IPMAT Indore 2022 · Quantitative Ability · Question 27

Step 1 — fix the signs and sizes on 0<\theta <\pi /4. On this interval

0<\sin \theta <\cos \theta <1

Consequently both logarithms are negative (logs of numbers below 1), and since \sin \theta <\cos \theta,

{\log}_{2}\sin \theta <{\log}_{2}\cos \theta <0

Both exponential factors {\left( \sin \theta \right)}^{\sin \theta} and {\left( \cos \theta \right)}^{\sin \theta} are positive, and because \sin \theta <\cos \theta with a common positive exponent,

{\left( \sin \theta \right)}^{\sin \theta}<{\left( \cos \theta \right)}^{\sin \theta}

Also, raising a number less than 1 to a larger power makes it smaller, and \cos \theta >\sin \theta, so

{\left( \sin \theta \right)}^{\cos \theta}<{\left( \sin \theta \right)}^{\sin \theta}

Step 2 — order the four quantities. Write P={\log}_{2}\cos \theta and Q={\log}_{2}\sin \theta, both negative with Q<P.

a={\left( \sin \theta \right)}^{\sin \theta}\,P

c={\left( \sin \theta \right)}^{\cos \theta}\,P

d={\left( \sin \theta \right)}^{\sin \theta}\,Q

b={\left( \cos \theta \right)}^{\sin \theta}\,Q

Since P is negative, multiplying by a smaller positive number gives a larger (less negative) result, so c>a. Since Q is negative and {\left( \cos \theta \right)}^{\sin \theta}>{\left( \sin \theta \right)}^{\sin \theta}, we get b<d. And comparing a with d: same positive factor, and P>Q, so a>d.

Putting it together:

b<d<a<c

Step 3 — take the median. With four values, the median is the average of the middle two, d and a:

\text{median}=\frac{a+d}{2}

Numerical check at \theta ={30}^{∘} (\sin =0.5, \cos \approx 0.866):

a\approx 0.707\left( -0.2075 \right)=-0.147, b\approx 0.931\left( -1 \right)=-0.931,

c\approx 0.5456\left( -0.2075 \right)=-0.113, d\approx 0.707\left( -1 \right)=-0.707

Order: b<d<a<c ✔ median =\frac{a+d}{2} ✔

Answer: Option 2 — \frac{a+d}{2}

Shortcut. Don’t reason symbolically — substitute \theta ={30}^{∘}. All four expressions become concrete numbers in under a minute, and the ordering b<d<a<c is immediate.

The substitution is safe because the question asks which pair forms the median, and the ordering of a,b,c,d is the same throughout 0<\theta <\pi /4 (no crossings occur).

If you prefer the structure: every term is (positive) × (negative log), so all four are negative, and the comparison reduces to two facts — \sin \theta <\cos \theta, and {x}^{p} decreases in p when 0<x<1. Answer a+d/2.