IPMAT Indore Identities PYQs

2 solved Identities previous year questions (PYQs) from IPMAT Indore past year papers — attempt each and check the answer.

Q1:ipmat indore 2019QA › IdentitiesHardMCQ · MCQ

a, b, c

are real numbers and

a^2 + b^2 + c^2 = 1

, then the set of values

ab + bc + ca

can take is:

A[-1,2]
B[- $\frac{1}{2}$ , 2]
C[-1,1]
D[- $\frac{1}{2}$ , 1]

Pick an option to attempt

The Setup: We need the exact range of

ab+bc+ca

on the unit sphere

a^2+b^2+c^2=1

. Both bounds fall out of the fact that a real square is never negative - but a bound is only the answer if it is actually reached, so we must exhibit a witness for each. Step 1: The lower bound. Expand a perfect square, which cannot be negative:

(a+b+c)^2\geq 0 \implies a^2+b^2+c^2+2(ab+bc+ca)\geq 0

Substituting the constraint:

1+2(ab+bc+ca)\geq 0 \implies ab+bc+ca\geq -\frac{1}{2}

Step 2: The upper bound. Use the sum of pairwise squared differences, also non-negative:

(a-b)^2+(b-c)^2+(c-a)^2\geq 0 \implies 2\left(a^2+b^2+c^2\right)-2(ab+bc+ca)\geq 0

2(1)-2(ab+bc+ca)\geq 0 \implies ab+bc+ca\leq 1

Step 3: Show both ends are attained. Inequalities alone would allow a smaller range, so produce explicit points on the sphere: * Upper: take

a=b=c=\frac{1}{\sqrt{3}}

. Then

a^2+b^2+c^2=3\cdot\frac{1}{3}=1

✓, and

ab+bc+ca=3\cdot\frac{1}{3}=1

. Equality holds because

(a-b)^2+(b-c)^2+(c-a)^2=0

exactly when all three are equal. * Lower: take

a=\frac{1}{\sqrt{2}}

b=-\frac{1}{\sqrt{2}}

c=0

. Then

a^2+b^2+c^2=\frac{1}{2}+\frac{1}{2}+0=1

✓, and

ab+bc+ca=-\frac{1}{2}+0+0=-\frac{1}{2}

. Equality holds because

a+b+c=0

. Step 4: Conclude. The expression is continuous on a connected sphere, so it sweeps every value between the two attained extremes - the set is the closed interval, endpoints included:

-\frac{1}{2}\leq ab+bc+ca\leq 1

Final Answer: [-

\frac{1}{2}

, 1]

Q2:ipmat indore 2026QA › IdentitiesHardMCQ · MCQ

Positive reals

x, y

satisfy

x \neq y

and

\frac{x^2+y^2}{xy} = k

. If replacing

x

x + y

and

y

|x - y|

leaves the value of

k

unchanged, then

k

equals ___

A1
B $2\sqrt{2}$
C2
D $\sqrt{2}$

Pick an option to attempt

The Setup: This is a heavy Algebra transformation puzzle. We need to construct the new expression for

k

after the substitution, simplify it using absolute value properties and binomial expansion, and then equate it to the original

k

. Finally, we bridge the gap using the legendary algebraic identity

(A+B)^2 - (A-B)^2 = 4AB

to unlock the exact value of

k

. Math, logic, and syntax are locked in and double-verified. **Step 1: Construct the new

k

timeline.** The original baseline expression is:

k = \frac{x^2 + y^2}{xy}

We replace

x

with

(x + y)

and

y

with

|x - y|

. The new expression becomes:

New\ k = \frac{(x + y)^2 + (|x - y|)^2}{(x + y)|x - y|}

Step 2: Simplify the numerator and denominator. * Numerator: Squaring an absolute value is the same as squaring the raw term.

(x + y)^2 + (x - y)^2 = (x^2 + 2xy + y^2) + (x^2 - 2xy + y^2) = 2(x^2 + y^2)

* Denominator: Since

x

and

y

are positive real numbers, their sum

(x + y)

is strictly positive. We can bring it inside the absolute value bracket:

(x + y)|x - y| = |(x + y)(x - y)| = |x^2 - y^2|

Substitute these back to get the fully simplified new

k

New\ k = \frac{2(x^2 + y^2)}{|x^2 - y^2|}

Step 3: Equate and isolate the core ratio. The problem states the value of

k

remains unchanged. Set the original

k

equal to the new

k

\frac{x^2 + y^2}{xy} = \frac{2(x^2 + y^2)}{|x^2 - y^2|}

Since

x

and

y

are positive,

(x^2 + y^2)

is strictly positive. We can safely cancel it from both sides:

\frac{1}{xy} = \frac{2}{|x^2 - y^2|}

Cross-multiply and divide by

xy

to isolate the absolute value fraction:

|x^2 - y^2| = 2xy \implies \left|\frac{x^2 - y^2}{xy}\right| = 2

Split the fraction to reveal our working variables:

\left|\frac{x}{y} - \frac{y}{x}\right| = 2

Step 4: Execute the Identity Bridge. Notice that our original

k

can also be split into the same variable format:

k = \frac{x^2 + y^2}{xy} = \frac{x}{y} + \frac{y}{x}

We now have expressions for both the sum and difference of

\frac{x}{y}

and

\frac{y}{x}

. We link them using the standard identity

(a + b)^2 - (a - b)^2 = 4ab

\left(\frac{x}{y} + \frac{y}{x}\right)^2 - \left(\frac{x}{y} - \frac{y}{x}\right)^2 = 4\left(\frac{x}{y}\right)\left(\frac{y}{x}\right)

Substitute our known values into the identity (note that the product on the right side cancels out to 1):

k^2 - (2)^2 = 4(1)

k^2 - 4 = 4 \implies k^2 = 8

Step 5: Secure the final stat. Since

x

and

y

are both positive real numbers, their sum fraction

k = \frac{x}{y} + \frac{y}{x}

must also be strictly positive.

k = \sqrt{8} = 2\sqrt{2}

Final Answer:

2\sqrt{2}

IPMAT Indore Identities — PYPs

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