IPMAT Indore Functions PYQs

Q1:ipmat indore 2019QA › FunctionsHardSA · TITA

A real-valued function

f

satisfies the relation

f(x)f(y) = f(2xy + 3) + 3f(x + y) - 3f(y) + 6y

, for all real numbers

x

and

y

, then the value of

f(8)

is

Enter your answer to attempt

The Setup: Functional equations are the final boss of algebra. The standard strat is to substitute a convenient value to collapse one variable, and here

x=0

does something better than simplify - it *proves* the function is linear rather than making us assume it. **Step 1: Substitute

x=0

.**

f(0)f(y)=f(3)+3f(y)-3f(y)+6y \implies f(0)f(y)=f(3)+6y

Write

c=f(0)

, so that

c\,f(y)=6y+f(3)

. **Step 2: Rule out

c=0

before dividing.** If

c

were 0, the left side would be 0 for every

y

, forcing

6y+f(3)=0

for all real

y

- impossible, since the left side varies with

y

and the right is fixed. So

c\neq 0

, and we may divide:

f(y)=\frac{6}{c}y+\frac{f(3)}{c}

This is a genuine deduction, not a guess:

f

must be linear. Write

f(x)=kx+c

, noting

f(0)=c

is consistent and that

k=\frac{6}{c}

, i.e.

ck=6

. Step 3: Substitute the linear form into the original relation.

\text{LHS}=(kx+c)(ky+c)=k^2xy+kcx+kcy+c^2

\text{RHS}=\left[k(2xy+3)+c\right]+3\left[k(x+y)+c\right]-3\left[ky+c\right]+6y=2kxy+3kx+6y+3k+c

The

3ky

and

-3ky

cancel, which is why no

y

-term survives on the right beyond the

6y

. Step 4: Match all four coefficients - not just one.

Term	LHS	RHS	Equation
$xy$	$k^2$	$2k$	$k^2=2k$
$x$	$kc$	$3k$	$kc=3k$
$y$	$kc$	$6$	$kc=6$
constant	$c^2$	$3k+c$	$c^2=3k+c$

From

k^2=2k

we get

k=0

or

k=2

;

k=0

contradicts

ck=6

, so

k=2

. Then

kc=6

gives

c=3

. Check the two remaining rows hold rather than assuming they do:

kc=6=3k

✓, and

c^2=9=3(2)+3

✓. All four agree, so:

f(x)=2x+3

Step 5: Execute.

f(8)=2(8)+3=19

Final Answer: 19

Q2:ipmat indore 2023QA › FunctionsMediumSA · TITA

If

f(1) = 1

and

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

for all integers

n > 1

, then the value of

f(2023)

is

Enter your answer to attempt

The Setup: A recursive sequence is defined as

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

with base case

f(1)=1

. We must find the value of

f(2023)

. Step 1: Generate the first few terms to identify a pattern.

f(1) = 1

f(2) = 3(2) - 1 = 5

f(3) = 3(3) - 5 = 4

f(4) = 3(4) - 4 = 8

f(5) = 3(5) - 8 = 7

f(6) = 3(6) - 7 = 11

Step 2: Formulate a generalized relationship. Let's find the difference between every second term (

f(n)

and

f(n-2)

):

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

Substitute the recursive definition for

f(n-1)

:

f(n) = 3n - (3(n-1) - f(n-2))

f(n) = 3n - 3n + 3 + f(n-2)

f(n) = f(n-2) + 3

This confirms that the odd-indexed terms and even-indexed terms form independent Arithmetic Progressions, each with a common difference of

3

. Step 3: Evaluate

f(2023)

. Since

2023

is odd, we use the sequence of odd-indexed terms:

1, 4, 7, 10, \dots

The generic formula for the odd index sequence

f(2k-1)

is an AP where the first term is

1

and the difference is

3

. Let

n = 2023

. The number of steps (or the term position in the separated odd sequence) is

\frac{n-1}{2}

.

\text{Term Index} = \frac{2023 - 1}{2} = 1011

f(2023) = \text{Base} + (\text{Term Index}) \times \text{Difference}

f(2023) = 1 + 1011 \times 3

f(2023) = 1 + 3033 = 3034

Final Answer: 3034

Q3:ipmat indore 2021QA › FunctionsMediumSA · TITA

If a function

f(a) = max (a, 0)

then the smallest integer value of

x

for which the equation

f(x - 3) + 2f(x + 1) = 8

holds true is _______.

Enter your answer to attempt

The Setup: This is a piecewise function check. The function

f(a)=\max(a,0)

basically acts like a bouncer: if

a

is negative, it gets zeroed out. We need to find the smallest integer

x

where

f(x-3)+2f(x+1)=8

. Step 1: Test the lowest tier (where both inputs are negative). If

x \le -1

, then both

x-3

and

x+1

are negative or zero. The bouncer zeroes them both:

0 + 2(0) = 0 \neq 8

That's a hard pass. Step 2: Level up to the middle tier (

-1 < x \le 3

). Here,

x+1

is strictly positive (gets VIP access), but

x-3

is still negative or zero (gets zeroed). The equation becomes:

0 + 2(x+1) = 8

x + 1 = 4

x = 3

This works perfectly and fits exactly at the upper boundary of our assumed domain. Step 3: Double-check the top tier just to be safe (

x>3

). Here, both inputs are strictly positive:

(x-3) + 2(x+1) = 8

3x - 1 = 8

3x = 9 \implies x = 3

Since

x=3

is the only valid root across the whole domain, that's our main character. Final Answer: 3

Q4:ipmat indore 2021QA › FunctionsMediumMCQ · MCQ

Suppose that a real-valued function

f(x)

of real numbers satisfies

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

) for all real

x, y,

and that

f(2020) = 1

. Compute

f(2021)

.

A $\frac{2021}{2020}$
B $\frac{2020}{2019}$
C1
D $\frac{2020}{2021}$

Pick an option to attempt

The Setup: We are dealing with functional equations, which is essentially algebra's version of reverse-engineering a game engine. We need to find the core behavior of

f(x)

by plugging in strategic dummy variables until the underlying algorithm reveals itself. Step 1: Test the waters by setting

y=0

to find the baseline.

f(x+0)=f(x)+f(0) \implies f(x)=f(x)+f(0) \implies f(0)=0

Step 2: Unlock the additive property. For any

x \neq 0

and any real

z

, just pick

y=z/x

- then

xy=z

, and the given equation morphs into:

f(x+z)=f(x)+f(z)

Together with Step 1 (which covers

x=0

),

f

is additive across the reals. Step 3: Stack that property. Applying it repeatedly gives

f(nt)=nf(t)

for every positive integer

n

. That is the *only* thing we need - so we never have to assume

f

is linear or even continuous, which would be an unearned leap. Step 4: Use the given stat

f(2020)=1

, taking

t=1

and

n=2020

.

2020f(1)=f(2020)=1 \implies f(1)=\frac{1}{2020}

Step 5: Calculate the final target with

n=2021

.

f(2021)=2021f(1)=\frac{2021}{2020}

Final Answer:

\frac{2021}{2020}

Q5:ipmat indore 2024QA › FunctionsMediumSA · TITA

Let

f

and

g

be two functions defined by

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

and

g(x) = \frac{1}{x}

for

x \neq 0

. If

f(a) + g(f(a)) = \frac{13}{6}

for some real

a

, then the maximum possible value of

f(g(a))

is:

Enter your answer to attempt

The Setup: We are given the functions

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

and

g(x)=1/x

for non-zero

x

. Based on the constraint

f(a)+g(f(a))=13/6

, we must find the maximum possible value of

f(g(a))

. Step 1: Analyze the piecewise domains of

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

. If

x \le 0

, then

|x| = -x \Rightarrow f(x) = |x - x| = 0

. If

x > 0

, then

|x| = x \Rightarrow f(x) = |x + x| = 2x

. Step 2: Ascertain the domain of

a

using the given composite equation. The given condition is

f(a) + g(f(a)) = \frac{13}{6}

. The function

g(x) = \frac{1}{x}

is fundamentally undefined at

0

. Consequently,

f(a)

cannot equal

0

. By our piecewise analysis, if

f(a) \neq 0

, then

a

must be strictly positive (

a > 0

), locking

f(a) = 2a

. Step 3: Formulate and solve the rational equation for

a

. Substitute

f(a) = 2a

into the equation:

2a + g(2a) = \frac{13}{6}

2a + \frac{1}{2a} = \frac{13}{6}

Let

u = 2a

to clarify the quadratic structure:

u + \frac{1}{u} = \frac{13}{6}

Multiply entirely by

6u

:

6u^2 + 6 = 13u \Rightarrow 6u^2 - 13u + 6 = 0

Factor the resulting quadratic:

(2u - 3)(3u - 2) = 0

Thus,

u = \frac{3}{2}

or

u = \frac{2}{3}

. Because

u = 2a

, we trace back to two valid positive candidates for

a

:

a = \frac{3}{4} \quad \text{or} \quad a = \frac{1}{3}

Step 4: Evaluate the maximization query for

f(g(a))

. Since

a > 0

,

g(a) = \frac{1}{a} > 0

. Any positive input to

f

triggers the

2x

piecewise condition.

f(g(a)) = f\left(\frac{1}{a}\right) = 2\left(\frac{1}{a}\right) = \frac{2}{a}

Evaluate against both derived candidates of

a

: If

a = \frac{3}{4} \Rightarrow \frac{2}{3/4} = \frac{8}{3} \approx 2.67

If

a = \frac{1}{3} \Rightarrow \frac{2}{1/3} = 6

The strict mathematical maximum is

6

. Final Answer: 6

Q6:ipmat indore 2022QA › FunctionsMediumMCQ · MCQ

The set of all possible values of

f(x)

for which

(81)^{x}+(81)^{f(x)}=3

is

A( $0.25,3)$
B $(-\infty, 4)$
C $(-\infty, 0.25)$
D $(3,4)$

Pick an option to attempt

The Setup: Given the equation

81^x + 81^{f(x)} = 3

, we need to determine the complete set of possible values for the output of the function

f(x)

. Step 1: Isolate the term containing

f(x)

.

81^{f(x)} = 3 - 81^x

Step 2: Analyze the domain and range constraints. For any real number

x

, an exponential function with a positive base is strictly positive:

81^x > 0

. Since

81^{f(x)}

is also strictly positive, its equal counterpart must be positive:

3 - 81^x > 0 \implies 81^x < 3

Because

81^x

is strictly bounded between

0

and

3

, the expression

(3 - 81^x)

maps continuously across the open interval

(0, 3)

. Step 3: Translate the range to

f(x)

. Because

81^{f(x)} = 3 - 81^x

, the function

81^{f(x)}

takes all values in the interval

(0, 3)

.

0 < 81^{f(x)} < 3

The lower bound

0 < 81^{f(x)}

holds true for all real values of

f(x)

(an exponential function has no lower limit restriction other than

0

). Solve the upper bound:

81^{f(x)} < 3 \implies (3^4)^{f(x)} < 3^1 \implies 3^{4f(x)} < 3^1

Because base

3 > 1

, we can equate the exponents directly:

4f(x) < 1 \implies f(x) < 0.25

Thus,

f(x)

covers all real numbers strictly less than

0.25

. Final Answer:

(-\infty, 0.25)

Q7:ipmat indore 2020QA › FunctionsEasyMCQ · MCQ

Given

f(x) = x^2 + \log_3 x

and

g(y) = 2y + f(y)

, then the value of

g(3)

equals

A16
B15
C25
D26

Pick an option to attempt

The Setup: Pure function evaluation. No theory required - just substitute from the outside in, respecting each definition exactly as written. Step 1: Set up the outer function. The definition is

g(y)=2y+f(y)

, so at

y=3

:

g(3)=2(3)+f(3)=6+f(3)

Note the variable name is a red herring:

f

is written in

x

and

g

in

y

, but a function does not care what its argument is called. Feeding 3 into

f

is the same operation either way. Step 2: Evaluate the nested function. Using

f(x)=x^2+\log_3 x

at

x=3

:

f(3)=3^2+\log_3 3

We have

3^2=9

, and any log of its own base is 1, since

\log_b b=1

means *what power of

b

gives

b

?* - namely 1. So

\log_3 3=1

and

x=3>0

keeps the log defined.

f(3)=9+1=10

Step 3: Recombine. Substitute back:

g(3)=6+10=16

Final Answer: 16

Q8:ipmat indore 2022QA › FunctionsMediumMCQ · MCQ

Let

A=\{1,2,3\}

and

B=\{a, b\}

. Assuming all relations from set

A

to set

B

are equally likely, what is the probability that a relation from

A

to

B

is also a function?

A $\frac{1}{8}$
B $\frac{1}{2}$
C1
D $\frac{3^{2}}{2^{6}}$

Pick an option to attempt

The Setup: Set

A = \{1, 2, 3\}

and Set

B = \{a, b\}

. If we pull randomly from the pool of all possible mathematical relations mapping A to B, we need the probability that the relation is specifically a valid function. Step 1: Calculate the size of the total sample space (all relations). A mathematical relation from

A

to

B

is defined as any possible subset of the Cartesian product

A \times B

. The cardinality of the Cartesian product is

|A| \times |B| = 3 \times 2 = 6

. The total number of unique relations (subsets) is

2^{|A \times B|} = 2^6 = 64

. Step 2: Calculate the size of the restricted target group (all functions). A function is a highly specific relation where every element in set

A

maps to exactly one valid element in set

B

. Each of the

3

elements in

A

has exactly

2

choices to map to in

B

. Total number of functions =

|B|^{|A|} = 2^3 = 8

. Step 3: Calculate the resulting probability.

\text{Probability} = \frac{\text{Target Outcomes}}{\text{Total Outcomes}} = \frac{8}{64} = \frac{1}{8}

Final Answer:

\frac{1}{8}

Q9:ipmat indore 2022QA › FunctionsEasyMCQ · MCQ

A set of all possible values the function

f(x)=\dfrac{x}{|x|}

, where

x \neq 0

, takes is

A $\{1\}$
B $\{1,-1\}$
C $\{1,0\}$
D $\{1,0,-1\}$

Pick an option to attempt

The Setup: We need to specify the range (the set of all possible outputs) for the absolute value ratio function

f(x)=\frac{x}{|x|}

. Step 1: Analyze the piecewise domains of the function. The function possesses a domain restricting

x \neq 0

. If

x > 0

(positive numbers), then

|x| = x

.

f(x) = \frac{x}{x} = 1

If

x < 0

(negative numbers), then

|x| = -x

.

f(x) = \frac{x}{-x} = -1

Step 2: Formulate the set of possible values. Since the function mathematically collapses into exactly two static outputs dependent on algebraic sign, its absolute range is limited exclusively to those two integers. The set of outputs is

\{1, -1\}

. Final Answer:

\{1,-1\}

Q10:ipmat indore 2022QA › FunctionsHardMCQ · MCQ

If

f\left(x^{2}+f(y)\right)=x f(x)+y

for all non-negative integers

x

and

y

, then the value of

[f(0)]^{2}+f(0)

equals _________.

A2
B0
C6
D1

Pick an option to attempt

The Setup: A functional mapping rule dictates

f(x^2 + f(y)) = x f(x) + y

for all non-negative integers

x, y

. We must solve for

f(0)

to evaluate

[f(0)]^2 + f(0)

. Step 1: Exploit substitution to define the function. Let

x = 0

. The equation becomes:

f(0^2 + f(y)) = 0 \cdot f(0) + y \implies f(f(y)) = y

This proves that the function is an involution (it acts identically as its own inverse). Consequently, substituting

y = 0

yields

f(f(0)) = 0

. Step 2: Set up a root evaluation using the involution property. Let

c = f(0)

. Substituting this into our involution proof implies

f(c) = 0

. Now, set

y = 0

in the original primary equation:

f(x^2 + f(0)) = x f(x)

Substitute

x = c

:

f(c^2 + c) = c \cdot f(c)

Because we proved

f(c) = 0

:

f(c^2 + c) = c(0) = 0

Step 3: Solve for

c

. Because the function maps

c^2 + c

to

0

, and we previously established that

f(c)

uniquely maps to

0

(since

f

acts as a strict bijection due to

f(f(x))=x

), the internal arguments must be rigidly equal:

c^2 + c = c \implies c^2 = 0 \implies c = 0

Thus,

f(0) = 0

. Step 4: Evaluate the requested expression. We must calculate

[f(0)]^2 + f(0)

.

0^2 + 0 = 0

Final Answer: 0

Q11:ipmat indore 2019QA › FunctionsHardMCQ · MCQ

The function

f(x) = \dfrac{x^3 - 5x^2 - 8x}{3}

is

Apositive and monotonically increasing for x $\in (-\infty, \frac{5-\sqrt{57}}{2})$ and x $\in (\frac{5+\sqrt{57}}{2}, +\infty)$
Bnegative and monotonically decreasing for x $\in (-\infty, \frac{5-\sqrt{57}}{2})$ and x $\in (\frac{5+\sqrt{57}}{2},+\infty)$
Cnegative and monotonically increasing for x $\in (-\infty, \frac{5-\sqrt{57}}{2})$ and positive and monotonically increasing for x $\in (\frac{5+\sqrt{57}}{2},+\infty)$
Dpositive and monotonically increasing for x $\in (-\infty, \frac{5-\sqrt{57}}{2})$ and negative and monotonically decreasing for x $\in (\frac{5+\sqrt{57}}{2},+\infty)$

Pick an option to attempt

The Setup: A two-part question: where the function is positive or negative, and where it is increasing or decreasing. The options pair those up, so both must be settled and then matched on the same intervals. Step 1: Find the roots - this fixes the sign. Factor out

x

:

f(x)=\frac{x\left(x^2-5x-8\right)}{3}=0

The roots are

x=0

together with the solutions of

x^2-5x-8=0

:

x=\frac{5\pm\sqrt{25+32}}{2}=\frac{5\pm\sqrt{57}}{2}

Write

\alpha=\frac{5-\sqrt{57}}{2}\approx -1.27

and

\beta=\frac{5+\sqrt{57}}{2}\approx 6.27

, so the three roots in order are

\alpha<0<\beta

. Being a cubic with positive leading coefficient,

f

alternates sign across them:

Interval	Sign of $f$
$x<\alpha$	negative
$\alpha<x<0$	positive
$0<x<\beta$	negative
$x>\beta$	positive

Step 2: Differentiate - this fixes the monotonicity.

f'(x)=\frac{3x^2-10x-8}{3}=\frac{(3x+2)(x-4)}{3}

The critical points are

x=-\frac{2}{3}

and

x=4

, and since the parabola opens upward,

f'>0

outside them:

f \text{ is increasing on } \left(-\infty,-\tfrac{2}{3}\right) \text{ and } (4,\infty), \text{ decreasing on } \left(-\tfrac{2}{3},4\right)

Step 3: Match the two pictures on the same intervals. The containments are what make the answer work, so check them numerically: * On

(-\infty,\alpha)

: here

\alpha\approx -1.27<-\frac{2}{3}\approx -0.67

, so this whole interval lies inside the increasing region

\left(-\infty,-\frac{2}{3}\right)

. From the table,

f

is negative there. So: negative and monotonically increasing. * On

(\beta,\infty)

: here

\beta\approx 6.27>4

, so this interval lies inside the increasing region

(4,\infty)

. From the table,

f

is positive there. So: positive and monotonically increasing. Both intervals are increasing, but they differ in sign - which is precisely what separates the correct option from the three that pair the signs the other way round or call one of them decreasing. Final Answer: negative and monotonically increasing for x

\in (-\infty, \frac{5-\sqrt{57}}{2})

and positive and monotonically increasing for x

\in (\frac{5+\sqrt{57}}{2},+\infty)

IPMAT Indore Functions — PYPs

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