2+\frac{1}{\text{x}^2}+2$
$\Big(\text{x}+\frac{1}{\text{x}}\Big)=4$ (given)
$\Rightarrow\text{x}^2+\frac{1}{\text{x}^2}=(4)^2-2=16-2=14\ ...(1)$
Squaring equation (1)
$\Big(\text{x}^2+\frac{1}{\text{x}^2}\Big)^2=(14)^2$
$\Rightarrow(\text{x}^2)^2+\Big(\frac{1}{\text{x}^2}\Big)^2+2(\text{x}^2)\frac{1}{\text{x}^2}=196$
$\Rightarrow\text{x}^4+\frac{1}{\text{x}^4}=196-2$
$\Rightarrow\text{x}^4+\frac{1}{\text{x}^4}=194$
Hence, correct option is (b). View full question & answer→MCQ 161 Mark
If $\text{x}^4+\frac{1}{\text{x}^4}=194,$ then $\text{x}^3+\frac{1}{\text{x}^3}=$
Answer$\text{x}^4+\frac{1}{\text{x}^4}=194$
Now $\Big(\text{x}^2+\frac{1}{\text{x}^2}\Big)^2=\text{x}^4+\frac{1}{\text{x}^4}+2$
$\Rightarrow\Big(\text{x}^2+\frac{1}{\text{x}^2}\Big)^2=194+2=196$
$\Rightarrow\text{x}^2+\frac{1}{\text{x}^2}=14\ ...(1)$
Now $\Big(\text{x}+\frac{1}{\text{x}}\Big)^2=\text{x}^2+\frac{1}{\text{x}^2}+2$ $\Big\{\text{x}^2+\frac{1}{\text{x}^2}=14\Big\}$
$\Rightarrow\Big(\text{x}+\frac{1}{\text{x}}\Big)^2=14+2=16 [$From $(1)]$
$\Rightarrow\text{x}+\frac{1}{\text{x}}=\sqrt{16}$
$\Rightarrow\text{x}+\frac{1}{\text{x}}=4\ ...(3)$
By identity $a^3 + b^3 $
$= (a + b)(a^2 + b^2 - ab)$
$\Rightarrow\text{x}^3+\frac{1}{\text{x}^3}=\Big(\text{x}+\frac{1}{\text{x}}\Big)\Big(\text{x}^2+\frac{1}{\text{x}^2}-1\Big)$
$=(4)(14-1)$
$=4\times13$
$=52$
View full question & answer→Question 171 Mark
If $\text{x}^3+\frac{1}{\text{x}^3}=110,$ then $\text{x}+\frac{1}{\text{x}}=$
- 5
- 10
- 15
- None of these.
Answer
- 5
Solution:
$\Big(\text{x}+\frac{1}{\text{x}}\Big)^3=\text{x}^3+\frac{1}{\text{x}^3}+3\Big(\text{x}+\frac{1}{\text{x}}\Big)$
$\Rightarrow\Big(\text{x}+\frac{1}{\text{x}}\Big)^3-3\Big(\text{x}+\frac{1}{\text{x}}\Big)=\text{x}^3+\frac{1}{\text{x}^3}$
$\Rightarrow\Big(\text{x}+\frac{1}{\text{x}}\Big)^3-3\Big(\text{x}+\frac{1}{\text{x}}\Big)=110$
Let $\text{x}+\frac{1}{\text{x}}=\text{t}$
$\Rightarrow\text{t}^3-3\text{t}-110=0$
t =5 is one of it's solution which is real, other two solutions are imaginary
$\Rightarrow\text{x}+\frac{1}{\text{x}}=5$
Hence, correct option is (a). View full question & answer→MCQ 181 Mark
If $\frac{\text{a}}{\text{b}}+\frac{\text{b}}{\text{a}}=-1,$ then $a^3 - b^3 =$
- A
$1$
- B
$-1$
- C
$\frac{1}{2}$
- ✓
$0$
Answer$\frac{\text{a}}{\text{b}}+\frac{\text{b}}{\text{a}}=-1$
$\Rightarrow\frac{\text{a}^2+\text{b}^2}{\text{ab}}=-1$
$\Rightarrow\text{a}^2+\text{b}^2+\text{ab}=0$
Now using identity
$a3 - b3$
$= (a - b)(a2 + b2 + ab)$
$=(a - b)(0) (\because a2 + b2 + ab = 0)$
$= 0$
View full question & answer→MCQ 191 Mark
If $\text{x}^3-\frac{1}{\text{x}^3}=14,$ then $\text{x}-\frac{1}{\text{x}}=$
Answer$\Big(\text{x}-\frac{1}{\text{x}}\Big)^3=\text{x}^3-\frac{1}{\text{x}^3}-3\not\text{x}\frac{1}{\not\text{x}}\Big(\text{x}-\frac{1}{\text{x}}\Big)$
$\text{x}^3-\frac{1}{\text{x}^3}=\Big(\text{x}-\frac{1}{\text{x}}\Big)^3+3\Big(\text{x}-\frac{1}{\text{x}}\Big)$
$\Rightarrow\Big(\text{x}-\frac{1}{\text{x}}\Big)^3+3\Big(\text{x}-\frac{1}{\text{x}}\Big)-\text{x}^3-\frac{1}{\text{x}^3}=0$
$\Rightarrow\Big(\text{x}-\frac{1}{\text{x}}\Big)^3+3\Big(\text{x}-\frac{1}{\text{x}}\Big)-14=0$
Let $\Rightarrow\text{x}-\frac{1}{\text{x}}=\text{t}$
$\Rightarrow t^3 + 3t - 14 = 0$
$\Rightarrow t^3 - 2t^2 + 2t^2 - 4t + 7t -14 = 0$
$\Rightarrow t(t - 2) + 2t(t - 2) + 7(t - 2) = 0$
$\Rightarrow (t - 2)(t + 2t + 7) = 0$
$\Rightarrow t^2 + 2t + 7 = 0$ has no real roots
So$, t = 2$ is a solution
$\Rightarrow\text{x}-\frac{1}{\text{x}}=2$
View full question & answer→Question 201 Mark
If $\text{x}^2+\frac{1}{\text{x}^2}=102,$ then $\text{x}-\frac{1}{\text{x}}=$
- 8
- 10
- 12
- 13
Answer
- 10
Solution:
$\Big(\text{x}-\frac{1}{\text{x}}\Big)^2=\text{x}^2+\frac{1}{\text{x}^2}-2(\text{x})\frac{1}{\text{x}}$
$\Big(\text{x}-\frac{1}{\text{x}}\Big)^2=\text{x}^2+\frac{1}{\text{x}^2}-2$
$=102-2$ $\Big\{\text{x}^2+\frac{1}{\text{x}^2}=102\Big\}$
$=100$
$\Rightarrow\text{x}-\frac{1}{\text{x}}=\sqrt{100}$
$\Rightarrow\text{x}-\frac{1}{\text{x}}=10$
Hence, correct option is (b). View full question & answer→MCQ 211 Mark
If $\text{a}^{\frac{1}{3}}+\text{b}^{\frac{1}{3}}+\text{c}^{\frac{1}{3}}= 0,$ then:
- A
$a + b + c = 0$
- ✓
$(a + b + c)^3 =27abc$
- C
$a + b + c = 3abc$
- D
$a^3 + b^3 + c^3 = 0$
AnswerCorrect option: B. $(a + b + c)^3 =27abc$
Let $\text{a}^{\frac{1}{3}}=\text{A},\ \text{b}^{\frac{1}{3}}=\text{B}$ and $\text{c}^{\frac{1}{3}}=\text{C}$
Now, $A + B + C = 0 ($given$)$
If$ A + B + C = 0,$
then $A^3 + B^3 + C^3 - 3ABC = 0$
$\Rightarrow A^3 + B^3 + C^3 - 3ABC = 0$
$\Rightarrow A^3 + B^3 + C^3 = 3ABC ...(1)$
$\begin{Bmatrix}\text{A}=\text{a}^{\frac{1}{3}},\ \text{B}=\text{b}^{\frac{1}{3}},\ \text{C}=\text{c}^{\frac{1}{3}}\\\text{A}^3=\text{a},\ \text{B}^3=\text{b},\ \text{C}^3=\text{c}\end{Bmatrix}$
Then, equation $(1)$ becomes
$\text{a}+\text{b}+\text{c}=3(\text{abc})^{\frac{1}{3}}$
Cubing both Sides of above equation, we get
$(a + b + c)^3 = 27abc$
View full question & answer→MCQ 221 Mark
If $a^2 + b^2 + c^2 - ab - bc - ca =0,$ then:
- A
$a + b + c$
- B
$b + c = a$
- C
$c + a = b$
- ✓
$a = b = c$
AnswerCorrect option: D. $a = b = c$
$a^2 + b^2 + c^2 - ab - bc - ca = 0$
Multiplying by $2$ on both the sides, we have
$2(a^2 + b^2 + c^2 - ab - bc - ca) = 0$
$2a^{2 }+ 2b^2 + 2c^2 - 2ab - 2bc - 2ca = 0$
$a^2 + a^2 + b^2 + b^{2 }+ c^{2 }+ c^2 - 2ab - 2bc - 2ca = 0$
$(a^2 + b^2 - 2ab) + (b^2 + c^2 - 2bc) + (a^2 + c^2 - 2ac) = 0$
$(a - b)^2 + (b - c)^2 + (a - c)^2 = 0$
$(a - b)^2 = 0, (b - c)^2 = 0, (a - c)^2 = 0$
$(a - b) = 0, (b - c) = 0, (a - c) = 0$
$a = b, b = c, a = c$
or we can say $a = b = c$
View full question & answer→MCQ 231 Mark
If $\text{x}+\frac{1}{\text{x}}=5,$ then $\text{x}^2+\frac{1}{\text{x}^2}=$
AnswerBy using identity $(a + b)^2 = a^2 + b^2 + 2ab.$
we have,
$\Big(\text{x}+\frac{1}{\text{x}}\Big)^2=\text{x}^2+\Big(\frac{1}{\text{x}}\Big)^2+2\times\not\text{x}\times\frac{1}{\not\text{x}}$
$\Rightarrow\Big(\text{x}+\frac{1}{\text{x}}\Big)^2=\text{x}^2+\frac{1}{\text{x}^2}+2$
$\Rightarrow(5)^2=\text{x}^2+\frac{1}{\text{x}^2}+2$ $\Big\{\text{x}+\frac{1}{\text{x}}=5$ given$\Big\}$
$\Rightarrow\text{x}^2+\frac{1}{\text{x}^2}=25-2$
$\Rightarrow\text{x}^2+\frac{1}{\text{x}^2}=23$
View full question & answer→MCQ 241 Mark
If $a + b + c = 9$ and $ab + bc + ca = 23,$ then $a^2 + b^2 + c^2 =$
AnswerWe know that $(a + b + c)^2 $
$= a^2 + b^2 + c^2 + 2(ab + bc + ca)$
Here$, a + b + c = 9, ab + bc + ca = 23$
Thus, we have
$(9)^2 = a^2 + b^2 + c^2 + 2(23)$
$81 = a^2 + b^2 + c^2 + 46$
$a^2 + b^2 + c^2 = 81 - 46$
$a^2 + b^2 + c^2 = 35$
View full question & answer→MCQ 251 Mark
If $\text{x}+\frac{1}{\text{x}}=2,$ then $\text{x}^3+\frac{1}{\text{x}^3}=$
AnswerBy using identity,
$(a + b)^3 = a^3 + b^3 + 3ab(a + b)$
$\Big(\text{x}+\frac{1}{\text{x}}\Big)^3=\text{x}^3+\frac{1}{\text{x}^3}+3(\not\text{x})\frac{1}{\not\text{x}}\Big(\text{x}+\frac{1}{\text{x}}\Big)$
$=\text{x}^3+\frac{1}{\text{x}^3}+3\Big(\text{x}+\frac{1}{\text{x}}\Big)$
Now $\text{x}+\frac{1}{\text{x}}=2$
$\Rightarrow(2)^3=\text{x}^3+\frac{1}{\text{x}^3}+3(2)$
$\Rightarrow\text{x}^3+\frac{1}{\text{x}^3}$
$=(2)^3-3\times2$
$=8-6=2$
$\Rightarrow\text{x}^2+\frac{1}{\text{x}^2}$
$=23$
View full question & answer→MCQ 261 Mark
$75 \times 75 + 2 \times 75 \times 25 + 25 \times 25$ is equal to:
- ✓
$10000$
- B
$6250$
- C
$7500$
- D
$3750$
AnswerCorrect option: A. $10000$
Given expression is $75 \times 75 + 2 \times 75 \times 25 + 25 \times 25$
Let $75 = a$ and $25 = b$
Then, we have
$a \times a + 2 \times a \times b + b \times b$
$= a^2 + 2ab + b^2$
$= (a + b)^2$
$= (75 + 25)^2$
$= (100)^2$
$= 10000$
View full question & answer→MCQ 271 Mark
The product $(a + b)(a - b)(a^2 - ab + b^2)(a^2 + ab + b^2)$ is equal to:
- A
$a^6 + b^6$
- ✓
$a^6 - b^6$
- C
$a^3 - b^3$
- D
$a^3{ }+ b^3$
AnswerCorrect option: B. $a^6 - b^6$
$(a + b)(a - b)(a^2 - ab + b^2)(a^2 + ab + b^2)$
$= (a^2 - b^2)(a^2 + b^2 - ab)(a^2 + b^2 - ab)$
$= (a^2 - b^2) \Big\{ (a^2 + b^2)^2 - (ab)^2 \Big\}$
$= (a^2 - b^2) {a^4 + b^4 + 2a^2b^2 - a^2b^2}$
$= (a^2 - b^2) {a^4 + b^4 + a^2b^2}$
$= {a^6 + a^2b^4 + a^4b^2 - b^2a^4 - b^6 - b^4a^2}$
$= a^6 - b^6$
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