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Maths Question 2 – JEE-MAIN 2025

If the domain of the function f(x)=log2(1log4(x29x+18)) is (α,β)(γ,δ), then α+β+γ+δ is equal to

For a logarithm loga(b) to be defined, its base a must be positive and not equal to 1, and its argument b must be positive.

🥷
Ninja StrategySum of Boundary Points

The boundary points of the domain intervals are the roots of the quadratic equations x29x+14=0 and x29x+18=0. The sum of these roots directly gives the required sum.

Step 1: Apply Outer Logarithm Domain Condition✦ Active

For f(x)=log2(1log4(x29x+18)) to be defined, the argument of the outer logarithm must be positive:

1log4(x29x+18)>0

This implies log4(x29x+18)<1. Since the base 4>1, we can remove the logarithm:

x29x+18<41 x29x+14<0 (x2)(x7)<0

This inequality holds for x(2,7). (Condition 1)

Step 2: Apply Inner Logarithm Domain Condition○ Expand

The argument of the inner logarithm must also be positive:

x29x+18>0 (x3)(x6)>0

This inequality holds for x(,3)(6,). (Condition 2)

Step 3: Find Intersection and Calculate Sum○ Expand

The domain of f(x) is the intersection of Condition 1 and Condition 2:

(2,7)((,3)(6,))=(2,3)(6,7)

Comparing this with the given domain (α,β)(γ,δ), we have α=2, β=3, γ=6, and δ=7. The required sum is:

α+β+γ+δ=2+3+6+7=18
💡 Teacher's Secret Hint

Ensure to correctly identify the intervals and their boundaries from the inequalities.

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