NIMCET 2007 — Mathematics PYQ

Step 1: Condition for the Logarithm Argument

The argument of the logarithm is $\frac{x-1}{x+5}$. For the log function to be defined:

\frac{x-1}{x+5} > 0

Using the wavy curve (sign-scheme) method, the critical points are $x=1$ and $x=-5$.

• The expression is positive when $x$ lies outside the interval $[-5,1]$.

• Thus, we get:

  $$x\in (-\infty ,-5)\cup (1,\infty )\text{— (Condition 1)}$$

Step 2: Condition for the Logarithm Value (Base < 1)

The base of the logarithm is $0.4$, which is less than $1$ (0 < \text{base} < 1).

When the base of a logarithm is less than $1$, the value of the log is non-negative ($\ge 0$) if and only if its argument is less than or equal to $1$:

$${\log }_{0.4}\frac{x-1}{x+5}\ge 0⟹\frac{x-1}{x+5}\le 1$$

Let's solve this inequality:

$$\frac{x-1}{x+5}-1\le 0$$

$$\frac{(x-1)-(x+5)}{x+5}\le 0$$

$$\frac{-6}{x+5}\le 0$$

For a fraction with a negative numerator to be less than or equal to $0$, its denominator must be strictly positive:

x + 5 > 0 \implies x > -5 \quad \text{--- (Condition 2)}

Step 3: Condition for the Rational Term Denominator

The expression has a fraction $\frac{1}{x^2-6}$ under the square root. Its denominator cannot be zero:

$$x^2-6\ne 0⟹x\ne \pm \sqrt{6}$$

Step 4: Intersection of All Conditions

Now, let's find the common interval that satisfies both Condition 1 and Condition 2:

• From Condition 1: $x\in (-\infty ,-5)\cup (1,\infty )$

• From Condition 2: $x\in (-5,\infty )$

Taking the intersection of these two sets:

$$x\in (1,\infty )$$

Finally, we remove the values where the denominator becomes zero ($x=\pm \sqrt{6}$). Since $-\sqrt{6}$ is already excluded from $(1,\infty )$, we only need to remove $\sqrt{6}\approx 2.45$:

\text{Domain} = \{x : x > 1, x \neq \sqrt{6}\}

Comparing this result with the given options:

• Option (c) states \{x : x > 1, x \neq 6\}, which contains an incorrect value ($6$ instead of $\sqrt{6}$).

Therefore, the correct choice is (d) none