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Functions question

2025 · 3 Apr · Shift 2 · Q42

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Functions
2025 · 3 Apr · Shift 2 · Q42

Functions question

2025 · 3 Apr · Shift 2 · Q42

JEE MainMathematicsFunctionsMCQ+4 / −1

If the domain of the function

f(x)=\log _7\left(1-\log _4\left(x^2-9 x+18\right)\right)

is

(\alpha, \beta) \cup(\gamma, o)

, then

\alpha+\beta+\gamma+\hat{o}

is equal to

A
17
B
15
C
16
D
18

View written solutionFree

Correct answer: D

We need the domain of $f(x)=\log_7\left(1-\log_4(x^2-9x+18)\right).$

For logarithms to be defined, we need:

the argument of $\log_4$ to be positive,
the argument of $\log_7$ to be positive.

So the conditions are: $x^2-9x+18>0$ and $1-\log_4(x^2-9x+18)>0.$

Solve the second condition: $1-\log_4(x^2-9x+18)>0$ $\Rightarrow \log_4(x^2-9x+18)<1.$

Since base $4>1$ , this gives $x^2-9x+18<4.$

Also, from the inner logarithm, $x^2-9x+18>0.$

Hence combined: $0<x^2-9x+18<4.$

Factor the quadratic: $x^2-9x+18=(x-3)(x-6).$

So we need $0<(x-3)(x-6)<4.$

This means solving two inequalities together.

First solve $(x-3)(x-6)>0.$ This holds for $x<3 \quad \text{or} \quad x>6.$

Now solve $(x-3)(x-6)<4.$ $x^2-9x+18<4$ $x^2-9x+14<0$ $(x-7)(x-2)<0.$ Thus, $2<x<7.$

Intersect the two solution sets:

From step 4: $(-\infty,3)\cup(6,\infty)$ From step 5: $(2,7)$

Their intersection is $(2,3)\cup(6,7).$

Thus the domain is of the form $(\alpha,\beta)\cup(\gamma,o)=(2,3)\cup(6,7).$ So, $\alpha=2,\; \beta=3,\; \gamma=6,\; o=7.$

Therefore, $\alpha+\beta+\gamma+o=2+3+6+7=18.$

Compare with stored answer:

Derived answer = $18$ . Stored correct answer = $18$ (Option D).

They agree.

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