CAT 2019 – Slot 1 – Quantitative Ability – For any positive integer n, let f(n) = n(n + 1) if n is even, and f(n) = n + 3 if n is odd

Q. 1: For any positive integer n, let f(n) = n(n + 1) if n is even, and f(n) = n + 3 if n is odd. If m is a positive integer such that 8f(m + 1) – f(m) = 2, then m equals

Case 1) If m is odd, then m+1 will be even

So f(m) = m+3 and f(m+1) = m(m+1)

Given, 8f(m + 1) – f(m) = 2

Or 8* (m+2)(m+1)- (m+3) = 2

8m^2 + 24m + 16 – m – 3 =2

8m^2 + 23m – 11 = 0

No integer solution so m is not odd means m is even.

Case 2) if m is even , m+1 = odd

So f(m+1) = m+3 and f(m) = m(m+1)

Given, 8f(m + 1) – f(m) = 2

Or 8(m+1 + 3) – m(m+1) =2

8m + 32 – m^2 –m =2

m^2 -7m -30 =0

(m+3)*(m-10)=0

M = 10 or -3

As m is even so m =10

 

 

Q. 2: A chemist mixes two liquids 1 and 2. One litre of liquid 1 weighs 1 kg and one litre of liquid 2 weighs 800 gm. If half litre of the mixture weighs 480 gm, then the percentage of liquid 1 in the mixture, in terms of volume, is
1. 70
2. 80
3. 75
4. 85 

Solution: weight of half litre of the liquid 1 and liquid 2 will be 500 and 400 gm respectively.

So required ratio of liquid 1 an 2 in mixture =(480n-400)/(500 -480) =80/20 = 4:1

So the percentage of liquid 1 in the mixture= 4/(4+1) *100 = 80

 

Q. 3: If a(base1)+a(base2)+a(base3)+….+ a(base n) = 3(2^(n+1) – 2), for every n≥1, then a(base11) equals ? 

Given, a(base1)+a(base2)+a(base3)+ a(base4) + ………..+ a(base n) = 3*(2^(n+1) -2)

Put n =1

a(base1) = 3*(2^2 -2) = 6

Put n =2

a(base1)+a(base2) = 3*(2^3 -2) = 18

Or a(base2) = 12

Put n =3

a(base1)+a(base2)+a(base3) = 3*(2^4 -2) = 42

a(base3) = 42 – a(base2) – a(base1) = 42 – 12 – 6 =24

we can see that a1, a2 , a3 form a GP with common ratio 2 so

a(base11) = a(base1)*2^10 = 6*1024 = 6144

 

Q. 4:Consider a function f satisfying f(x+y) = f(x) f(y) where x, y are positive integers, and f(1) = 2 if f(a+1) + f(a+2)+…+f(a+n)+16 (2^n – 1) then a is equal to? 

Given, f(x+y) = f(x)*f(y), and f(1) =2

f(a+1) + f(a+2) + f(a+3) + …..+f(a+n) = 16 (2^n -1)————–1)

putting n =1

f(a+1) = 16

f(a)*f(1) = 16

or f(a) = 8 as f(1) =2

Putting n =2 in eq 1)

F(a+1) + f(a+2) =16*(2^2 -1)

16 + f(a)*f(2) = 16*3 = 48

8*(f2) = 48 -16 = 32

Or f(2) = 4

By putting n = 3 , in eq 1 we get,

f(a+1) + f(a+2) + f(a+3) = 16*(2^3 -1)

16 + 32 + f(a)*f(3) = 16*7 = 112

F(a)*f(3) = 112 – 16 – 32 = 64

8*f(3) = 64

F(3) = 8

As we know f(a) = 8 so a = 3

Q. 5: The number of the real roots of the equation 2cos (x ( x + 1 ) ) = 2^x + 2^-x is?
1. 1
2. 0
3. 2
4. infinite

As we know maximum value of cos A = 1 at A =0

So maximum value of 2cos (x ( x + 1 ) ) = 2*1 =2

Minimum value of 2^x + 2^(-x) = 2 ( as the minimum value of k + 1/k = 2 at k = 1) at x = 0

Thus there is only one possible case when 2cos (x ( x + 1 ) ) = 2^x + 2^-x at x =0

So no of real solution =1

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