Algebra - Functions & Graphs - Previous Year CAT/MBA Questions

Answer: Option D

Explanation :

(x + 4)(x + 6)(x + 8) ⋯ (x + 98) < 0

Critical points are -98, -96, -94, … -8. -6 and -4

For x > -4, the given expression will be positive. Hence, we will reject all values of x > -4.

For -6 < x < -4, the given expression will be negative. The only integral value for x in this range is -5.

For -8 < x < -6, the given expression will be positive.

For -10 < x < -8, the given expression will be negative. The only integral value for x in this range is -9.

The given expression will be negative for x = -5, -9, -13, and so on

Now the only integral value of x between -98 and -96 is -97
∴ f(-97) = -93 × - 91 × -89 × … × -1 × 1

Here we have product of 47 negative terms and 1 positive term, hence the product will be negative.

∴ For -98 < x < -96, the given expression will be negative. The only integral value for x in this range is -97.

Finally, for x < -98, f(x) is product of 48 negative terms, which will give us a positive number. Hence, we will reject all value of x < -98.

The given expression will be negative for x = -5, -9, -13, and so on till -97

Number of possible values of x for which f(x) < 0 = (-5 + 97)/4 + 1 = 24

Hence, option (d).

Answer: Option B

Explanation :

Q(9, 4) = (9)² – (9)(4) + (4)²

= 81 – 36 + 16 = 61

∴ P(7,Q(9,4)) = P(7,61)

= (7)² + (7)(61) + (61)²

= 49 + 427 + 3721 = 4197

Hence, option 2.

Answer: Option C

Explanation :

f(x) = ax + b

∴ f(f(x) = a(ax + b) + b

f(f(x) = a²x + ab + b 

⇒ a²x + b(a + 1) = 9x + 8

∴ a² = 9 ⇒ a = 3. and

b(a + 1) = 8

⇒ b(3 + 1) = 8

⇒ b = 2

Thus, a + b = 5 

Hence, option (c).

Answer: Option B

Explanation :

f(5) = 1/(1 + 5) = 1/6 

f[f(5)] = f(1/6) = 1/[1 + (1/6)] = 6/7

f[f{f(5)}] = f(6/7) = 1/[1 + (6/7)] = 7/13

Hence, option 2.

Answer: Option D

Explanation :

Given function can be rearranged as,

f(x) – f(x – 1) = 2x

⇒ f(2) – f(1) = 2 × 2
⇒ f(3) – f(2) = 2 × 3
⇒ f(4) – f(3) = 2 × 4
…
⇒ f(31) – f(30) = 2 × 31

Adding all these equations we get

f(31) – f(1) = 2 × (2 + 3 + 4 + … + 31)

f(31) – 1 = 2 × ((31×32)/2-1) = 990

⇒ f(31) = 991

Alternately,
From the given function ;

f(2) = 4 + f(1) = 5 = 22 + 1

f(3) = 6 + f(2) = 6 + 5 = 11 = 32 +2

f(4) = 8 + f(3) = 8 + 11 = 19 = 42 + 3

Hence we can see that f(x) = x2 + (x-1)

Hence f(31) = 312 + 30 = 991.

Hence, option (d).

Answer: Option E

Explanation :

From the graph, observe that at y = 0, x
= –19
Only x = 2y² + 3y – 19 satisfies the above criteria.
Hence, option 5.

Answer: Option C

Explanation :

Substitute x = 0 in f(x² – 1) = x⁴ – 7x² + k₁
⇒ f(–1) = k₁ … (i)
Substitute x = 1 in f(x₃ – 2) = x⁶ – 9x³ + k₂
⇒ f(–1) = 1 – 9 + k₂ … (ii)
From (i) and (ii), k₁ = –8 + k₂
∴ k₂ – k₁ = 8
Hence, option 3.

Answer: Option E

Explanation :

f(1) = 4
f(2) = f(1+1) = f(1 × 1) = f(1) = 4
…
Assume that f(n – 1) = 4
f(n) = f(1 + (n – 1)) = f(1 × (n – 1)) = f(n – 1) = 4
Thus, by induction principle, f(n) = 4, for all n.
∴ f(1003) = k = 4
Hence, option 5.