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(y1)+x(y1)+y3=0
x real so D > =0
(y1)^24*(y1)(y2) > =0
y1)(y14y+12) > =0
(y1)(3y+11) > =0
(y1)(3y11)
so (1,11/3]Method 2 :
y=1+((2/(x^2+x+1))
now x^2+x+1=x^2+x+1/41/4+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+2x7)=y
(1y)x^2+ (342y)x 71+7y=0
b^24ac > =0 for real values therefore
(342y)^24*(1y)(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) = 1x^n
so f(4) = 14^3
= 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