Solved ? - Not Yet!

Topic - Quant Mixed Bag

Source - Elite's Grid CAT Preparation Forum ]]>

Solved ? - Not Yet!

Topic - Quant Mixed Bag

Source - Elite's Grid CAT Preparation Forum ]]>

(a) 364

(b) 231

(c) 455

(d) 472

(e) none ]]>

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 ]]>

Here f(1) = 4 = > f(x) = 4^x so f(3) = 4^3 = 64 ]]>

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

a) 2

b) 3

c) 4

d) none ]]>

(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 ]]>

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 ]]>

a) −7

b) 6

c) 1

d) 7 ]]>

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 ]]>

a) (-1, 2)

b) (-1, 2) U (3, ∞)

c) (∞, -1) U (2, 3)

d) (-1, 2) U (2, 3) ]]>

so sum of first 10 terms will be = (2 + 4 + 8 + 16 + - - - - - ) - (1 + 2 +3 +4 +5 +6 +7 - - -)

2046 - 55 = 1991 ]]>

a) 0 ≤ w ≤ 2

b) -1/2 ≤ w ≤ 1/2

c) -2/3 ≤ w ≤ 1/3

d) -2/3 ≤ w ≤ 2/3 ]]>

(2x - p)^2 + y^2 + 32 = 0

3x^2 + y^2 - 16 = 0 ]]>