MCQ 11 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
$x^8+y^8$
- D
$x^16+y^16$
AnswerCorrect option: B. $x^8-y^8$
View full question & answer→MCQ 21 Mark
Which of the following is a factor of $(x+y)^3-\left(x^3+y^3\right)$ ?
- A
$x^2+2 x y+y^2$
- B
$x^2-x y+y^2$
- C
$x y^2$
- ✓
$3 x y(x+y)$
AnswerCorrect option: D. $3 x y(x+y)$
View full question & answer→MCQ 31 Mark
The value of $\frac{(a+b)^2}{(b-c)(c-a)}+\frac{(b+c)^2}{(a-b)(c-a)}+\frac{(c+a)^2}{(a-b)(b-c)}$ is
Answer(a)
We find that
$ \begin{array}{l} \frac{(a+b)^2}{(b-c)(c-a)}+\frac{(b+c)^2}{(a-b)(c-a)}+\frac{(c+a)^2}{(a-b)(b-c)}=\frac{(a-b)(a+b)^2+(b-c)(b+c)^2+(c-a)(c+a)^2}{(a-b)(b-c)(c-a)} \\
=\frac{(a+b)\left(a^2-b^2\right)+(b+c)\left(b^2-c^2\right)+(c+a)\left(c^2-a^2\right)}{(a-b)(b-c)(c-a)}=\frac{a^2 b-a b^2+b^2 c-b c^2+c^2 a-c a^2}{(a-b)(b-c)(c-a)} \\
=\frac{\left(b^2 c-b^2 a\right)+\left(a^2 b-b c^2\right)+\left(c^2 a-c a^2\right)}{(a-b)(b-c)(c-a)}=\frac{b^2(c-a)-b\left(c^2-a^2\right)+c a(c-a)}{(a-b)(b-c)(c-a)}=\frac{(c-a)\left(b^2-b(c+a)+c a\right)}{(a-b)(b-c)(c-a)} \\
=\frac{(c-a)\left\{\left(b^2-b c\right)+(c a-b a)\right\}}{(a-b)(b-c)(c-a)}=\frac{(c-a)\{b(b-c)-a(b-c)\}}{(a-b)(b-c)(c-a)}=\frac{(c-a)(b-c)(b-a)}{(a-b)(b-c)(c-a)}=-1 . \end{array} $
View full question & answer→MCQ 41 Mark
The value of $249^2-248^2$ is
View full question & answer→MCQ 51 Mark
The square root of the expression $(x y+x z-y z)^2-4 x y z(x-y)$ is
- A
$x y+y z-2 x y z$
- B
$x+y-2 x y z$
- C
$x y+z-y$
- ✓
$x y+y z-z x$
AnswerCorrect option: D. $x y+y z-z x$
(d)
We have, $(x y+x z-y z)^2-4 x y z(x-y)$
$=(x y+x z-y z)^2-4(x y)(x z)+4(x y)(y z)$
$=(x y)^2+(x z)^2+(y z)^2-2(x y)(y z)-2(x z)(y z)-4(x y)(x z)+4(x y)(y z)+2(x y)(x z)$
$=(x y)^2+(x z)^2+(y z)^2+2(x y)(y z)-2(x y)(x z)-2(x z)(y z)=(x y-x z+y z)^2$
So, required square root is $(x y-x z+y z)$
View full question & answer→MCQ 61 Mark
The square root of the expression $\frac{1}{a b c}\left(a^2+b^2+c^2\right)+2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)$ is
- A
$\frac{a+b+c}{a b c}$
- B
$\sqrt{a}+\sqrt{b}+\sqrt{c}$
- C
$\sqrt{\frac{b c}{a}}+\sqrt{\frac{c a}{b}}+\sqrt{\frac{a b}{c}}$
- ✓
$\sqrt{\frac{a}{b c}}+\sqrt{\frac{b}{c a}}+\sqrt{\frac{c}{a b}}$
AnswerCorrect option: D. $\sqrt{\frac{a}{b c}}+\sqrt{\frac{b}{c a}}+\sqrt{\frac{c}{a b}}$
(d)
We have
$\frac{1}{a b c}\left(a^2+b^2+c^2\right)+2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=\frac{1}{a b c}\left(a^2+b^2+c^2\right)+2\left(\frac{a b+b c+c a}{a b c}\right)$
$=\frac{1}{a b c}\left(a^2+b^2+c^2+2 a b+2 b c+2 c a\right)=\frac{1}{a b c}(a+b+c)^2=\left(\frac{a+b+c}{\sqrt{a b c}}\right)^2=\left(\sqrt{\frac{a}{b c}}+\sqrt{\frac{b}{c a}}+\sqrt{\frac{c}{a b}}\right)^2$
So, the square root of the given expression is $\sqrt{\frac{a}{b c}}+\sqrt{\frac{b}{c a}}+\sqrt{\frac{c}{a b}}$.
View full question & answer→MCQ 71 Mark
The square root of $\frac{x^2}{9}+\frac{9}{4 x^2}-\frac{x}{3}-\frac{3}{2 x}+\frac{5}{4}$ is
- A
$\frac{2 x}{3}+\frac{3}{2 x}-\frac{1}{2}$
- B
$\frac{x}{3}-\frac{3}{2 x}+1$
- C
$\frac{3}{x}+\frac{2}{3 x}-\frac{1}{2}$
- ✓
$\frac{x}{3}+\frac{3}{2 x}-\frac{1}{2}$
AnswerCorrect option: D. $\frac{x}{3}+\frac{3}{2 x}-\frac{1}{2}$
(d)
$\frac{x^2}{9}+\frac{9}{4 x^2}-\frac{x}{3}-\frac{3}{2 x}+\frac{5}{4}$
=$\left(\frac{x}{3}\right)^2+\left(\frac{3}{2 x}\right)^2+\left(\frac{1}{2}\right)^2+2\left(\frac{x}{3}\right)\left(\frac{1}{2}\right)+2\left(\frac{3}{2 x}\right)\left(\frac{1}{2}\right)+2\left(\frac{x}{3}\right)\left(\frac{3}{2 x}\right)=\left(\frac{x}{3}+\frac{3}{2 x}-\frac{1}{2}\right)^2$
Hence, the required square root is $\frac{x}{3}+\frac{3}{2 x}-\frac{1}{2}$.
View full question & answer→MCQ 81 Mark
The square root of $\frac{a^2}{4}+\frac{1}{a^2}-\frac{1}{a}+\frac{a}{2}-\frac{3}{4}$ is
- ✓
$\frac{a}{2}-\frac{1}{a}+\frac{1}{2}$
- B
$\frac{a}{2}+\frac{2}{a}-1$
- C
$\frac{a}{2}+\frac{1}{a}-\frac{1}{2}$
- D
$\frac{a}{2}-\frac{2}{a}-\frac{1}{2}$
AnswerCorrect option: A. $\frac{a}{2}-\frac{1}{a}+\frac{1}{2}$
(a)
$\frac{a^2}{4}+\frac{1}{a^2}-\frac{1}{a}+\frac{a}{2}-\frac{3}{4}$
$=\left(\frac{a}{2}\right)^2+\left(-\frac{1}{a}\right)^2+\left(\frac{1}{2}\right)^2+2\left(\frac{a}{2}\right)\left(\frac{1}{2}\right)+2\left(-\frac{1}{a}\right)\left(\frac{1}{2}\right)+2\left(\frac{a}{2}\right)\left(-\frac{1}{a}\right)=\left(\frac{a}{2}-\frac{1}{a}+\frac{1}{2}\right)^2$
Hence, required square root is $\left(\frac{a}{2}-\frac{1}{a}+\frac{1}{2}\right)$
View full question & answer→MCQ 91 Mark
The square root of $a+\frac{1}{a}-2$ is
AnswerCorrect option: C. $\pm\left(\sqrt{a}-\frac{1}{\sqrt{a}}\right)$
(c)
We find that
$a+\frac{1}{a}-2=(\sqrt{a})^2+\left(\frac{1}{\sqrt{a}}\right)^2-2 \times \sqrt{a} \times \frac{1}{\sqrt{a}}=\left(\sqrt{a}-\frac{1}{\sqrt{a}}\right)^2$
$\sqrt{a+\frac{1}{a}-2}= \pm\left(\sqrt{a}-\frac{1}{\sqrt{a}}\right)$
View full question & answer→MCQ 101 Mark
The square root of $a^2+\frac{1}{a^2}+2$ is
- ✓
$a+\frac{1}{a}$
- B
$a-\frac{1}{a}$
- C
$a^2+\frac{1}{a^2}$
- D
$a^2-\frac{1}{a^2}$
AnswerCorrect option: A. $a+\frac{1}{a}$
(a)
We find that
$a^2+\frac{1}{a^2}+2=a^2+\frac{1}{a^2}+2 \times a \times \frac{1}{a}=\left(a-\frac{1}{a}\right)^2$
So, the square root of $a^2+\frac{1}{a^2}+2$ is $a+\frac{1}{a}$.
View full question & answer→MCQ 111 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. $x^6-1$
View full question & answer→MCQ 121 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. $a^6-b^6$
View full question & answer→MCQ 131 Mark
The expression $(4 a+5 b+5 c)^2-(5 a+4 b+4 c)^2+9 a^2$ is a perfect square of the expression
- A
$\sqrt{3}(b+c)$
- B
$3(b+c-a)$
- ✓
$3(b+c)$
- D
$3(-b+c-a)$
AnswerCorrect option: C. $3(b+c)$
(c)
$(4 a+5 b+5 c)^2-(5 a+4 b+4 c)^2+9 a^2$
$=(4 a+5 b+5 c+5 a+4 b+4 c)(4 a+5 b+5 c-5 a-4 b-4 c)+9 a^2$
$=9(a+b+c)(-a+b+c)+9 a^2$
$=9\{(b+c)+a\}\{(b+c)-a\}+9 a^2=9\left\{(b+c)^2-a^2\right\}+9 a^2=9(b+c)^2=\{3(b+c)\}^2$
View full question & answer→MCQ 141 Mark
The expression $(3 a+2 b+3 c)^2-(2 a+3 b+2 c)^2+5 b^2$ is perfect square of the expression
- A
$\sqrt{5}(a+b+c)$
- B
$\sqrt{5}(a+b)$
- ✓
$\sqrt{5}(a+c)$
- D
$\sqrt{5}(a+c-b)$
AnswerCorrect option: C. $\sqrt{5}(a+c)$
(c)
$(3 a+2 b+3 c)^2-(2 a+3 b+2 c)^2+5 b^2$
$=(3 a+2 b+3 c+2 a+3 b+2 c)(3 a+2 b+3 c-2 a-3 b-2 c)+5 b^2$
$=(5 a+5 b+5 c)(a-b+c)+5 b^2=5|(a+c)+b|\left\{(a+c)-b \mid+5 b^2\right.$
$=5\left\{(a+c)^2-b^2\right\}+5 b^2=5(a+c)^2=\{\sqrt{5}(a+c)\}^2$
View full question & answer→MCQ 151 Mark
The coefficient of $x$ in $(x+3)^3$ is
View full question & answer→MCQ 161 Mark
$\sqrt{(a+b+c)^2+(a+b-c)^2+2\left(c^2-a^2-b^2-2 a b\right)}$ is equal to
Answer(a)
We find that,
$(a+b+c)^2+(a+b-c)^2+2\left(c^2-a^2-b^2-2 a b\right)$
$=2\left(a^2+b^2+c^2+2 a b\right)+2\left(c^2-a^2-b^2-2 a b\right)=2\left(2 c^2\right)=4 c^2$
$\therefore \quad \sqrt{(a+b+c)^2+(a+b-c)^2+2\left(c^2-a^2-b^2-2 a b\right)}=2 c$
View full question & answer→MCQ 171 Mark
One of the factors of $(5 x+1)^2-(5 x-1)^2$ is
View full question & answer→MCQ 181 Mark
If $x+y=2$ and $x y=1$, then $x^4+y^4=$
View full question & answer→MCQ 191 Mark
If $x-\frac{1}{x}=\frac{15}{4}$, then $x+\frac{1}{x}=$
- A
- ✓
$\frac{17}{4}$
- C
$\frac{13}{4}$
- D
$\frac{1}{4}$
AnswerCorrect option: B. $\frac{17}{4}$
View full question & answer→MCQ 201 Mark
If $x+\frac{1}{x}=7$, then $x^3-\frac{1}{x^3}=$
- A
$9 \sqrt{5}$
- ✓
$144 \sqrt{5}$
- C
$135 \sqrt{5}$
- D
$\sqrt{5}$
AnswerCorrect option: B. $144 \sqrt{5}$
(b)
We have,$x+\frac{1}{x}=7$
$\Rightarrow \quad\left(x+\frac{1}{x}\right)^2=7^2 \Rightarrow x^2+\frac{1}{x^2}+2=49 \Rightarrow x^2+\frac{1}{x^2}=47$
$\Rightarrow \quad x^2+\frac{1}{x^2}-2=45 \Rightarrow\left(x-\frac{1}{x}\right)^2=(3 \sqrt{5})^2 \Rightarrow x-\frac{1}{x}=3 \sqrt{5}$
$\therefore \quad x^3-\frac{1}{x^3}=\left(x-\frac{1}{x}\right)\left(x^2+\frac{1}{x^2}+1\right)=3 \sqrt{5}(47+1)=144 \sqrt{5}$
View full question & answer→MCQ 211 Mark
If $x+\frac{1}{x}=5$, then $x^2+\frac{1}{x^2}=$
View full question & answer→MCQ 221 Mark
If $x+\frac{1}{x}=4$, then $x^4+\frac{1}{x^4}=$
View full question & answer→MCQ 231 Mark
If $x+\frac{1}{x}=3$, then $x^6+\frac{1}{x^6}=$
View full question & answer→MCQ 241 Mark
If $x+\frac{1}{x}=2$, then $x^3+\frac{1}{x^3}=$
View full question & answer→MCQ 251 Mark
If $x^4+\frac{1}{x^4}=623$, then $x+\frac{1}{x}=$
- A
- B
- ✓
$3 \sqrt{3}$
- D
$-3 \sqrt{3}$
AnswerCorrect option: C. $3 \sqrt{3}$
View full question & answer→MCQ 261 Mark
If $x^4+\frac{1}{x^4}=194$, then $x^3+\frac{1}{x^3}=$
View full question & answer→MCQ 271 Mark
If $x^3-\frac{1}{x^3}=14$, then $x-\frac{1}{x}=$
View full question & answer→MCQ 281 Mark
If $x^3+\frac{1}{x^3}=110$, then $x+\frac{1}{x}=$
View full question & answer→MCQ 291 Mark
If $x^2+y^2+x y=1$ and $x+y=2$, then $x y=$
View full question & answer→MCQ 301 Mark
If $x^2+\frac{1}{x^2}=102$, then $x-\frac{1}{x}=$
View full question & answer→MCQ 311 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
AnswerCorrect option: B. $3, x-3, x+3$
View full question & answer→MCQ 321 Mark
If $\left(a+\frac{1}{a}+2\right)^2=4$, then $a^2+\frac{1}{a^2}=$
Answer(c)
We have,
$\left(a+\frac{1}{a}+2\right)^2=4$
$\Rightarrow \quad a+\frac{1}{a}+2= \pm 2 \Rightarrow a+\frac{1}{a}=0$ or, $a+\frac{1}{a}=-4$
Now, $\quad a+\frac{1}{a}=0 \Rightarrow a^2+1=0$, which is impossible. Therefore, $a+\frac{1}{a} \neq 0$
$\therefore \quad a+\frac{1}{a}=-4 \Rightarrow\left(a+\frac{1}{a}\right)^2=16 \Rightarrow a^2+\frac{1}{a^2}+2=16 \Rightarrow a^2+\frac{1}{a^2}=14$
View full question & answer→MCQ 331 Mark
If $\frac{x}{y}+\frac{y}{x}=-1(x, y \neq 0)$, the value of $x^3-y^3$ is
View full question & answer→MCQ 341 Mark
If $\frac{a}{b}+\frac{b}{a}=2$, then $\left(\frac{a}{b}\right)^{10}-\left(\frac{b}{a}\right)^{10}$ is equal to
Answer(c)
We have, $\frac{a}{b}+\frac{b}{a}=2$
$\Rightarrow \quad\left(\sqrt{\frac{a}{b}}\right)^2+\left(\sqrt{\frac{b}{a}}\right)^2-2 \sqrt{\frac{a}{b}} \times \sqrt{\frac{b}{a}}=0$
$\Rightarrow \quad\left(\sqrt{\frac{a}{b}}-\sqrt{\frac{b}{a}}\right)^2=0 \Rightarrow \sqrt{\frac{a}{b}}=\sqrt{\frac{b}{a}} \Rightarrow \frac{a}{b}=\frac{b}{a} \Rightarrow\left(\frac{a}{b}\right)^{10}=\left(\frac{b}{a}\right)^{10} \Rightarrow\left(\frac{a}{b}\right)^{10}-\left(\frac{b}{a}\right)^{10}=0$.
View full question & answer→MCQ 351 Mark
If $\frac{a}{b}+\frac{b}{a}=1$, then $a^3+b^3=$
View full question & answer→MCQ 361 Mark
If $\frac{a}{b}+\frac{b}{a}=-1$, then $a^3-b^3=$
Answer(d)
We have,
$\frac{a}{b}+\frac{b}{a}=-1 \Rightarrow a^2+b^2+a b=0$
$\therefore \quad a^3-b^3=(a-b)\left(a^2+a b+b^2\right)=(a-b) \times 0=0$
View full question & answer→MCQ 371 Mark
If $a, b, c$ are natural numbers such that $a^2+b^2+c^2=29$ and $a b+b c+c a=26$, and $a+b+c=$ _____
View full question & answer→MCQ 381 Mark
If $a+b+c=9$ and $a b+b c+c a=23$, then $a^3+b^3+c^3-3 a b c=$
View full question & answer→MCQ 391 Mark
If $a+b+c=9$ and $a b+b c+c a=23$, then $a^2+b^2+c^2=$
View full question & answer→MCQ 401 Mark
If $a b c=6$ and $a+b+c=6$, then $\frac{1}{a c}+\frac{1}{a b}+\frac{1}{b c}=$
Answer(b)
We have, $a+b+c=6$ and $a b c=6$
$\Rightarrow \quad \frac{a+b+c}{a b c}=\frac{6}{6} \Rightarrow \frac{1}{b c}+\frac{1}{c a}+\frac{1}{a b}=1$
View full question & answer→MCQ 411 Mark
If $a+b+c=0$, then $\frac{a^2}{b c}+\frac{b^2}{c a}+\frac{c^2}{a b}=$
View full question & answer→MCQ 421 Mark
If $a-b=-8$ and $a b=-12$, then $a^3-b^3=$
View full question & answer→MCQ 431 Mark
If a + b = 8 and ab = 12 then $a^3+b^3=$
Answer(b)
We have, a + b = 8 and ab = 12
Now, $\quad a+b=8 \Rightarrow(a+b)^2=64 \Rightarrow a^2+b^2+2 a b=64 \Rightarrow a^2+b^2+2 \times 12=64 \Rightarrow a^2+b^2=40$
$ \therefore \quad a^3+b^3=(a+b)\left(a^2-a b+b^2\right)=8(40-12)=8 \times 28=224
$
View full question & answer→MCQ 441 Mark
If $a+b=3$ and $a b=2$, then $a^3+b^3=$
View full question & answer→MCQ 451 Mark
If $a^2+b^2+c^2-a b-b c-c a=0$, then
View full question & answer→MCQ 461 Mark
If $a^{1 / 3}+b^{1 / 3}+c^{1 / 3}=0$, then
- A
- ✓
$(a+b+c)^3=27 a b c$
- C
$a+b+c=3 a b c$
- D
$a^3+b^3+c^3=0$
AnswerCorrect option: B. $(a+b+c)^3=27 a b c$
View full question & answer→MCQ 471 Mark
If $9 x^2-b=\left(3 x+\frac{1}{2}\right)\left(3 x-\frac{1}{2}\right)$, then the value of $b$ is
- A
$0$
- B
$\frac{1}{\sqrt{2}}$
- ✓
$\frac{1}{4}$
- D
$\frac{1}{2}$
AnswerCorrect option: C. $\frac{1}{4}$
View full question & answer→MCQ 481 Mark
If $49 a^2-b=\left(7 a+\frac{1}{2}\right)\left(7 a-\frac{1}{2}\right)$, then the value of $b$ is
- A
$0$
- ✓
$\frac{1}{4}$
- C
$\frac{1}{\sqrt{2}}$
- D
$\frac{1}{2}$
AnswerCorrect option: B. $\frac{1}{4}$
View full question & answer→MCQ 491 Mark
If $3 x+\frac{2}{x}=7$, then $\left(9 x^2-\frac{4}{x^2}\right)=$
View full question & answer→MCQ 501 Mark
If $2 x+\frac{y}{3}=12$ and $x y=30$, then $8 x^3+\frac{y^3}{27}=$
View full question & answer→