Algebra - Functions & Graphs - Previous Year IPM/BBA Questions

Answer: 3034

Text Explanation :

Given, f(1) = 1, f(n) = 3n - f(n - 1)

Substituting,
n = 2, f(2) = 3 × 2 - f(1) = 5
n = 3, f(3) = 3 × 3 - f(2) = 4
n = 4, f(4) = 3 × 4 - f(3) = 8
n = 5, f(5) = 3 × 5 - f(4) = 7
n = 6, f(6) = 3 × 6 - f(5) = 11
n = 7, f(7) = 3 × 7 - f(6) = 10

Here, f(1), f(3), f(5), and so on, i.e., f(odd number)  form an arithmetic progression with common difference of 3.

We need to find f(2023). Now, how many odd numbers are there from 1 till 2023 = (2023 + 1)/2 = 1012

∴ f(2023) = f(1) + (1012- 1) × 3
⇒ f(2023) = 1 + 1011 × 3 = 3034

Hence, 3034.

Answer: 199

Text Explanation :

Given, f(n) = 1 + 2 + 3 + ... + (n+1) = $\frac{(\mathrm{n}+1)(\mathrm{n}+2)}{2}$

Now,

$\underset{k=1}{\overset{k=n}{\sum \frac{1}{f\left(k\right)}}}$ = $\frac{1}{\mathrm{f}\left(1\right)}$ + $\frac{1}{\mathrm{f}\left(2\right)}$ + $\frac{1}{\mathrm{f}\left(3\right)}$ + ... + $\frac{1}{\mathrm{f}\left(\mathrm{n}\right)}$

= $\frac{2}{2\times 3}$ + $\frac{2}{3\times 4}$ + $\frac{2}{4\times 5}$ + ... + $\frac{2}{\left(\mathrm{n}\right)\times (\mathrm{n}+1)}$ + $\frac{2}{(\mathrm{n}+1)\times (\mathrm{n}+2)}$

= $2\left[\frac{1}{2\times 3}+\frac{1}{3\times 4}+\frac{1}{4\times 5}+...+\frac{1}{\mathrm{n}\times (\mathrm{n}+1)}+\frac{1}{(\mathrm{n}+1)\times (\mathrm{n}+2)}\right]$

= $2\left[\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{\mathrm{n}}-\frac{1}{\mathrm{n}+1}+\frac{1}{(\mathrm{n}+1)}-\frac{1}{\mathrm{n}+2}\right]$

= $2\left[\frac{1}{2}-\frac{1}{\mathrm{n}+2}\right]$

= $2\left[\frac{\mathrm{n}}{2(\mathrm{n}+2)}\right]$

= $\frac{\mathrm{n}}{(\mathrm{n}+2)}$

Now, this should be greater than 99/100

⇒ $\frac{\mathrm{n}}{(\mathrm{n}+2)}$ > $\frac{99}{100}$

⇒ 100n > 99n +198

⇒ n > 198

∴ Least possible integer value of n = 199.

Hence, 199.

Answer: 26

Text Explanation :

Given, f(x) = |x| + 2|x−1| + |x−2| + |x−4| + |x−6| + 2|x−10|

f(x) can be written as sum of h(x) and g(x), where
h(x) = |x| + |x−1| + |x−2| + |x−4| + |x−6| + |x−10| and
g(x) = |x−1| + |x−10|

For h(x) critical points are 0, 1, 2, 4, 6 and 10.
∴ h(x) will be least when x is between 2 and 4.

For g(x) critical points are 1 and 10.
∴ g(x) will be least when x is between 1 and 10.

∴ Both h(x) and g(x) will be least when x is between 2 and 4.

Let us take x = 3.

Least value of h(x) = 3 + 2 + 1 + 1 + 3 + 7 = 17
Least value of g(x) = 2 + 7 = 9

⇒ Least value of f(x) = 17 + 9 = 26.

Hence, 26.

Answer: Option C

Text Explanation :

(81)^x + (81)^f(x) = 3 

⇒ (3)^4x + (3)^4f(x) = 3 

⇒ (3)^4x-1 + (3)^4f(x)-1 = 1   ...(1)

Now, power of a positive number is always greater than 0.

∴ (3)^4x-1 > 0 and (3)^4f(x)-1 > 0

From (1): Sum of two positive numbers is 1. This is only possible when both number are less than 1.

⇒ (3)^4f(x)-1 < 1

⇒ 4f(x) - 1 < 0

⇒ f(x) < 1/4 or 0.25

∴ f(x) ∈ (-∞, 0.25)

Hence, option (c).

Answer: Option B

Text Explanation :

Given, f(x) = $\frac{\mathrm{x}}{\left|\mathrm{x}\right|}$

Case 1: x > 0
⇒ |x| = x
∴ f(x) = $\frac{\mathrm{x}}{\mathrm{x}}$ = 1

Case 2: x < 0
⇒ |x| = - x
∴ f(x) = $\frac{\mathrm{x}}{-\mathrm{x}}$ = - 1

∴ f(x) = 1 when x > 0 while f(x) = -1 when x < 0.

⇒ f(x) can take only two values {1, -1}.

Hence, option (b).

Answer: Option B

Text Explanation :

If f(x² + f(y)) = xf(x) + y for all non-negative integers x and y, then the value of [f(0)]² + f(0) equals _________.

Given, f(x² + f(y)) = xf(x) + y

Subsituting x = 1 and y = 0, we get

f(1 + f(0)) = f(1)

Now, if f(x) = f(y) it implies that x = y

Hence, 1 + f(0) = 1

⇒ f(0) = 0

∴ [f(0)]² + f(0) = [0]² + 0 = 0

Hence, option (b).

Answer: 3

Text Explanation :

Case 1: x > 3
⇒ f(x - 3) = max(x - 3, 0) = x - 3
⇒ f(x + 1) = max(x + 1, 0) = x + 1
Given, f(x - 3) + 2f(x + 1) = 8
⇒ x - 3 + 2(x + 1) = 8
⇒ 3x = 9
⇒ x = 3

Case 2: -1 < x ≤ 3
⇒ f(x - 3) = max(x - 3, 0) = 0
⇒ f(x + 1) = max(x + 1, 0) = x + 1
Given, f(x - 3) + 2f(x + 1) = 8
⇒ 0 + 2(x + 1) = 8
⇒ 2x = 6
⇒ x = 3

Case 3: x < -1
⇒ f(x - 3) = max(x - 3, 0) = 0
⇒ f(x + 1) = max(x + 1, 0) = 0
Given, f(x - 3) + 2f(x + 1) = 8
⇒ 0 + 0 = 8
Not possible

∴ x = 3 is the only value that satisfies the given function.

Hence, 3.

Answer: Option A

Text Explanation :

f(x + xy) = f(x) + f(xy)

Substituting y = 1
⇒ f(x + x) = f(x) + f(x)
⇒ f(2x) = 2f(x)

Substituting y = 2
⇒ f(x + 2x) = f(x) + f(2x)
⇒ f(3x) = f(x) + 2f(x)
⇒ f(3x) = 3f(x)

Hence, we can say, f(ax) = a × f(x)   ...(1)

Substituting a = 2020 and x = 1
⇒ f(2020) = 2020 × f(1)
⇒ 1 = 2020 × f(1)
⇒ f(1) = 1/2020

Now, putting x = 1 and a = 2021 in (1) we get,
f(2021) = 2021 × f(1) = 2021/2020

Hence, option (a).

Answer: Option B

Text Explanation :

g(x) = -2x passes through (a, b)
⇒ b = -2a   ...(1)

Now, f(x) = ax² + bx + c
⇒ f(x) = ax² - 2ax + c   [from (1)]

f(x) passes through (2, 0)
⇒ a(-2)² - 2a(-2) + c = 0
⇒ 8a + c = 0
⇒ c = -8a

∴ f(x) = ax² - 2ax - 8a

Now, we have to find the least value of f(x) + 9a + 1
= ax² - 2ax - 8a + 9a + 1
= ax² - 2ax + a + 1
= a(x² - 2x + 1) + 1
= a(x - 1)² + 1

Least value of this expression will be 1, when x = 1.

Hence, option (b).

Answer: 19

Text Explanation :

Answer: Option C

Text Explanation :