# Quant Boosters - Hemant Malhotra - Set 12

• Number of Questions - 30
Solved ? - Not Yet!
Topic - Quant Mixed Bag
Source - Elite's Grid CAT Preparation Forum

• Q1) A function V(a, b) is defined for positive integers a, b and satisfies V(a, a) = a, V(a, b) = V(b, a), V(a, a+b) = (1 + a/b) V(a, b). The value represented by V(66, 14) is ?
(a) 364
(b) 231
(c) 455
(d) 472
(e) none

• Here we know value of V(a, a + b)
so 1st term should be less than 2nd term, but in V(66, 14) 66 > 14
so change this we know that V(a, b) = V(b, a)
so V(66, 14) = V(14, 66)
now proceed
V(66,14 ) = V(14,66= (33/26) * V(14,52)
= (33/26) * (26/19) * V(14, 38 )
= (33/19) * (19/12) * V(14, 24)
= (33/12) * (12/5) * V(14,10)
= (33/5) * (7/2) * V(10, 4)
=(231/10) * (5/3) * V(4, 6)
= (77/2) * 3 * V(4, 2)
= (231/2) * 2 * V(2, 2)
= 231 * 2
= 462

• Q2) f (x + y) = f(x) * f(y) then find f(3) if f(1) = 4

• If f(x + y) = f(x) * f(y) then f(x)= (f(1))^x always
Here f(1) = 4 = > f(x) = 4^x so f(3) = 4^3 = 64

• Q3) Find range of the function (x^2 + x + 3)/(x^2 + x + 1)

• Method 1 :
y=x^2+x+3/(x^2+x+1))
yx^2+yx+y=x^2+x+3
x^2(y-1)+x(y-1)+y-3=0
x real so D > =0
(y-1)^2-4*(y-1)(y-2) > =0
y-1)(y-1-4y+12) > =0
(y-1)(-3y+11) > =0
(y-1)(3y-11)
so (1,11/3]

Method 2 :
y=1+((2/(x^2+x+1))
now x^2+x+1=x^2+x+1/4-1/4+1
(x+1/2)^2+3/4
so min value is 3/4 and maax infinity
so y=1+2/inf
y=1
and y=1+2/(3/4))
y=1+8/3=11/3

• Q4) How many integral values from 1 to 15 can the expression (x^2 + 34x -71)/(x^2 + 2x - 7) not take?
a) 2
b) 3
c) 4
d) none

• (x^2 +34x -71)/(x^2+2x-7)=y
(1-y)x^2+ (34-2y)x -71+7y=0
b^2-4ac > =0 for real values therefore
(34-2y)^2-4*(1-y)(-71+7y) > =0
now on simplifying
y^2- 14y +45 > =0
that is y < = 5 and y > = 9 so y can't take values
6,7,8 so three integral values

• Q5) f(x) * f(1/x) = f(x) + f(1/x) for all real x. If f(3) = -26 find f(4)

• If f(x) * f(1/x) = f(x) + f(1/x)
then f(x) = 1+-x^n
now f(3) = -26
so 1+-(x)^n = -26
so +-(x)^n = -27
so +-(3)^n = -27 (negative case will be considered bcz in positive case 3^n=-27 not possible)
so -3^n = -27
so n = 3
and f(x) = 1-x^n
so f(4) = 1-4^3
= -63

• Q6) If f(x − 1) + f(x + 1) = f(x) and f(2) = 6, f(0) = 1, then what is the value of f(50) ?
a) −7
b) 6
c) 1
d) 7

• Since we know both f(0) and f(2), we can find f(1).
f(1) = f(0) + f(2) = 7
f(2) = f(1) + f(3)
f(3) = f(2) – f(1)
= 6 – 7
= −1
Also, f(3) = f(2) + f(4)
f(4) = f(3) – f(2) = −7
Continuing in a similar way, we can find out
f(0) = 1
f(1) = 7
f(2) = 6
f(3) = −1
f(4) = −7
f(5) = −6
f(6) = 1
f(7) = 7 and so on
After every 6 integral values of x, f(x) repeats itself.
f(6) = f(12) = f(18 ) = f(24) = f(30) = f(36) = f(42) = f(48 ) = 1
f(49) = 7
f(50) = 6

• Q7) What is the maximum value for the equation -2x^2 + 12x + 15

• Q8) For the equation x^2 + 6x + |n| to have real roots, how many values can n take ?

• Q9) Find the solution set of the equation x^3 - 4x^2 + x + 6 > 0
a) (-1, 2)
b) (-1, 2) U (3, ∞)
c) (∞, -1) U (2, 3)
d) (-1, 2) U (2, 3)

• Q10) Find the largest value of t such that x^t + 1 divides 1 + x + x^2 + x^3 + ... + x^143

• Q11) If x + 1/x = 3 then find the value of x^4 + 1/x^4

• Q12) A father and his son are waiting at a bus stop in the evening. There is a lamp post behind them. The lamp post, the father and his son stand on the same straight line. The father observes that the shadows of his head and his son's head are incident at the same point on the ground. If the heights of the lamp post, the father and his son are 6 metres, 1.8 metres and 0.9 metres respectively, and the father is standing 2.1 metres away from the post then how far (in metres) is son standing form his father?

• Q13) For how many values of y, y(y + 4)(y + 6)(y + 8) < 300 is satisfied

64

61

34

63

61

45

61