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# Algebra Questions for MAH-CET

Question 1: What is the coefficient of $x^2$ in the expansion of $\left(5-\frac{x^2}{3}\right)^3$?

a) -25

b) $-\frac{25}{3}$

c) 25

d) $-\frac{5}{3}$

Solution:

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

= $\left(5-\frac{x^2}{3}\right)\left(25+\frac{x^4}{9}-\frac{10x^2}{3}\right)$

= $125+\frac{5x^4}{9}-\frac{50x^2}{3}-\frac{25x^2}{3}-\frac{x^6}{27}+\frac{10x^4}{9}$

= $-\frac{x^6}{27}+\frac{15x^4}{9}-\frac{75x^2}{3}+125$

= $-\frac{x^6}{27}+\frac{5x^4}{3}-25x^2+125$

The coefficient of $x^2$ in the expansion = -25

Hence, the correct answer is Option A

Question 2: Given that $x^8 – 34x^4 + 1 = 0, x > 0$. What is the value of $(x^3 – x^{-3})$?

a) 14

b) 12

c) 18

d) 16

Solution:

$x^8-34x^4+1=0$

$x^8+1=34x^4$

$x^4+\frac{1}{x^4}=34$

$x^4+\frac{1}{x^4}+2=36$

$\left(x^2+\frac{1}{x^2}\right)^2=36$

$x^2+\frac{1}{x^2}=6$

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

$\left(x-\frac{1}{x}\right)^2=4$

$x-\frac{1}{x}=2$……..(1)

$\left(x-\frac{1}{x}\right)^3=8$

$x^3-\frac{1}{x^3}-3.x.\frac{1}{x}\left(x-\frac{1}{x}\right)=8$

$x^3-\frac{1}{x^3}-3\left(2\right)=8$

$x^3-\frac{1}{x^3}-6=8$

$x^3-\frac{1}{x^3}=14$

Hence, the correct answer is Option A

Question 3: If $x^4 – 62 x^2 + 1 = 0$, where $x > 0$, then the value of $x^3 + x^{-3}$ is:

a) 500

b) 512

c) 488

d) 364

Solution:

$x^4-62x^2+1=0$

$x^4+1=62x^2$

$x^2+\frac{1}{x^2}=62$

$x^2+\frac{1}{x^2}+2=64$

$\left(x+\frac{1}{x}\right)^2=64$

$x+\frac{1}{x}=8$…….(1)

$\left(x+\frac{1}{x}\right)^3=512$

$x^3+\frac{1}{x^3}+3.x.\frac{1}{x}\left(x+\frac{1}{x}\right)=512$

$x^3+\frac{1}{x^3}+3\left(8\right)=512$

$x^3+\frac{1}{x^3}+24=512$

$x^3+\frac{1}{x^3}=488$

Hence, the correct answer is Option C

Question 4: If $x + \frac{1}{x} = \frac{17}{4}, x > 1$, then what is the value of $x – \frac{1}{x}?$

a) $\frac{9}{4}$

b) $\frac{3}{2}$

c) $\frac{8}{3}$

d) $\frac{15}{4}$

Solution:

$x+\frac{1}{x}=\frac{17}{4}$

$\left(x+\frac{1}{x}\right)^2=\frac{289}{16}$

$x^2+\frac{1}{x^2}+2=\frac{289}{16}$

$x^2+\frac{1}{x^2}=\frac{289}{16}-2$

$x^2+\frac{1}{x^2}=\frac{257}{16}$

$x^2+\frac{1}{x^2}-2=\frac{257}{16}-2$

$\left(x-\frac{1}{x}\right)^2=\frac{257-32}{16}$

$\left(x-\frac{1}{x}\right)^2=\frac{225}{16}$

$x-\frac{1}{x}=\frac{15}{4}$

Hence, the correct answer is Option D

Question 5: If $2x^2 – 7x + 5 = 0$, then what is the value of $x^3 + \frac{125}{8x^3}$?

a) $12\frac{5}{8}$

b) $16\frac{5}{8}$

c) $10\frac{5}{8}$

d) $18\frac{5}{8}$

Solution:

$2x^2-7x+5=0$

$2x^2-2x-5x+5=0$

$2x\left(x-1\right)-5\left(x-1\right)=0$

$\left(x-1\right)\left(2x-5\right)=0$

$x-1=0$ or $2x-5=0$

$x=1$ or $x=\frac{5}{2}$

When $x=1$,

$x^3+\frac{125}{8x^3}=\left(1\right)^3+\frac{125}{8\left(1\right)^3}=1+\frac{125}{8}=\frac{133}{8}=16\frac{5}{8}$

Hence, the correct answer is Option B

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Question 6: If $x – \frac{1}{x} = 1$, then what is the value of $x^8 + \frac{1}{x^8}?$

a) 3

b) 119

c) 47

d) -1

Solution:

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

Squaring on both sides,

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

$x^2+\frac{1}{x^2}=3$

Squaring on both sides,

$x^4+\frac{1}{x^4}+2=9$

$x^4+\frac{1}{x^4}=7$

Squaring on both sides,

$x^8+\frac{1}{x^8}+2=49$

$x^8+\frac{1}{x^8}=47$

Hence, the correct answer is Option C

Question 7: If $x^4 + \frac{1}{x^4} = 727, x > 1$, then what is the value of $\left(x – \frac{1}{x}\right)?$

a) 6

b) -6

c) -5

d) 5

Solution:

$x^4+\frac{1}{x^4}=727$

$x^4+\frac{1}{x^4}+2=729$

$\left(x^2+\frac{1}{x^2}\right)^2=729$

$x^2+\frac{1}{x^2}=27$

$x^2+\frac{1}{x^2}-2=25$

$\left(x-\frac{1}{x}\right)^2=25$

Since $x>1$,

$x-\frac{1}{x}=5$

Hence, the correct answer is Option D

Question 8: If $2x^2 – 8x – 1 = 0$, then what is the value of $8x^3 – \frac{1}{x^3}?$

a) 560

b) 540

c) 524

d) 464

Solution:

$2x^2-8x-1=0$

$2x^2-1=8x$

$2x-\frac{1}{x}=8$……..(1)

Cubing on both sides,

$8x^3-\frac{1}{x^3}-3.2x.\frac{1}{x}\left(2x-\frac{1}{x}\right)=512$

$8x^3-\frac{1}{x^3}-6\left(8\right)=512$  [From (1)]

$8x^3-\frac{1}{x^3}-48=512$

$8x^3-\frac{1}{x^3}=560$

Hence, the correct answer is Option A

Question 9: If $y = 2x + 1$, then what is the value of $(8x^3 – y^3 + 6xy)$?

a) 1

b) -1

c) 15

d) -15

Solution:

$y=2x+1$

$2x-y=-1$…….(1)

Cubing on both sides, we get

$8x^3-y^3-3.2x.y\left(2x-y\right)=-1$

$8x^3-y^3-6xy\left(-1\right)=-1$ [From (1)]

$8x^3-y^3+6xy=-1$

Hence, the correct answer is Option B

Question 10: If $x – \frac{2}{x} = 15$, then what is the value of $\left(x^2 + \frac{4}{x^2}\right)$?

a) 229

b) 227

c) 221

d) 223

Solution:

$x-\frac{2}{x}=15$

Squaring on both sides,

$x^2+\frac{4}{x^2}-2.x.\frac{2}{x}=225$

$x^2+\frac{4}{x^2}-4=225$

$x^2+\frac{4}{x^2}=229$

Hence, the correct answer is Option A

Question 11: If $2x + 3y + 1 = 0$, then what is the value of $\left(8x^3 + 8 + 27y^3 – 18xy \right)$?

a) -7

b) 7

c) -9

d) 9

Solution:

$2x+3y+1=0$

$2x+3y=-1$……..(1)

Cubing on both sides,

$8x^3+27y^3+3.2x.3y\left(2x+3y\right)=-1$

$8x^3+27y^3+18xy\left(-1\right)=-1$

$8x^3+27y^3-18xy+8=-1+8$

$8x^3+27y^3-18xy+8=7$

Hence, the correct answer is Option B

Question 12: If $x + \frac{1}{x} = 7$, then $x^2 + \frac{1}{x^2}$ is equal to:

a) 47

b) 49

c) 61

d) 51

Solution:

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

Squaring on both sides,

$x^2+\frac{1}{x^2}+2.x.\frac{1}{x}=49$

$x^2+\frac{1}{x^2}+2=49$

$x^2+\frac{1}{x^2}=47$

Hence, the correct answer is Option A

Question 13: If $(2x + y)^3 – (x – 2y)^3 = (x + 3y)[Ax^2 + By^2 + Cxy]$, then what is the value of $(A + 2B + C)?$

a) 13

b) 14

c) 7

d) 10

Solution:

$(2x+y)^3-(x-2y)^3=(x+3y)[Ax^2+By^2+Cxy]$

$\left[2x+y-\left(x-2y\right)\right]\left[\left(2x+y\right)^2+\left(2x+y\right)\left(x-2y\right)+\left(x-2y\right)^2\right]=(x+3y)[Ax^2+By^2+Cxy]$

$\left[x+3y\right]\left[4x^2+y^2+4xy+2x^2-3xy-2y^2+x^2+4y^2-4xy\right]=(x+3y)[Ax^2+By^2+Cxy]$

$\left(x+3y\right)\left[7x^2+3y^2-3xy\right]=(x+3y)[Ax^2+By^2+Cxy]$

Comparing both sides,

A = 7, B = 3 and C = -3

$A+2B+C\ =\ 7+2\left(3\right)-3$ = 10

Hence, the correct answer is Option D

Question 14: If $9(a^2 + b^2) + c^2 + 20 = 12(a + 2b)$, then the value of $\sqrt{6a + 9b + 2c}$ is:

a) 4

b) 3

c) 6

d) 2

Solution:

$9(a^2+b^2)+c^2+20=12(a+2b)$

$9a^2+9b^2+c^2+20=12a+24b$

$9a^2-12a+9b^2-24b+c^2+20=0$

$9a^2-12a+4-4+9b^2-24b+16-16+c^2+20=0$

$\left(3a-2\right)^2-4+\left(3b-4\right)^2-16+c^2+20=0$

$\left(3a-2\right)^2+\left(3b-4\right)^2+c^2=0$

$3a-2=0,\ 3b-4=0,\ c=0$

$a=\frac{2}{3},\ b=\frac{4}{3},\ c=0$

$\sqrt{6a+9b+2c}=\sqrt{6\left(\frac{2}{3}\right)+9\left(\frac{4}{3}\right)+2\left(0\right)}$

= $\sqrt{4+12}$

= $\sqrt{16}$

= 4

Hence, the correct answer is Option A

Question 15: If $x + \frac{1}{x} = 2\sqrt{5}$, then what is the value of $\frac{\left(x^4 + \frac{1}{x^2}\right)}{x^2 + 1}$?

a) 14

b) 17

c) 20

d) 23

Solution:

$x+\frac{1}{x}=2\sqrt{5}$………..(1)

$\left(x+\frac{1}{x}\right)^3=40\sqrt{5}$

$x^3+\frac{1}{x^3}+3.x.\frac{1}{x}\left(x+\frac{1}{x}\right)=40\sqrt{5}$

$x^3+\frac{1}{x^3}+3\left(2\sqrt{5}\right)=40\sqrt{5}$  [From (1)]

$x^3+\frac{1}{x^3}+6\sqrt{5}=40\sqrt{5}$

$x^3+\frac{1}{x^3}=34\sqrt{5}$………(2)

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

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

$=\frac{34\sqrt{5}}{2\sqrt{5}}$

$=17$

Hence, the correct answer is Option B

Question 16: If $x^4+x^2y^2+y^4=21$ and $x^2+xy+y^2=3$, then what is the value of $\left(-xy\right)$?

a) -1

b) 2

c) 1

d) -2

Solution:

$x^4+x^2y^2+y^4=21$……(1)

$x^2+xy+y^2=3$

$x^2+y^2=3-xy$

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

$x^4+y^4+2x^2y^2=9+x^2y^2-6xy$

$x^4+y^4+x^2y^2=9-6xy$

$21=9-6xy$  [From (1)]

$-6xy=12$

$-xy=2$

Hence, the correct answer is Option B

Question 17: If $(x+6)^3+(2x+3)^3+(3x+5)^3=(3x+18)(2x+3)(3x+5)$, then what is the value of x?

a) $-\frac{5}{3}$

b) $\frac{5}{3}$

c) $-\frac{7}{3}$

d) $\frac{7}{3}$

Solution:

$(x+6)^3+(2x+3)^3+(3x+5)^3=(3x+18)(2x+3)(3x+5)$

$(x+6)^3+(2x+3)^3+(3x+5)^3=\left[3\left(x+6\right)\right](2x+3)(3x+5)$

$(x+6)^3+(2x+3)^3+(3x+5)^3-3\left(x+6\right)(2x+3)(3x+5)=0$

This is in the form of $a^3+b^3+c^3-3abc=0$, where $a\ne b\ne c$ then $a+b+c=0$

$\Rightarrow$  $\left(x+6\right)+\left(2x+3\right)+\left(3x+5\right)=0$

$\Rightarrow$  $6x+14=0$

$\Rightarrow$  $x=-\frac{7}{3}$

Hence, the correct answer is Option C

Question 18: If $x + y + z = 3, xy + yz + zx = -12$ and $xyz = -16$, then the value of $\sqrt{x^3 + y^3 + z^3 + 13}$ is:

a) 9

b) 8

c) 10

d) 11

Solution:

$x+y+z=3$

$x+y=3-z$……..(1)

$\left(x+y\right)^3=\left(3-z\right)^3$

$x^3+y^3+3xy\left(x+y\right)=27-z^3-3.3.z\left(3-z\right)$

$x^3+y^3+3xy\left(3-z\right)=27-z^3-9z\left(x+y\right)$  [From (1)]

$x^3+y^3+9xy-3xyz=27-z^3-9xz-9yz$

$x^3+y^3+z^3=27-9xy-9xz-9yz+3xyz$

$x^3+y^3+z^3=27-9\left(xy+yz+zx\right)+3xyz$

$x^3+y^3+z^3=27-9\left(-12\right)+3\left(-16\right)$

$x^3+y^3+z^3=27+108-48$

$x^3+y^3+z^3=87$…….(2)

$\sqrt{x^3+y^3+z^3+13}=\sqrt{87+13}$

$=\sqrt{100}$

$=10$

Hence, the correct answer is Option C

Question 19: What is the coefficient of x in the expansion of $(3x – 4)^3$?

a) 108

b) -108

c) 144

d) -144

Solution:

$(3x – 4)^3$ = $(3x – 4)(3x – 4)^2$

= $(3x – 4)(9x^2+16-24x)$

= $27x^3+48x-72x^2-36x^2-64+96x$

= $27x^3-108x^2+144x-64$

The coefficient of x in the expansion = 144

Hence, the correct answer is Option C

Question 20: If $x – y = 4$ and $x^3 – y^3 = 316, y > 0$ then the value of $x^4 – y^4$ is:

a) 2500

b) 2320

c) 2401

d) 2482

Solution:

$x-y=4$………..(1)

$\left(x-y\right)^3=64$

$x^3-y^3-3xy\left(x-y\right)=64$

$316-3xy\left(4\right)=64$

$12xy=252$

$xy=21$……….(2)

$x-y=4$

$\left(x-y\right)^2=4^2$

$x^2+y^2-2xy=16$

$x^2+y^2-2\left(21\right)=16$

$x^2+y^2=58$……….(3)

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

$\left(x+y\right)^2=58+2\left(21\right)$

$\left(x+y\right)^2=100$

$x+y=10$……….(4)

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

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

$=\left(58\right)\left(10\right)\left(4\right)$

$=2320$

Hence, the correct answer is Option B