The domain of f(x) = x+1(x - 2) / e^2 x x - (2x + 3), x R is: — Solution | JEE Mathematics

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The domain of $f(x) = \log_{x+1}(x - 2) / e^{2\log_x x - (2x + 3)}, \: x \in \mathbb{R}$ is:

Answer

Correct answer:B

Standard Method

Given: $f(x) = \log_{x+1}(x - 2) / e^{2\log_x x - (2x + 3)}$ with $x \in \mathbb{R}$ .

Find: The domain of the function.

For the logarithmic part to be defined:

x - 2 > 0

x > 2

Also, the base of the logarithm must satisfy

x + 1 > 0, \quad x + 1 \neq 1

which gives

x > -1, \quad x \neq 0

From the expression shown in the solution, the denominator restriction gives

x^2 - 2x - 3 \neq 0

that is,

(x - 3)(x + 1) \neq 0

x \neq 3, -1

Combining all valid conditions, the strongest interval condition is

x > 2

and within this interval we must exclude

x = 3

Therefore, the domain is $(2, \infty) - \{3\}$ . Hence, the correct option is B.

The solution contains contradictory option labeling at one place, but its worked conclusion clearly gives the domain as $(2, \infty) - \{3\}$ .

Ignoring the condition on the logarithm argument and allowing $x \le 2$ . This is wrong because $\log_{x+1}(x-2)$ requires $x-2>0$ . Always start by enforcing positivity of the log argument.
Checking only $x+1>0$ for the logarithm base and forgetting that the base cannot be $1$ . This is wrong because a logarithm base must satisfy $a>0$ and $a\ne 1$ . Here that means $x+1>0$ and $x\ne 0$ .
Missing the excluded value $x=3$ from the denominator condition. This is wrong because any denominator equal to zero makes the function undefined. Factor the expression completely before writing the final domain.

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