MCQ 11 Mark
If $\text{x}+\frac{1}{\text{x}}=3,$ then $\text{x}^6+\frac{1}{\text{x}^6}=$
Answer$\Big(\text{x}+\frac{1}{\text{x}}\Big)^2=\text{x}^2+\frac{1}{\text{x}^2}+2$
$\text{x}+\frac{1}{\text{x}}=3 \ ($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 $(a + b)(a - b) = a^2 - b^2$]
$\Rightarrow\Big(7\text{a}+\frac{1}{2}\Big)\Big(7\text{a}-\frac{1}{2}\Big)=49\text{a}^2-\frac{1}{4}$
$\Rightarrow49\text{a}^2-\text{b}=49\text{a}^2-\frac{1}{4}$
$\Rightarrow\text{b}=\frac{1}{4}$
Hence, correct option is $(b).$
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
$(a-b)^3+(b-c)^3+(c-a)^3$
$=(a-\not b+\not b-\not+\not-\not a)^3 -3(a-\not b+\not b-c)(b-c t+c-a)(c-\not a+\not a-b)$
$\Rightarrow(a-b)^3+(b-c)^3+(c-a)^3$
$=0-3(a-c)(b-a)(c-b)$
$\Rightarrow(a-b)^3+(b-c)^3+(c-a)^3$
$=3(a-b)(b-c)(c-a)$
Hence, correct option is $(c).$
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 $(1)]$
$=3$
Hence, correct option is $(d).$
View full question & answer→MCQ 51 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
$a^3 - b^3 = (a - b)(a^2 + b^2 + ab)$
$= (-8)(28)$
$= -224$
Hence, correct option is $(c).$
View full question & answer→MCQ 61 Mark
If $\text{x}^4+\frac{1}{\text{x}^4}=623,$ then $\text{x}+\frac{1}{\text{x}}=$
- A
$27$
- B
$25$
- ✓
$3\sqrt3$
- D
$-3\sqrt3$
AnswerCorrect option: C. $3\sqrt3$
$\Big(\text{x}+\frac{1}{\text{x}}\Big)^2=\text{x}^2+\frac{1}{\text{x}^2}+2.\text{x}.\frac{1}{\text{x}}=\text{x}^2+\frac{1}{\text{x}^2}+2$
$\Rightarrow\text{x}^2+\frac{1}{\text{x}^2}=\Big\{\Big(\text{x}+\frac{1}{\text{x}}\Big)^2-2\Big\}$
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 71 Mark
If the volume of a cuboid is $3 x^2-27$, then its possible dimensions are:
- A
$3, x^2,-27 x$
- ✓
$3, x-3, x+3$
- C
$3, x^2, 27 x$
- D
$3,3,3$
AnswerCorrect option: B. $3, x-3, x+3$
Volume of a cuboid of side $a, b$ and $c=a b c$
Now, Volume $=3 x^2-27 ($given$)$
$a b c=3\left(x^2-9\right)$
$a b c=3(x-3)(x+3)$
So, possible dimensions are $3, x-3$ and $x+3$
Hence, correct option is $(b).$
View full question & answer→MCQ 81 Mark
The product $\left(x^2-1\right)\left(x^4+x^2+1\right)$ 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 $\left(x^2-1\right)\left(x^4+x^2+1\right)$
Let $x ^2= A$ and $1= B$
Then, we have
$(A-B)\left(A^2+A B+B^2\right)$
$=A^3-B^3$
$=\left(x^2\right)^3-(1)^3$
$=X^6-1$
Hence, correct option is $(c).$
View full question & answer→MCQ 91 Mark
If $\text{x}-\frac{1}{\text{x}}=\frac{15}{4},$ then $\text{x}+\frac{1}{\text{x}}=$
- A
$4$
- ✓
$\frac{17}{4}$
- C
$\frac{13}{4}$
- D
$\frac{1}{4}$
AnswerCorrect option: B. $\frac{17}{4}$
$\Big(\text{x}-\frac{1}{\text{x}}\Big)^2=\text{x}^2+\frac{1}{\text{x}^2}-2\ ...(1)$
$\Big(\text{x}+\frac{1}{\text{x}}\Big)^2=\text{x}^2+\frac{1}{\text{x}^2}+2\ ...(2)$
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 101 Mark
$(x-y)(x+y)\left(x^2+y^2\right)\left(x^4+y^4\right)$ is equal to:
- A
$x^{16} - y^{16}$
- ✓
$x^8 - y^8$
- C
$x^8 + y8$
- D
$x^{16} + y^{16}$
AnswerCorrect option: B. $x^8 - y^8$
$(x-y)(x+y)=x^2-y^2\left[\text { by identity }(a+b)(a-b)=a^2-b^2\right]$
$\left(x^2-y^2\right)\left(x^2+y^2\right)=x^4-y^4$
$\left(x^4-y^4\right)\left(x^4+y^4\right)=x^8-y^8$
Now,
$(x-y)(x+y)\left(x^2+y^2\right)\left(x^4+y^4\right)$
$=\left(x^2-y^2\right)\left(x^2+y^2\right)\left(x^4+y^4\right)$
$=\left(x^4-y^4\right)\left(x^4+y^4\right)$
$=x^8-y^8$
Hence, correct option is $(b).$
View full question & answer→MCQ 111 Mark
$\frac{(\text{a}^2-\text{b}^2)^3+(\text{b}^2-\text{c}^2)^3+(\text{c}^2-\text{a}^2)^3}{(\text{a}-\text{b})^3+(\text{b}-\text{c})^3+(\text{c}-\text{a})^3}=$
- A
$3(a + b)( b+ c)(c + a)$
- B
$3(a - b)(b - c)(c - a)$
- C
$(a - b)(b - c)(c - a)$
- ✓
AnswerIf $a + b + c = 0$ 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
$\frac{(\text{a}^2-\text{b}^2)^3+(\text{b}^2-\text{c}^2)^3+(\text{c}^2-\text{a}^2)^3}{(\text{a}-\text{b})^3+(\text{b}-\text{c})^3+(\text{c}-\text{a})^3}$
$=\frac{3(\text{a}^2-\text{b}^2)(\text{b}^2-\text{c}^2)(\text{c}^2-\text{a}^2)}{3(\text{a}-\text{b})(\text{b}-\text{c})(\text{c}-\text{a})}$
$=\frac{(\text{a}-\text{b})(\text{a}+\text{b})(\text{b}-\text{c})(\text{b}+\text{c})(\text{c}-\text{a})(\text{c}+\text{a})}{(\text{a}-\text{b})(\text{b}-\text{c})(\text{c}-\text{a})}$
$=(\text{a}+\text{b})(\text{b}+\text{c})(\text{c}+\text{a})$
Hence, correct option is $(d).$
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 $a^3 + b^3 = 0$
Hence, correct option is $(d).$
View full question & answer→MCQ 131 Mark
If $3\text{x}+\frac{2}{\text{x}}=7,$ then $\Big(9\text{x}^2-\frac{4}{\text{x}^2}\Big)=$
Answer$\Big(3\text{x}+\frac{2}{\text{x}}\Big)^2=9\text{x}^2+\frac{4}{\text{x}^2}+12\ ...(1)$
$\Big(3\text{x}-\frac{2}{\text{x}}\Big)^2=9\text{x}^2+\frac{4}{\text{x}^2}-12\ ...(2)$
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 $a b+b c+c a=23$, then $a^3+b^3+c^3-3 a b c=$
AnswerGiven, $a + b + c =9$
Hence, $(a+b+c)^2=81$
So, $a^2+b^2+c^2+2 a b+2 b c+2 c a=81$
i.e. $a^2+b^2+c^2+2(a b+b c+c a)=81$
i.e. $a^2+b^2+c^2+2(23)=81$
i.e. $a^2+b^2+c^2=81-46=35$
Now, $a^3+b^3+c^3-3 a b c$
$=(a+b+c)\left(a^2+b^2+c^2-a b-b c-c a\right)$
$=(a+b+c)\left[\left(a^2+b^2+c^2\right)-(a b+b c+c a)\right]$
$=(9)[35-23]$
$=9 \times 12$
$=108$
Hence, correct option is $(a).$
View full question & answer→MCQ 151 Mark
If $\text{x}+\frac{1}{\text{x}}=4,$ then $\text{x}^4+\frac{1}{\text{x}^4}=$
Answer$\Big(\text{x}+\frac{1}{\text{x}}\Big)^2=\text{x}^2+\frac{1}{\text{x}^2}+2$
$\Big(\text{x}+\frac{1}{\text{x}}\Big)=4 \ ($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 $a^3 + b^3$ = $(a + b)(a^2 + b^2 - ab)$
$\Rightarrow\text{x}^3+\frac{1}{\text{x}^3}=\Big(\text{x}+\frac{1}{\text{x}}\Big)\Big(\text{x}^2+\frac{1}{\text{x}^2}-1\Big)$
$=(4)(14-1)$
$=4\times13$
$=52$
Hence, correct option is $(b).$
View full question & answer→MCQ 171 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
$a \times a+2 \times a \times b+b \times b$
$=a^2+2 a b+b^2$
$=(a+b)^2$
$=(75+25)^2$
$=(100)^2$
$=10000$
Hence, correct option is $(a).$
View full question & answer→MCQ 181 Mark
If $\text{x}^3+\frac{1}{\text{x}^3}=110,$ then $\text{x}+\frac{1}{\text{x}}=$
Answer$\Big(\text{x}+\frac{1}{\text{x}}\Big)^3=\text{x}^3+\frac{1}{\text{x}^3}+3\Big(\text{x}+\frac{1}{\text{x}}\Big)$
$\Rightarrow\Big(\text{x}+\frac{1}{\text{x}}\Big)^3-3\Big(\text{x}+\frac{1}{\text{x}}\Big)=\text{x}^3+\frac{1}{\text{x}^3}$
$\Rightarrow\Big(\text{x}+\frac{1}{\text{x}}\Big)^3-3\Big(\text{x}+\frac{1}{\text{x}}\Big)=110$
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 191 Mark
If $a^2+b^2+c^2-a b-b c-c a=0$, then:
- A
$a+b+c$
- B
$b+c=a$
- C
$c+a=b$
- ✓
$a=b=c$
AnswerCorrect option: D. $a=b=c$
$a^2+b^2+c^2-a b-b c-c a=0$
Multiplying by $2$ on both the sides, we have
$2\left(a^2+b^2+c^2-a b-b c-c a\right)=0$
$2 a^2+2 b^2+2 c^2-2 a b-2 b c-2 c a=0$
$a^2+a^2+b^2+b^2+c^2+c^2-2 a b-2 b c-2 c a=0$
$\left(a^2+b^2-2 a b\right)+\left(b^2+c^2-2 b c\right)+\left(a^2+c^2-2 a c\right)=0$
$(a-b)^2+(b-c)^2+(a-c)^2=0$
$(a-b)^2=0,(b-c)^2=0,(a-c)^2=0$
$(a-b)=0,(b-c)=0,(a-c)=0$
$a=b, b=c, a=c$
or we can say $a = b = c$
Hence, correct option is $(d).$
View full question & answer→MCQ 201 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
$a^3-b^3$
$=(a-b)\left(a^2+b^2+a b\right)$
$=(a-b)(0)\left(\because a^2+b^2+a b=0\right)$
$=0$
Hence, correct option is $(d).$
View full question & answer→MCQ 211 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 $\Rightarrow\text{x}-\frac{1}{\text{x}}=\text{t}$
$\Rightarrow t^3+3 t-14=0$
$\Rightarrow t^3-2 t^2+2 t^2-4 t+7 t-14=0$
$\Rightarrow t(t-2)+2 t(t-2)+7(t-2)=0$
$\Rightarrow(t-2)(t+2 t+7)=0$
$\Rightarrow t^2+2 t+7=0 \text { has no real roots }$
So, $t=2$ is a solution
$\Rightarrow\text{x}-\frac{1}{\text{x}}=2$
Hence, correct option is $(d).$
View full question & answer→MCQ 221 Mark
If $\text{x}^2+\frac{1}{\text{x}^2}=102,$ then $\text{x}-\frac{1}{\text{x}}=$
Answer$\Big(\text{x}-\frac{1}{\text{x}}\Big)^2=\text{x}^2+\frac{1}{\text{x}^2}-2(\text{x})\frac{1}{\text{x}}$
$\Big(\text{x}-\frac{1}{\text{x}}\Big)^2=\text{x}^2+\frac{1}{\text{x}^2}-2$
$=102-2$ $\Big\{\text{x}^2+\frac{1}{\text{x}^2}=102\Big\}$
$=100$
$\Rightarrow\text{x}-\frac{1}{\text{x}}=\sqrt{100}$
$\Rightarrow\text{x}-\frac{1}{\text{x}}=10$
Hence, correct option is $(b)$.
View full question & answer→MCQ 231 Mark
If $\text{a}^{\frac{1}{3}}+\text{b}^{\frac{1}{3}}+\text{c}^{\frac{1}{3}}= 0,$ then:
- A
$a+b+c=0$
- ✓
$(a+b+c)^3=27 a b c$
- C
$a+b+c=3 a b c$
- D
$a^3+b^3+c^3=0$
AnswerCorrect option: B. $(a+b+c)^3=27 a b c$
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 $A^3+B^3+C^3-3 A B C=0$
$\Rightarrow A^3+B^3+C^3-3 A B C=0$
$\Rightarrow A^3+B^3+C^3=3 A B C \ldots(1)$
$\begin{Bmatrix}\text{A}=\text{a}^{\frac{1}{3}},\ \text{B}=\text{b}^{\frac{1}{3}},\ \text{C}=\text{c}^{\frac{1}{3}}\\\text{A}^3=\text{a},\ \text{B}^3=\text{b},\ \text{C}^3=\text{c}\end{Bmatrix}$
Then, equation $(1)$ becomes
$\text{a}+\text{b}+\text{c}=3(\text{abc})^{\frac{1}{3}}$
Cubing both Sides of above equation, we get
$(a + b + c)^3 = 27abc$
Hence, correct option is $(b).$
View full question & answer→MCQ 241 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,
$\Big(\text{x}+\frac{1}{\text{x}}\Big)^2=\text{x}^2+\Big(\frac{1}{\text{x}}\Big)^2+2\times\not\text{x}\times\frac{1}{\not\text{x}}$
$\Rightarrow\Big(\text{x}+\frac{1}{\text{x}}\Big)^2=\text{x}^2+\frac{1}{\text{x}^2}+2$
$\Rightarrow(5)^2=\text{x}^2+\frac{1}{\text{x}^2}+2$
$\Big\{\text{x}+\frac{1}{\text{x}}=5\text{ given}\Big\}$
$\Rightarrow\text{x}^2+\frac{1}{\text{x}^2}=25-2$
$\Rightarrow\text{x}^2+\frac{1}{\text{x}^2}=23$
Hence, correct option is $(c).$
View full question & answer→MCQ 251 Mark
If $a+b+c=9$ and $a b+b c+c a=23$, then $a^2+b^2+c^2=$
AnswerWe know that $(a+b+c)^2=a^2+b^2+c^2+2(a b+b c+c a)$
Here, $a+b+c=9, a b+b c+c a=23$
Thus, we have
$(9)^2=a^2+b^2+c^2+2(23)$
$81=a^2+b^2+c^2+46$
$a^2+b^2+c^2=81-46$
$a^2+b^2+c^2=35$
Hence, correct option is $(a).$
View full question & answer→MCQ 261 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 $\text{x}+\frac{1}{\text{x}}=2$
$\Rightarrow(2)^3=\text{x}^3+\frac{1}{\text{x}^3}+3(2)$
$\Rightarrow\text{x}^3+\frac{1}{\text{x}^3}=(2)^3-3\times2=8-6=2$
$\Rightarrow\text{x}^2+\frac{1}{\text{x}^2}=23$
Hence, correct option is $(d).$
View full question & answer→MCQ 271 Mark
The product $(a+b)(a-b)\left(a^2-a b+b^2\right)\left(a^2+a b+b^2\right)$ 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. $a^6-b^6$
$(a+b)(a-b)\left(a^2-a b+b^2\right)\left(a^2+a b+b^2\right)$
$=\left(a^2-b^2\right)\left(a^2+b^2-a b\right)\left(a^2+b^2-a b\right)$
$=\left(a^2-b^2\right)\left\{\left(a^2+b^2\right)^2-(a b)^2\right\}$
$=\left(a^2-b^2\right)\left\{a^4+b^4+2 a^2 b^2-a^2 b^2\right\}$
$=\left(a^2-b^2\right)\left\{a^4+b^4+a^2 b^2\right\}$
$=\left\{a^6+a^2 b^4+a^4 b^2-b^2 a^4-b^6-b^4 a^2\right\}$
$=a^6-b^6$
Hence, correct option is $(b).$
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