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# Algebra Questions for MAH-CET [PDF]

Question 1: If $\sqrt{x}-\frac{1}{\sqrt{x}}=\sqrt{7}$, then the value of $x^2 + \frac{1}{x^2}$ is:

a) 81

b) 60

c) 79

d) 75

Solution:

$\sqrt{x}-\frac{1}{\sqrt{x}}=\sqrt{7}$

$\left(\sqrt{x}-\frac{1}{\sqrt{x}}\right)^2=\left(\sqrt{7}\right)^2$

$x+\frac{1}{x}-2=7$

$x+\frac{1}{x}=9$

$\left(x+\frac{1}{x}\right)^2=9^2$

$x^2+\frac{1}{x^2}+2=81$

$x^2+\frac{1}{x^2}=79$

Hence, the correct answer is Option C

Question 2: If $(56\sqrt{7}x^3-2\sqrt{2}y^3)\div(2\sqrt{7}x-\sqrt{2}y)=Ax^2+By^2-Cxy$, then find the value of $A + B – \sqrt{14}C$.

a) 38

b) 10

c) 19

d) 58

Solution:

$(56\sqrt{7}x^3-2\sqrt{2}y^3)\div(2\sqrt{7}x-\sqrt{2}y)=Ax^2+By^2-Cxy$

$\frac{\left(2\sqrt{7}x-\sqrt{2}y\right)\left(28x^2+2\sqrt{14}xy+2y^2\right)}{\left(2\sqrt{7}x-\sqrt{2}y\right)}=Ax^2+By^2-Cxy$

$28x^2+2\sqrt{14}xy+2y^2=Ax^2+By^2-Cxy$

Comparing both sides,

A = 28, B = 2, C = $-2\sqrt{14}$

$A+B-\sqrt{14}C=28+2-\sqrt{14}\left(-2\sqrt{14}\right)$

$=30+28$

$=58$

Hence, the correct answer is Option D

Question 3: If $\frac{x}{y} + \frac{y}{x} = 2, (x, y \neq 0)$, then the value of $(x – y)$ is:

a) 1

b) 0

c) 2

d) -2

Solution:

$\frac{x}{y}+\frac{y}{x}=2$

$\frac{x^2+y^2}{xy}=2$

$x^2+y^2=2xy$

$x^2+y^2-2xy=0$

$\left(x-y\right)^2=0$

$x-y=0$

Hence, the correct answer is Option B

Question 4: If $\left(2a+\frac{3}{a}-1\right)=11$, what is the value of $\left(4a^2 + \frac{9}{a^2}\right)?$

a) 121

b) 148

c) 132

d) 110

Solution:

$\left(2a+\frac{3}{a}-1\right)=11$

$2a+\frac{3}{a}=12$

$4a^2+\frac{9}{a^2}+2.2a.\frac{3}{a}=144$

$4a^2+\frac{9}{a^2}+12=144$

$4a^2+\frac{9}{a^2}=132$

Hence, the correct answer is Option C

Question 5: If $a^3 – b^3 = 2349$ and $(a – b) = 9$, then $(a + b)^2 – ab$ is equal to:

a) 280

b) 244

c) 261

d) 229

Solution:

$(a-b)=9$………….(1)

$(a-b)^3=729$

$a^3-b^3-3ab\left(a-b\right)=729$

$2349-3ab\left(9\right)=729$

$27ab=1620$

$ab=60$…………..(2)

$(a-b)=9$

$(a-b)^2=81$

$a^2+b^2-2ab=81$

$a^2+b^2-2\left(60\right)=81$

$a^2+b^2-120=81$

$a^2+b^2=201$……….(3)

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

$=a^2+b^2+ab$

$=201+60$

$=261$

Hence, the correct answer is Option C

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Question 6: If $x – \frac{1}{x} = \sqrt{77}$, then one of the values of $x^3 + \frac{1}{x^3}$ is:

a) $80\sqrt{77}$

b) -702

c) $77\sqrt{77}$

d) $3\sqrt{77}$

Solution:

$x – \frac{1}{x} = \sqrt{77}$

$\left(x-\frac{1}{x}\right)^2=77$

$x^2+\frac{1}{x^2}-2=77$

$x^2+\frac{1}{x^2}=79$

$x^2+\frac{1}{x^2}+2=81$

$\left(x+\frac{1}{x}\right)^2=81$

$x+\frac{1}{x}=9$ or $x+\frac{1}{x}=-9$

When $x+\frac{1}{x}=-9$

$x^3+\frac{1}{x^3}+3.x.\frac{1}{x}\left(x+\frac{1}{x}\right)=-729$

$x^3+\frac{1}{x^3}+3\left(-9\right)=-729$

$x^3+\frac{1}{x^3}-27=-729$

$x^3+\frac{1}{x^3}=-702$

Hence, the correct answer is Option B

Question 7: If $\sqrt{x}=\sqrt{3}-\sqrt{5}$, then the value of $x^2-16x+6$ is:

a) 4

b) 0

c) 2

d) -2

Solution:

Given,  $\sqrt{x}=\sqrt{3}-\sqrt{5}$

$\Rightarrow$  $x=\left(\sqrt{3}-\sqrt{5}\right)^2$

$\Rightarrow$  $x=3+5-2\sqrt{15}$

$\Rightarrow$  $x=8-2\sqrt{15}$ ……………(1)

$\Rightarrow$  $x^2=\left(8-2\sqrt{15}\right)^2$

$\Rightarrow$  $x^2=64+60-32\sqrt{15}$

$\Rightarrow$  $x^2=124-32\sqrt{15}$ ………..(2)

$\therefore\$ $x^2-16x+6=124-32\sqrt{15}-16\left(8-2\sqrt{15}\right)+6$

$=124-32\sqrt{15}-128+32\sqrt{15}+6$

$=130-128$

$=2$

Hence, the correct answer is Option C

Question 8: If $x=\frac{\sqrt{3}}{2}$, then the value of $\frac{\sqrt{1+x}+\sqrt{1-x}}{\sqrt{1+x}-\sqrt{1-x}}$ is equal to:

a) $\sqrt 2$

b) $\sqrt 3$

c) 3

d) 2

Solution:

Given, $x=\frac{\sqrt{3}}{2}$

$\frac{\sqrt{1+x}+\sqrt{1-x}}{\sqrt{1+x}-\sqrt{1-x}}=\frac{\sqrt{1+x}+\sqrt{1-x}}{\sqrt{1+x}-\sqrt{1-x}}\times\frac{\sqrt{1+x}+\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}$

$=\frac{1+x+1-x+2\left(\sqrt{1+x}\right)\left(\sqrt{1-x}\right)}{1+x-\left(1-x\right)}$

$=\frac{2+2\left(\sqrt{1-x^2}\right)}{2x}$

$=\frac{1+\sqrt{1-x^2}}{x}$

$=\frac{1+\sqrt{1-\left(\frac{\sqrt{3}}{2}\right)^2}}{\frac{\sqrt{3}}{2}}$

$=\frac{1+\sqrt{1-\frac{3}{4}}}{\frac{\sqrt{3}}{2}}$

$=\frac{1+\frac{1}{2}}{\frac{\sqrt{3}}{2}}$

$=\frac{3}{2}\times\frac{2}{\sqrt{3}}$

$=\sqrt{3}$

Hence, the correct answer is Option B

Question 9: If $a^3+b^3=62$ and a + b = 2, then the value of ab is:

a) -6

b) 9

c) 6

d) -9

Solution:

Given,  $a+b=2$

$a^3+b^3=62$

$\Rightarrow$  $\left(a+b\right)\left(a^2-ab+b^2\right)=62$

$\Rightarrow$  $\left(2\right)\left(a^2+2ab+b^2-3ab\right)=62$

$\Rightarrow$  $\left(a+b\right)^2-3ab=31$

$\Rightarrow$  $\left(2\right)^2-3ab=31$

$\Rightarrow$  $4-3ab=31$

$\Rightarrow$  $3ab=-27$

$\Rightarrow$  $ab=-9$

Hence, the correct answer is Option D

Question 10: If $a-b=18$ and $a^3-b^3=324$, then find ab.

a) 105

b) -102

c) -104

d) 103

Solution:

Given,  $a-b=18$ and

$a^3-b^3=324$

$\Rightarrow$  $\left(a-b\right)\left(a^2+ab+b^2\right)=324$

$\Rightarrow$  $\left(18\right)\left(a^2-2ab+b^2+3ab\right)=324$

$\Rightarrow$  $\left(a-b\right)^2+3ab=18$

$\Rightarrow$  $\left(18\right)^2+3ab=18$

$\Rightarrow$  $324+3ab=18$

$\Rightarrow$  $3ab=-306$

$\Rightarrow$  $ab=-102$

Hence, the correct answer is Option B

Question 11: If $x^4+\frac{1}{x^4}=14159$, then the value of $x+\frac{1}{x}$ is:

a) 11

b) 12

c) 9

d) 10

Solution:

Given,  $x^4+\frac{1}{x^4}=14159$

$\Rightarrow$  $x^4+\frac{1}{x^4}+2=14159+2$

$\Rightarrow$  $\left(x^2+\frac{1}{x^2}\right)^2=14161$

$\Rightarrow$  $\left(x^2+\frac{1}{x^2}\right)^2=\left(119\right)^2$

$\Rightarrow$  $x^2+\frac{1}{x^2}=119$

$\Rightarrow$  $x^2+\frac{1}{x^2}+2=119+2$

$\Rightarrow$  $\left(x+\frac{1}{x}\right)^2=121$

$\Rightarrow$   $x+\frac{1}{x}=11$

Hence, the correct answer is Option A

Question 12: If $p+\frac{1}{p}=112$, find $(p-112)^{15}+\frac{1}{p^{15}}$

a) 1

b) 15

c) 10

d) 0

Solution:

Given, $p+\frac{1}{p}=112$

$\Rightarrow$  $p-112=-\frac{1}{p}$

$(p-112)^{15}+\frac{1}{p^{15}}=\left(-\frac{1}{p}\right)^{15}+\frac{1}{p^{15}}$

$=-\frac{1}{p^{15}}+\frac{1}{p^{15}}$

$=0$

Hence, the correct answer is Option D

Question 13: If $\frac{4}{3}\left(x^2+\frac{1}{x^2}\right)=110\frac{2}{3}$, find $\frac{1}{9} \left(x^3 – \frac{1}{x^3}\right)$, where $x$ > 0.

a) 74

b) 76

c) 84

d) 85

Solution:

Given, $\frac{4}{3}\left(x^2+\frac{1}{x^2}\right)=110\frac{2}{3}$

$\Rightarrow$  $\frac{4}{3}\left(x^2+\frac{1}{x^2}\right)=\frac{332}{3}$

$\Rightarrow$  $x^2+\frac{1}{x^2}=83$

$\Rightarrow$  $x^2+\frac{1}{x^2}-2=83-2$

$\Rightarrow$  $\left(x-\frac{1}{x}\right)^2=81$

$\Rightarrow$  $x-\frac{1}{x}=9$

$\Rightarrow$  $\left(x-\frac{1}{x}\right)^3=9^3$

$\Rightarrow$  $x^3-\frac{1}{x^3}-3.x.\frac{1}{x}\left(x-\frac{1}{x}\right)=729$

$\Rightarrow$  $x^3-\frac{1}{x^3}-3\left(9\right)=729$

$\Rightarrow$  $x^3-\frac{1}{x^3}-27=729$

$\Rightarrow$  $x^3-\frac{1}{x^3}=756$

$\Rightarrow$  $\frac{1}{9}\left(x^3-\frac{1}{x^3}\right)=\frac{1}{9}\left(756\right)$

$\Rightarrow$  $\frac{1}{9}\left(x^3-\frac{1}{x^3}\right)=84$

Hence, the correct answer is Option C

Question 14: If $x^3+y^3=16$ and $x+y=4$, then the value of $x^4+y^4$ is:

a) 48

b) 32

c) 64

d) 30

Solution:

Given,  $x^3+y^3=16$ and $x+y=4$

$\Rightarrow$  $\left(x+y\right)\left(x^2-xy+y^2\right)=16$

$\Rightarrow$  $\left(4\right)\left(x^2+2xy+y^2-3xy\right)=16$

$\Rightarrow$  $\left(x+y\right)^2-3xy=4$

$\Rightarrow$  $\left(4\right)^2-3xy=4$

$\Rightarrow$  $3xy=16-4$

$\Rightarrow$  $3xy=12$

$\Rightarrow$  $xy=4$

$\therefore\$ $x^4+y^4=x^4+y^4+2x^2y^2-2x^2y^2$

$=\left[x^2+y^2\right]^2-2\left(xy\right)^2$

$=\left[x^2+y^2+2xy-2xy\right]^2-2\left(4\right)^2$

$=\left[\left(x+y\right)^2-2xy\right]^2-32$

$=\left[\left(4\right)^2-2\left(4\right)\right]^2-32$

$=\left[16-8\right]^2-32$

$=64-32$

$=32$

Hence, the correct answer is Option B

Question 15: If $a^2+\frac{2}{a^2}=16$, then find the value of $\frac{72a^2}{a^4+2+8a^2}$

a) 2

b) 4

c) 1

d) 3

Solution:

Given,  $a^2+\frac{2}{a^2}=16$

$\frac{72a^2}{a^4+2+8a^2}=\frac{72a^2}{a^2\left(a^2+\frac{2}{a^2}+8\right)}$

$=\frac{72}{a^2+\frac{2}{a^2}+8}$

$=\frac{72}{16+8}$

$=\frac{72}{24}$

$=3$

Hence, the correct answer is Option D

Question 16: If a + 3b = 12 and ab = 9, then the value of (a – 3b) is:

a) 9

b) 8

c) 6

d) 4

Solution:

Given, $a+3b=13$

$\Rightarrow$  $\left(a+3b\right)^2=12^2$

$\Rightarrow$  $a^2+9b^2+6ab=144$

$\Rightarrow$  $a^2+9b^2+6\left(9\right)=144$

$\Rightarrow$  $a^2+9b^2+54-6ab+6ab=144$

$\Rightarrow$  $a^2+9b^2-6ab+6ab=90$

$\Rightarrow$  $\left(a-3b\right)^2+6ab=90$

$\Rightarrow$  $\left(a-3b\right)^2+6\left(9\right)=90$

$\Rightarrow$  $\left(a-3b\right)^2+54=90$

$\Rightarrow$  $\left(a-3b\right)^2=36$

$\Rightarrow$  $a-3b=6$

Hence, the correct answer is Option C

Question 17: If $1 + 9r^2 + 81r^4 = 256$ and $1 + 3r + 9r^2 = 32$, then find the value of $1 – 3r + 9r^2$.

a) 4

b) 8

c) 16

d) 12

Solution:

Given that $1 + 3r + 9r^2 = 32$

and let $1 – 3r + 9r^2$ be K

Multiply $1 + 3r + 9r^2$ and $1 – 3r + 9r^2$

we get the product as $1 + 9r^2 + 81r^4$ = 32K

but $1 + 9r^2 + 81r^4$= 256
substitute this value in the product

256 = 32k

K=8

Question 18: If $x^3 — 6x^2 + ax + b$ is divisible by $(x^2 — 3x + 2)$, then the values of a and b are:

a) a = -6 and b = -11

b) a = -11 and b = 6

c) a = 11 and b = -6

d) a = 6 and b = 11

Solution:

Given, $x^3—6x^2+ax+b$ is divisible by $(x^2 — 3x + 2)$

Let the quotient when $x^3 — 6x^2 + ax + b$ is divisible by $(x^2 — 3x + 2)$ be $x-p$

$\Rightarrow$ $(x^2—3x+2)\left(x-p\right)=x^3—6x^2+ax+b$

$\Rightarrow$  $x^3-3x^2+2x-px^2+3px-2p=x^3—6x^2+ax+b$

$\Rightarrow$  $x^3-\left(3+p\right)x^2+\left(2+3p\right)x-2p=x^3—6x^2+ax+b$

Comparing both sides,

$-\left(3+p\right)=-6$

$\Rightarrow$  $p=3$

$2+3p=a$

$\Rightarrow$  $2+3\left(3\right)=a$

$\Rightarrow$  $a=11$

$-2p=b$

$\Rightarrow$  $-2\left(3\right)=b$

$\Rightarrow$  $b=-6$

$\therefore\$ a = 11 and b = -6

Hence, the correct answer is Option C

Question 19: If $x – \frac{1}{x} = 13,$ then the value of $x^2 + \frac{1}{x^2}$ is:

a) 165

b) 171

c) 167

d) 169

Solution:

Given, $x-\frac{1}{x}=13$

$\Rightarrow$ $\left(x-\frac{1}{x}\right)^2=13^2$

$\Rightarrow$ $x^2-2.x.\frac{1}{x}+\frac{1}{x^2}=169$

$\Rightarrow$  $x^2+\frac{1}{x^2}-2=169$

$\Rightarrow$  $x^2+\frac{1}{x^2}=171$

Hence, the correct answer is Option B

Question 20: If $A = \frac{x – 1}{x + 1}$, then the value of $A – \frac{1}{A}$ is:

a) $\frac{-4(2x – 1)}{x^2 – 1}$

b) $\frac{x^2 – 1}{-4(2x – 1)}$

c) $\frac{x^2 – 1}{-4(2x + 1)}$

d) $\frac{-4x}{x^2 – 1}$

Solution:

Given, $A=\frac{x-1}{x+1}$

$A-\frac{1}{A}$ = $\frac{x-1}{x+1}-\frac{x+1}{x-1}$

$=\frac{\left(x-1\right)^2-\left(x+1\right)^2}{x^2-1}$

$=\frac{x^2-2x+1-\left(x^2+2x+1\right)^{ }}{x^2-1}$

$=\frac{x^2-2x+1-x^2-2x-1}{x^2-1}$

$=\frac{-4x}{x^2-1}$

Hence, the correct answer is Option D