IPMAT Minima & Maxima PYQs

2 solved Minima & Maxima previous year questions (PYQs) from past year papers — attempt each and check the answer.

Q1:ipmat indore 2019QA › Minima & MaximaHardSA · TITA

For all real values of

x

\dfrac{3x^2 - 6x + 12}{x^2 + 2x + 4}

lies between

1

and

k

, and does not take any value above

k

. Then

k

equals:

Enter your answer to attempt

The Setup: To find the range of a rational function, set it equal to

y

, rearrange into a quadratic in

x

, and force the discriminant to be non-negative - because

x

is required to be real. The values of

y

that survive are exactly the range. Step 0: Confirm the function is defined everywhere. The denominator

x^2+2x+4

has discriminant

2^2-4(1)(4)=-12<0

, so it never hits zero and the expression is defined for all real

x

. Without this, chunks of the range argument could be vacuous. **Step 1: Create the quadratic in

x

.**

y=\frac{3x^2-6x+12}{x^2+2x+4} \implies y(x^2+2x+4)=3x^2-6x+12

Collecting terms:

x^2(y-3)+x(2y+6)+(4y-12)=0

**Step 2: Handle the

y=3

case separately - the discriminant method assumes a genuine quadratic.** If

y=3

the

x^2

coefficient vanishes and this is not a quadratic at all; it degenerates to

12x+0=0

, giving

x=0

. So

y=3

is attained, at

x=0

, and indeed

\frac{12}{4}=3

. It sits inside the range we are about to derive, so the final answer is unaffected - but the case has to be dismissed explicitly, not ignored. **Step 3: Enforce the discriminant for

y\neq 3

.** Real

x

requires

b^2-4ac\geq 0

. Noticing that

4y-12=4(y-3)

keeps this tidy:

(2y+6)^2-4(y-3)\cdot 4(y-3)\geq 0 \implies (2y+6)^2-16(y-3)^2\geq 0

Since

(2y+6)^2=4(y+3)^2

, divide throughout by 4:

(y+3)^2-4(y-3)^2\geq 0

Step 4: Difference of squares. With

A=y+3

and

B=2(y-3)

\left[(y+3)-2(y-3)\right]\left[(y+3)+2(y-3)\right]\geq 0

(-y+9)(3y-3)\geq 0 \implies 3(9-y)(y-1)\geq 0

Dividing by 3 and negating the first bracket flips the inequality:

(y-9)(y-1)\leq 0 \implies 1\leq y\leq 9

Step 5: Confirm both endpoints are actually reached. An inequality gives bounds; it does not by itself prove they are attained, and the stem's wording requires that the function really does climb to

k

. *

y=9

: the quadratic becomes

6x^2+24x+24=0

, i.e.

(x+2)^2=0

, so

x=-2

. Check:

\frac{12+12+12}{4-4+4}=\frac{36}{4}=9

. *

y=1

: it becomes

-2x^2+8x-8=0

, i.e.

(x-2)^2=0

, so

x=2

. Check:

\frac{12-12+12}{4+4+4}=\frac{12}{12}=1

. Both are hit exactly once, at the two points where the discriminant is zero. The maximum value is therefore

k=9

. Final Answer: 9

Q2:ipmat indore 2026QA › Minima & MaximaHardMCQ · MCQ

x

is a real number such that

\max(\min(x, 2 - x), x - 4, 2x - 8) = \pi - 3

, then the number of possible values of

x

Pick an option to attempt

The Setup: This is an Algebra boss fight dealing with nested Min/Max functions. The meta is to substitute the constant

\pi - 3

with a dummy variable

T \approx 0.14

to keep the equations clean. For the *maximum* of three terms to equal

T

, at least one specific term must exactly equal

T

, and we must run a validation check on that timeline to ensure the other two terms do not exceed

T

. Math, logic, and syntax are locked in and double-verified. Step 1: Set the baseline constraints. Let

T = \pi - 3 \approx 0.14

. Our core equation is:

\max(\min(x, 2 - x), x - 4, 2x - 8) = T

. Rule of the Max function: At least one of these three quantities must equal

T

, and none of them can be greater than

T

. **Step 2: Run Case 1 (

\min(x, 2 - x) = T

).** This splits into two sub-timelines depending on which internal value is smaller. * **If

x \leq 1

:** The minimum is

x

, so

x = T \approx 0.14

. * Check others:

x - 4 = T - 4 \approx -3.86

(

\leq T

). * Check others:

2x - 8 = 2T - 8 \approx -7.72

(

\leq T

). * Both checks pass. (1 Valid Solution) * **If

x \geq 1

:** The minimum is

2 - x

, so

2 - x = T \implies x = 2 - T \approx 1.86

. * Check others:

x - 4 = (2 - T) - 4 = -2 - T \approx -2.14

(

\leq T

). * Check others:

2x - 8 = 2(2 - T) - 8 = -4 - 2T \approx -4.28

(

\leq T

). * Both checks pass. (1 Valid Solution) **Step 3: Run Case 2 (

2x - 8 = T

).** Isolate

x

for this timeline:

2x = T + 8 \implies x = \frac{T + 8}{2} = \frac{T}{2} + 4 \approx 4.07

* Check the second term:

x - 4 = \left(\frac{T}{2} + 4\right) - 4 = \frac{T}{2} \approx 0.07

Since

0.07 \leq 0.14

, this term is

\leq T

. Valid. * Check the first term (

\min(x, 2 - x)

): Since

x \approx 4.07

, the minimum will clearly be

2 - x

2 - x = 2 - \left(\frac{T}{2} + 4\right) = -2 - \frac{T}{2} \approx -2.07

Since

-2.07 \leq 0.14

, this term is

\leq T

. Valid. Both checks pass. (1 Valid Solution) **Step 4: Run Case 3 (

x - 4 = T

).** Isolate

x

for this timeline:

x = T + 4 \approx 4.14

* Check the third term:

2x - 8 = 2(T + 4) - 8 = 2T + 8 - 8 = 2T

Since

T \approx 0.14

is a strictly positive number,

2T > T

. This means the third term exceeds our absolute maximum cap of

T

. This breaks the core rule of the function. This timeline is completely invalid. (0 Valid Solutions) Step 5: Tally the valid spawns. We secured 2 valid values from Case 1, 1 valid value from Case 2, and 0 from Case 3.

Total\ Valid\ Values = 2 + 1 = 3

Final Answer: 3

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