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

Answer: 44

Explanation :

$\left[\frac{1}{5}+\frac{n}{25}\right]$ = 0 when $\frac{1}{5}+\frac{n}{25}$ < 1 ⇒ n < 20

$\left[\frac{1}{5}+\frac{n}{25}\right]$ = 1 when 1 ≤ $\frac{1}{5}+\frac{n}{25}$ < 2 ⇒ 20 ≤ n < 45

$\sum _{n=1}^{44}\left[\frac{1}{5}+\frac{n}{25}\right]$ = $\sum _{n=1}^{19}\left[\frac{1}{5}+\frac{n}{25}\right]$ + $\sum _{n=20}^{44}\left[\frac{1}{5}+\frac{n}{25}\right]$ = 0 + (1 + 1 + 1 + … + 1 (25 times)) = 25

∴ N = 44

Hence, 44.

Answer: 12

Explanation :

f(x² – x) = 5
put x = 0 ⇒ f(0) = 5

Given f(x) + f(x – 1) – 1 = 0
⇒ f(x) = 1 – f(x - 1)   …(1)

put x = 1 in (1)
⇒ f(1) = 1 – f(0)
⇒ f(1) = 1 – 5 = -4

Put x = 2 in (1)
⇒ f(2) = 1 – f(1) = 1 – (-4) = 5

∴ f(odd value of x) = -4 & 
f(even value of x) = 5

Now, we have f(g(5)) + g(f(5))
⇒ f(g(5)) + g(f(5)) = f(52) + g(-4)
⇒ f(g(5)) + g(f(5)) = f(25) + 16
⇒ f(g(5)) + g(f(5)) = -4 + 16
⇒ f(g(5)) + g(f(5)) = 12

Hence, 12.

Answer: Option D

Explanation :

Case 1: x < r
⇒ f(x) = r
⇒ f(f(x)) = f(f(r)) = 2r – r = r
∴ f(x) = f(f(x))

Case 2: x = r
⇒ f(x) = 2r – r = r
⇒ f(f(x)) = f(f(r)) = 2r – r = r
∴ f(x) = f(f(x))

Case 3: x > r
⇒ f(x) = 2x – r > r
⇒ f(f(x)) = f(f(2x - r)) = 2(2x - r) – r = 4x - 3r
∴ f(x) ≠ f(f(x))

⇒ f(x) = f(f(x) when x ≤ r

Hence, option (d).

Answer: Option C

Explanation :

Given, f(x) = $\frac{x^2+2x-15}{x^2-7x-18}$ < 0

⇒ $\frac{(x+5)(x-3)}{(x-9)(x+2)}$  < 0

Here, the critical points are -5, -2, 3 and 9.

∴ f(x) will be positive when -5 < x < -2 or 3 < x < 9.

Hence, option (c).

Answer: Option B

Explanation :

f(x) = x² – 7x and g(x) = x + 3

⇒ f(g(x)) – 3x = (g(x))² – 7g(x) – 3x

⇒ f(g(x)) - 3x = (x + 3)² – 7(x + 3) - 3x 

⇒ f(g(x)) - 3x = x^{​​​​​​​2} + 6x + 9 – 7x – 21 – 3x

⇒ f(g(x)) = x² - 4x – 12

f(g(x)) is a quadratic equation. Least value of ax² + bx + c occurs at x = -b/2a

∴ Minimum value of x² - 4x – 12 occurs at x = -(-4)/2 = 2

∴ Minimum value of f(g(x)) = 4 - 8 - 12 = -16.

Hence, option (b).

Answer: Option D

Explanation :

Given, x₀ = max(x₁, x₂, …., x₁₂)

x₀ will be minimum when we distribute students all 100 students as equally among the 12 months as possible.

This can be done in the following way: 8, 8, 8, 8, 8, 8, 8, 8, 9, 9, 9, 9 (Adding these 12 we will get 100)

∴ x₀ = max(8, 8, 8, 8, 8, 8, 8, 8, 9, 9, 9, 9).

⇒ Minimum value of x₀ = 9

Hence, option (d).

Answer: Option 20

Explanation :

Given f(5 + x) = f(5 – x)

Assuming one of the roots is (5 + α).

⇒ f(5 + α) = 0

Now, we know, f(5 + α) = f(5 - α) = 0

⇒ If one of the roots is 5 + α, the other root will be 5 – α.

∴ If (5 + α) and (5 + β) are two of the roots then (5 – α) and (5 – β) will be the other two roots.

⇒ Sum of the roots = (5 + α) + (5 + β) + (5 – α) + (5 – β) = 20

Hence, 20.

Answer: 3

Explanation :

Given, 

|x| - y ≤ 1
⇒ y ≥ |x| - 1

y ≥ 0 and y ≤ 1.

The following diagram can be drawn from the given information.

​​​​​​​

The required area is highlighted in orange i.e., a trapezium whose parallel sides are 4 units and 2 units and height is 1 unit.

∴ Area of the trapezium = 1/2 × (2 + 4) × 1 = 3 square units.

Hence, 3.

Answer: Option C

Explanation :

Given, g(x) = f(x + 1) – f(x – 1).

f(x) = x² + ax + b

∴ g(x) = (x + 1)² + a(x + 1) + b – [(x - 1)² + a(x - 1) + b]

⇒ g(x) = x2 + 2x + 1 + ax + a + b - x2 + 2x - 1 - ax + a – b

⇒ g(x) = 4x + 2a

Given, g(20) = 72

⇒ 4 × 20 + 2a = 72

⇒ a = -4

Also given, f(x) ≥ 0

⇒ x² - 4x + b ≥ 0

This is possible when discriminant is less than or equal to 0.

⇒ 16 – 4b ≤ 0

⇒ b ≥ 4

∴ Least possible value of b is 4.

Hence, option (c).

Answer: Option B

Explanation :

Given, f(x + y) = f(x) × f(y)

Substitute y = 0

⇒ f(x + 0) = f(x) × f(0)

f(x) = f(x) × f(0)

[f(x) cannot be ‘0’ since it is given that f(5) = 4]

∵ f(0) = 1

f(5) = 4 (Given)

f(10) = f(5 + 5) = f(5) × f(5) = 4 × 4 = 16

f(10) = 16

We know that f(0) = 1

f(0) = f(10 - 10) = f(10 + (-10)) = f(10) × f(-10) = 1

16 × f(-10) = 1

f(-10) = 1/16 = 0.0625

f(10) - f(-10) = 16 - 0.0625 = 15.9375.

Hence, option (b).

Answer: 10

Explanation :

Case I: m is odd.

So, (m + 1) is even.  

∴ 8[(m + 1)(m + 2)] − (m + 3) = 2

∴ 8m² + 23m + 11 = 0.

Both roots of this equation are negative as sum of the roots (−23/8) is negative and the product (11/8) is positive. But it is given that m is a positive integer. Hence this case is discarded.

Case II: m is even.

So, (m + 1) is odd.

∴ 8(m + 3 + 1) − m(m + 1) = 2.

∴ m² − 7m − 30 = 0

Solving this equation, we get; m = 10 or −3.

Since m is positive, m = 10.  

Hence, 10.

Answer: 3

Explanation :

Given, f(1) = 2

Now,

f(2) = f(1 + 1) = f(1) × f(1) = 2 × 2 = 4 = 2². 

f(3) = f(1 + 2) = f(1) × f(2) = 2 × 2² = 2³. 

f(3) = f(1 + 3) = f(1) × f(3) = 2 × 2³ = 2⁴.   

So, f(n) = 2ⁿ. 

f(a + 1) + f(a + 2) +…+ f(a + n) = 16(2ⁿ – 1) 

∴ 2^{(a + 1)} + 2^{(a + 2)} + ... + 2^{(a + n)} = 2⁴(2ⁿ – 1) 

∴ 2^{(a + 1)}[1 + 2 + ... 2⁽ⁿ⁻¹⁾] = 2⁴(2ⁿ – 1) 

∴ 2^{(a + 1)}(2ⁿ – 1) = 2⁴(2ⁿ – 1) 

∴ 2^{(a + 1)} = 2⁴

So, a + 1 = 4

∴ a = 3.

Hence, 3.

Answer: 12

Explanation :

f(mn) = f(m) × f(n)

f(1), f(2) & f(3) are positive integers.

Now we know, f(24) = 54

So, f(2) × f(3) × f(4) = f(2) × f(3) × f(2)² = 54

f(3) × f(2)³ = 54

Also, we know, 54 = 2 × 3³

Therefore, f(2) = 3, f(3) = 2

Now, f(18) = f(3)² × f(2)

f(18) = 2² × 3 = 12

Hence, 12.

Answer: 54

Explanation :

f(x + 2) = f(x) + f(x + 1)

Subtituting x = 13

f(15) = f(14) + f(13)

= [f(13) + f(12)] + f(13)

= 2 × f(13) + f(12)

= 2[f(12) + f(11)] + f(12)

= 3 × f(12) + 2 × f(11)

= 3[f(11) + f(10)] + 2 × f(11)

= 5 × f(11) + 2 × f(10)

∴ 617 = 5 × 91 + 3 × f(10)

Solving this we get, f(10) = 54

Hence, 54.

Answer: 32

Explanation :

For positive value of x, (2x²) is increasing.

As value of x increases, value of (52 – 5x) decreases.

Since, these 2 functions are continuously increasing and decreasing respectively, the max value of f(x) will occur when both these functions are equal.

i.e. 2x² = 52 − 5x, for x = 4 and − 6.5

At x = 4, function attains maximum value for positive real value of x.

∴ at x = 4, f(x) = 32

Hence, 32.

Answer: 20

Explanation :

The graph of the function 52 – 2x² will be of ‘inverted U’ shape, while the graph of the function y = 5x will be a straight line with a positive slope.

Therefore, the minimum value of the required function will be obtained at a point of intersection of y = 52 – 2x² and y = 5x.

Therefore, 52 – 2x² = 5x or 2x² + 5x – 52 = 0.

∴ (x - 4)(2x + 13) = 0

∴ x = 4 or x = $-\frac{13}{2}$

Since x is a positive real number, x = 4.

At x = 4, 5x = 52 – 2x² = 20

Hence, 20.

Answer: Option A

Explanation :

The graph of y = |x – 1| + |x + 1| is shown above.

The shortest distance of (1/2, 1) from the graph is 1.

Hence, option (a).

Answer: Option A

Explanation :

f(x) = (5x+2)/(3x-5), g(x) = x² – 2x – 1

f(3) = $\frac{5\times 3+2}{3\times 3-5}=\frac{17}{4}$

f(17/4) = $\frac{5\times {\displaystyle \frac{17}{4}}+2}{3\times {\displaystyle \frac{17}{4}}-5}$ = $\frac{85+8}{51-20}$ = $\frac{93}{31}$ = 3

g(3) = 3² – 2 × 3 – 1 = 2.

Hence, option (a).

Answer: Option C

Explanation :

At x = 1, f[g(x)] = f[g(1)]
Now g(1) = 2¹ = 2
f[g(1)] = f(2) = 4

At x = 1, g[f(x)] = g[f(1)]
Now f(1) = 1 and
g[f(1)] = g(1) = 2¹ = 2

Now at x = 1
f[f(g(x)) + g(f(x))] = f(4 + 2) = f(6) = 36

Hence, option (c).

Answer: 1

Explanation :

Let f(1) = x

Suppose f(1 × 1) = f(1) × f(1)
∴ f(1) = f(1) × f(1)
⇒ x = x × x
⇒ x² = x
⇒ x = 1 or x = 0

But the highest value of x = 1

Hence, 1.

Answer: Option D

Explanation :

Now

|f(x) + g(x)| = |f(x)| + |g(x)|

This is true only if both f(x) and g(x) are both negative or both positive or both are zero

Case 1: Now if both f(x) and g(x) are greater than or equal to zero.

f(x) = 2x - 5 ≥ 0 or x ≥ $\frac{5}{2}$

g(x) = 7 - 2x ≥ 0 or x ≤ $\frac{7}{2}$

∴$\frac{5}{2}\le x\le \frac{7}{2}$

Case 2: Now if both f(x) and g(x) are less than or equal to zero.

f(x) = 2x - 5 ≤ 0 or x ≤ $\frac{5}{2}$

g(x) = 7 - 2x ≤ 0 or x ≥ $\frac{7}{2}$

This means x ≥ $\frac{7}{2}$ and x ≤ $\frac{5}{2}$.

However, this is not possible.

Hence, option (d).

Answer: Option E

Explanation :

The roots at f(x) = 0 are 3 and −4

∴ The equation can be written as (x – 3)(x + 4) = 0

Or, x² − x + 12 = 0

The co-efficient of x² is 1 here, but all equations which are multiple of this equation will also have same roots.

For example,

 10(x² − x + 12) = 0 will also have same roots

∴ (a + b + c) cannot be determined uniquely.

Hence, option (e).

Answer: Option B

Explanation :

Given, f(x) × f(y) = f(xy)

Let x = 1 and y = 2

∴ f(1) × f(2) = f(2)
∴ f(1) = 1

Now, let x = 1/2 and y = 2

∴ f(1/2) × f(2) = f(1/2 × 2) = f(1)

∴ f(1/2) × 4 = 1

∴ f(1/2) = 1/4

Hence, option (b).

Answer: Option A

Explanation :

f(1) + f(2) + f (3) + … + f(n −1) + f(n) = n²f(n)  ... (i)

Similarly, f(1) + f(2) + f (3) + … + f(n − 1) = (n − 1)² f(n −1)  ... (ii)

⇒ f(n) = n² f(n) – (n – 1)²f(n − 1) ... (i) – (ii)

⇒ (n² – 1)f(n) = (n – 1)²f(n – 1)

⇒ f(n) = $\frac{{(\mathrm{n}-1)}^2\mathrm{f}(\mathrm{n}-1)}{({\mathrm{n}}^2-1)}$

⇒ f(n) = $\frac{(\mathrm{n}-1)\mathrm{f}(\mathrm{n}-1)}{(\mathrm{n}+1)}$

Now, putting

n = 2 we get f(2) = $\frac{1}{3}\times \mathrm{f}\left(1\right)$
n = 3 we get f(2) = $\frac{2}{4}\mathrm{f}\left(2\right)$ = $\frac{2}{4}\times \frac{1}{3}\times \mathrm{f}\left(1\right)$
...
n = 9 we get f(9) = $\frac{8}{10}\mathrm{f}\left(8\right)$ = $\frac{8}{10}\times \frac{7}{9}\times \frac{6}{8}\times \frac{5}{7}\times \frac{4}{6}\times \frac{3}{5}\times \frac{2}{4}\times \frac{1}{3}\times \mathrm{f}\left(1\right)$

∴  f(9) = $\frac{8}{10}\times \frac{7}{9}\times \frac{6}{8}\times \frac{5}{7}\times \frac{4}{6}\times \frac{3}{5}\times \frac{2}{4}\times \frac{1}{3}\times 3600$

= $\frac{2}{10\times 9}\times 3600$ = 80

Hence, option (a).

Answer: Option D

Explanation :

All the given graphs are drawn to the same scale.

We can see that the line makes an angle which is more than 45° with the horizontal axis.

∴ The slope of the line is greater than 1.

Let the slope be k.

∴ (y – x) = k(y + x)     {∵ k > 1}

∴ y – x = ky + kx

$\therefore y=\frac{x(k+1)}{(1-k)}$

$\frac{(k+1)}{(1-k)}$ is negative and $\left|\frac{(k+1)}{(1-k)}\right|>1$

∴ The graph of y against x will be such that when x is positive, y is negative and |x| < |y|, except at (0, 0).

Hence, option (d).