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Question 1: If $\sqrt{x}=\sqrt{3}-\sqrt{5}$, then the value of $x^2-16x+6$ is:

a) 4

b) 0

c) 2

d) -2

Question 2: 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

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

a) -6

b) 9

c) 6

d) -9

Question 4: If $a-b=18$ and $a^3-b^3=324$, then find ab.

a) 105

b) -102

c) -104

d) 103

Question 5: 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

Question 6: If a + 3b = 12 and ab = 9, then the value of (a – 3b) is:

a) 9

b) 8

c) 6

d) 4

Question 7: 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

Question 8: If a + b + c + d = 2, then the maximum value of (1 + a)(1 + b)(1 + c)(1 + d) is

a) $\frac{91}{9}$

b) $\frac{63}{22}$

c) $\frac{54}{13}$

d) $\frac{81}{16}$

Question 9: If $x + \frac{1}{x} = 4,$ then the value of $x^4 + \frac{1}{x^4}$ is :

a) 16

b) 196

c) 194

d) 14

Question 10: If x – y = 13 and xy = 25, then the value of $x^2 – y^2$ = ?

a) $13 \sqrt 240$

b) $13 \sqrt 229$

c) $13 \sqrt 269$

d) $13 \sqrt 210$

Question 11: 2x — 3y is a factor of:

a) $4x^2 + 2x – 3y + 9y^2 – 12xy$

b) $4x^2 + 9y^2 + 12xy$

c) $8x^3 + 27y^3$

d) $4x^2 + 2x – 3y + 36y^2 + 12xy$

Question 12: If a and b are two positive real numbers such that a + b = 20 and ab = 4, then the value of $a^3 + b^3$ is:

a) 7760

b) 8000

c) 8240

d) 240

Question 13: If x + y = 15 and xy = 14, then the value of x — y is:

a) 13

b) 12

c) 11

d) 14

Question 14: If a = 355, b = 356, c = 357,then find the value of $a^3 + b^3 + c^3 – 3abc$.

a) 3204

b) 3206

c) 3202

d) 3208

Question 15: If $x + y = 14; x^3 + y^3 = 1064$, then the value of $(x – y)^2$ is:

a) 36

b) 64

c) 81

d) 100

Question 16: If $a + b = 8$ and $a + a^2b + b + ab^2 = 128$ then the positive value of $a^3 + b^3$ is:

a) 344

b) 96

c) 224

d) 152

Question 17: The coefficient of y in the expansion of $(2y – 5)^3$, is:

a) 150

b) 50

c) -30

d) -150

Question 18: The given table represents the revenue (in ₹ crores) of a company from the sale of four products A, B, C and D in 6 years. Study the table carefully and answer the question that follows. By what percentage is the total revenue of the company from the sale of products A, B and D in 2012 and 2013 more than the total revenue from the sale of product B in 2013 to 2016?(Correct to one decimal place)

a) 44.5

b) 31.2

c) 43.6

d) 45.4

Question 19: If $a + b + c = 19, ab + bc + ca = 120$, then what is the value of $a^3 + b^3 + c^3 – 3abc$?

a) 18

b) 23

c) 31

d) 19

Question 20: Solve the following:
(a + b + c)(ab + bc + ca) – abc = ?

a) (a + b)(b + c)(c – a)

b) (a + b)(b – c)(c + a)

c) (a + b)(b + c)(c + a)

d) (a – b)(b – c)(c – a)

Question 21: If $x^6 – 512 y^6 = (x^2 + Ay^2) (x^4 – Bx^2 y^2 + Cy^4)$, then what is the value of $(A + B – C)$?

a) -80

b) -72

c) 72

d) 48

Question 22: If $x, y, z$ are three integers such that $x + y = 8, y + z = 13$ and $z + x = 17$, then the value of $\frac{x^2}{yz}$ is:

a) $\frac{7}{5}$

b) $\frac{18}{11}$

c) 1

d) 0

Question 23: On simplification, $\frac{x^3 – y^3}{x[(x + y)^2 – 3xy]} \div \frac{y[(x – y)^2 + 3xy]}{x^3 + y^3} \times \frac{(x + y)^2 – (x – y)^2}{x^2 – y^2}$ is equal to:

a) 4

b) 1

c) $\frac{1}{2}$

d) $\frac{1}{4}$

Question 24: If $\left(x^3 + \frac{1}{x^3} – k\right)^2 + \left(x + \frac{1}{x} – p\right)^2 = 0$ where k and p are real numbers and x ≠ 0, then $\frac{k}{p}$ is equal to:

a) $P^2+1$

b) $P^2+3$

c) $P^2-1$

d) $P^2-3$

Question 25: For real a, b, c if $a^2 + b^2 + c^2 = ab + bc + ca$, the value of $\frac{(a+b)}{c}$

a) 3

b) 1

c) 2

d) 0

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

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

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

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

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

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

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

Given, a + b + c + d = 2

We know that, AM $\ge\$ GM

$\Rightarrow$ Arithmetic mean of (1 + a),(1 + b),(1 + c),(1 + d) $\ge\$ Geometric mean of (1 + a),(1 + b),(1 + c),(1 + d)

$\Rightarrow$ $\frac{\left(1+a\right)+\left(1+b\right)+\left(1+c\right)+\left(1+d\right)}{4}\ge\left[\ \left(1+a\right)\left(1+b\right)\left(1+c\right)\left(1+d\right)\right]^{\frac{1}{4}}$

$\Rightarrow$ $\frac{4+a+b+c+d}{4}\ge\left[\left(1+a\right)\left(1+b\right)\left(1+c\right)\left(1+d\right)\right]^{\frac{1}{4}}$

$\Rightarrow$ $\frac{4+2}{4}\ge\left[\left(1+a\right)\left(1+b\right)\left(1+c\right)\left(1+d\right)\right]^{\frac{1}{4}}$

$\Rightarrow$ $\frac{6}{4}\ge\left[\ \left(1+a\right)\left(1+b\right)\left(1+c\right)\left(1+d\right)\right]^{\frac{1}{4}}$

$\Rightarrow$ $\left[\left(1+a\right)\left(1+b\right)\left(1+c\right)\left(1+d\right)\right]^{\frac{1}{4}}\le\ \frac{3}{2}$

$\Rightarrow$ $\left(1+a\right)\left(1+b\right)\left(1+c\right)\left(1+d\right)\le\ \left(\frac{3}{2}\right)^4$

$\Rightarrow$ $\left(1+a\right)\left(1+b\right)\left(1+c\right)\left(1+d\right)\le\ \frac{81}{16}$

$\therefore\$Maximum value of (1 + a)(1 + b)(1 + c)(1 + d) = $\frac{81}{16}$

Hence, the correct answer is Option D

Given,  $x + \frac{1}{x} = 4$

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

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

$=$>  $x^2+\frac{1}{x^2}+2=16$

$=$>  $x^2+\frac{1}{x^2}=14$

$=$>  $\left(x^2+\frac{1}{x^2}\right)^2=14^2$

$=$>  $x^4+\frac{1}{x^4}+2.x^2.\frac{\ 1}{x^2}=196$

$=$>  $x^4+\frac{1}{x^4}+2=196$

$=$>  $x^4+\frac{1}{x^4}=196-2$

$=$>  $x^4+\left(\frac{1}{x}\right)^4=194$

Hence, the correct answer is Option C

Given,

$x – y = 13$ and  $xy = 25$

$=$>  $\left(x-y\right)^2=13^2$

$=$>  $x^2+y^2-2xy=169$

$=$>  $x^2+y^2+2xy-4xy=169$

$=$>  $\left(x+y\right)^2-4xy=169$

$=$>  $\left(x+y\right)^2-4\left(25\right)=169$

$=$>  $\left(x+y\right)^2-100=169$

$=$>  $\left(x+y\right)^2=269$

$=$>  $x+y=\sqrt{269}$

$\therefore\$ $x^2-y^2=\left(x+y\right)\left(x-y\right)=\left(\sqrt{269}\right)\left(13\right)=13\sqrt{269}$

Hence, the correct answer is Option C

$4x^2 + 2x – 3y + 9y^2 – 12xy=4x^2+9y^2-12xy+2x-3y$

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

$=\left(2x-3y\right)\left(2x-3y+1\right)$

$\therefore\$ $2x+3y$ is factor of $4x^2 + 2x – 3y + 9y^2 – 12xy$

Hence, the correct answer is Option A

Given,  $a+b=20$  and  $ab=4$

$=$>  $\left(a+b\right)^3=20^3$

$=$>  $a^3+b^3+3ab\left(a+b\right)=8000$

$=$>  $a^3+b^3+3\left(4\right)\left(20\right)=8000$

$=$>  $a^3+b^3+240=8000$

$=$>  $a^3+b^3=8000-240$

$=$>  $a^3+b^3=7760$

Hence, the correct answer is Option A

Given, $x+y\ =15$  and $xy=14$

$=$> $\left(x+y\right)^2=15^2$

$=$> $x^2+y^2+2xy=225$

$=$> $x^2+y^2+2xy-2xy+2xy=225$

$=$> $x^2+y^2-2xy+4xy=225$

$=$> $\left(x-y\right)^2+4\left(14\right)=225$

$=$> $\left(x-y\right)^2+=225-56$

$=$> $\left(x-y\right)^2+=169$

$=$>  $x-y=13$

Hence, the correct answer is Option A

Given, $a$ = 355, $b$ = 356, $c$ = 357

$a^3+b^3+c^3—3abc=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)$

$=\frac{1}{2}\left(a+b+c\right)\left(2a^2+2b^2+2c^2-2ab-2bc-2ca\right)$

$=\frac{\left(a+b+c\right)}{2}\left[\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\right]$

$=\frac{\left(355+356+357\right)}{2}\left[\left(355-356\right)^2+\left(356-357\right)^2+\left(357-355\right)^2\right]$

$=\frac{1068}{2}\left[1+1+4\right]$

$=\frac{1068}{2}\left[6\right]$

$=3204$

Hence, the correct answer is Option A

Given, $x + y = 14$

$x^3 + y^3 = 1064$

$=$>  $\left(x+y\right)\left(x^2+y^2-xy\right)=1064$

$=$>  $14\left(x^2+y^2+2xy-3xy\right)=1064$

$=$>  $\left(x+y\right)^2-3xy=76$

$=$>  $\left(14\right)^2-3xy=76$

$=$>  $196-3xy=76$

$=$>  $3xy=120$

$=$>  $xy=40$

$\therefore\$ $(x – y)^2=x^2+y^2-2xy$

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

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

$=\left(14\right)^2-4\left(40\right)$

$=196-160$

$=36$

Hence, the correct answer is Option A

Given, $a + b = 8$

$a + a^2b + b + ab^2 = 128$

$=$>  $a\left(1+ab\right)+b\left(1+ab\right)=128$

$=$>  $\left(1+ab\right)\left(a+b\right)=128$

$=$>  $\left(1+ab\right)8=128$

$=$>  $1+ab =\frac{128}{8}$

$=$>  $1+ab=16$

$=$>  $ab=15$

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

$= 8 (a^2+b^2+2ab-3ab)$

$=8 \left(\left(a+b\right)^2-3ab\right)$

$=8 \left(8^2-3\left(15\right)\right)$

$=8 \left(64-45\right)$

$=152$

Hence, the correct answer is Option D

$(2y – 5)^3$

= $8y^3 – 125 +150y – 60y^2$

$(\because(a-b)^3 = a^3 – b^3 – 3a^2b + 3ab^2)$

The coefficient of y = 150

Total revenue of the company from the sale of products A, B and D in 2012 and 2013 = 98 + 74 + 74 + 94 + 96 + 102 = 538

Total revenue from the sale of product B in 2013 to 2016 = 96 + 92 + 84 + 98 = 370

Required percentage = $\frac{538 – 370}{370} \times 100$ = 45.4%

$(a + b + c)^2 = a^2 +b^2 + c^2 + 2ab + 2bc + 2ca$

$19^2 = a^2 +b^2 + c^2 + 2(120)$

$a^2 +b^2 + c^2 = 361 – 240 = 121$

$a^3 + b^3 + c^3 – 3abc$

=$(a + b + c)(a^2 + b^2 + c^2 -ab -bc – ca)$

= 19(121$\times -120) = 19$

(a + b + c)(ab + bc + ca) – abc

= $a^2b + a^2c + ab^2 + cb^2 + bc^2 + ac^2 + 2abc$

= $a^2(b +c)+bc(b+c)+a(b^2+c^2+2bc)$

= $a^2(b+c)+bc(b+c)+a(b+c)^2$

=$(b+c)(a^2+bc+ab+ac)$

= (b + c)[a(a + b) + c(a + b)]

= (a + b)(b + c)(c + a)

$x^6 – 512 y^6 = (x^2 + Ay^2) (x^4 – Bx^2 y^2 + Cy^4)$

$(x^2)^3 + (-8y^2)^3 = (x^2 + Ay^2) (x^4 – Bx^2 y^2 + Cy^4)$

$(\because a^3 + b^3 = (a + b)(a^2 – ab + b^2))$

by comparison –

$-8y^2 = Ay^2$

A = -8

$Bx^2y^2 = x^2 \times -8y^2$

B = -8

$(-8y^2)^2 = Cy^4$

C = 64

The value of $(A + B – C)$ = -8 – 8 – 64 = -80

x + y = 8 —-(1)

y + z = 13 —-(2)

z + x = 17 —-(3)

Eq (1) + (2) + (3),

2(x + y + z) = 8 + 13 + 17

x + y + z = 38/2 = 19 —-(4)

From eq (3) and (4),

x + 13 = 19

x = 6

From eq(3),

z + x = 17

z + 6 = 17

z = 11

From eq(2),

y + z = 13

y + 11 = 13

y = 2

$\frac{x^2}{yz}$

= $\frac{6^2}{2 \times 11}$

= 36/22 = 18/11

$\frac{x^3 – y^3}{x[(x + y)^2 – 3xy]} \div \frac{y[(x – y)^2 + 3xy]}{x^3 + y^3} \times \frac{(x + y)^2 – (x – y)^2}{x^2 – y^2}$

= $\frac{x^3 – y^3}{x[(x + y)^2 – 3xy]} \times \frac{x^3 + y^3}{y[(x – y)^2 + 3xy]} \times \frac{(x + y)^2 – (x – y)^2}{x^2 – y^2}$

= $\frac{(x – y)(x^2 + xy + y^2)}{x[(x + y)^2 – 3xy]} \times \frac{(x + y)(x^2 – xy + y^2)}{y[(x – y)^2 + 3xy]} \times \frac{(x^2 + y^2 + 2xy) – (x^2 + y^2 – 2xy)}{(x + y)(x – y)}$

= $\frac{x^2 + xy + y^2}{x[x^2 – xy + y^2]} \times \frac{x^2 – xy + y^2}{y[x^2 + xy + y^2]} \times 4xy$ = 4

$\left(x^3 + \frac{1}{x^3} – k\right)^2 + \left(x + \frac{1}{x} – p\right)^2 = 0$

It will be zero, if the individual terms will be zero.

So, $(x^3 + \frac{1}{x^3} – k)^2=0$ and $(x + \frac{1}{x} – p)^2$

So, $k=x^3 + \frac{1}{x^3}$ and $(x + \frac{1}{x} – p)^2=0$

$k=x^3 + \frac{1}{x^3}$ and $(x + \frac{1}{x} )=p$

Now, $(x + \frac{1}{x} )=p$ taking cube of both side,

$\Rightarrow (x + \frac{1}{x} )^3=p^3$

$\Rightarrow x^3+\dfrac{1}{x^3}+3(x+\dfrac{1}{x})=p^3$

Now substituting the values in the above,

$\Rightarrow k+3p=p^3$

$\Rightarrow k=p^3-3p$

$\Rightarrow \dfrac{k}{p}=p^2-3$

Given,    $a^2+b^2+c^2=ab+bc+ca$

Multiplying both sides by “2”, it becomes

$2\left(a^2+b^2+c^2\right)=2\left(ab+bc+ca\right)$

$2a^2+2b^2+2c^2=2ab+2bc+2ca$

$a^2+a^2+b^2+b^2+c^2+c^2-2ab-2bc-2ca=0$

$\left(a^2+b^2-2ab\right)+\left(b^2+c^2-2ca\right)+\left(c^2+a^2-2ca\right)=0$

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

Sum of squares is zero so each term should be zero

$=$>    $\left(a-b\right)^2=0$, $\left(b-c\right)^2=0$, $\left(c-a\right)^2=0$

$=$>      $a-b=0$,       $b-c=0$,       $c-a=0,$

$=$>         $a=b$,               $b=c$,                $c=a$

$=$>         $a=b=c$

Therefore     $\ \frac{\ a+b}{c}=\ \frac{\ a+a}{a}=\ \frac{\ 2a}{a}=2$