Logarithms Inequality PYQ | JEE Main 2025 | Domain + Range Trick

 

❓ Question

If

$$\log_{(x+3)}(x^2-x)<1$$

is satisfied for

$$x\in(a,b),$$

then find the value of

$$a+b.$$

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✍️ Solution

For the logarithm to be defined,

Domain conditions

→ Argument must be positive:

$$x^2-x>0$$
$$x(x-1)>0$$
$$\boxed{x\in(-\infty,0)\cup(1,\infty)}$$

→ Base must satisfy

$$x+3>0,\qquad x+3\ne1$$
$$\boxed{x>-3,\quad x\ne-2}$$

Since $x+3>1$ for $x>-2$, the logarithm is increasing on the required interval.

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Solve the inequality

$$\log_{x+3}(x^2-x)<1$$

As the base is greater than 1,

$$x^2-x<x+3$$
$$x^2-2x-3<0$$
$$(x-3)(x+1)<0$$
$$\boxed{-1<x<3}$$

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Intersect with the domain

Domain:

$$(-3,0)\cup(1,\infty)$$

Required interval:

$$(-1,3)$$

Hence,

$$x\in(-1,0)\cup(1,3).$$

The interval of the form $(a,b)$ is

$$(a,b)=(-1,3).$$

Therefore,

$$a+b=-1+3=2.$$

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✅ Final Answer

$$\boxed{2}$$