**Correct Answer: C**

**Step 1: Analyze Statement 1 alone**

[

\frac{x - 1}{x^2 - 1} > 0

]

Factor the denominator:

[

x^2 - 1 = (x - 1)(x + 1)

]

So:

[

\frac{x - 1}{(x - 1)(x + 1)} > 0

]

For (x \ne 1), simplify:

[

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

\Rightarrow x + 1 > 0

\Rightarrow x > -1

]

Also note (x \ne 1) (expression undefined at (x=1)).

Thus:

[

x > -1,\quad x \ne 1

]

Check possibilities:

* (x = 2) → satisfies, and (x > 1)

* (x = 0) → satisfies, but (x < 1)

Both possible.

**Statement 1 is NOT sufficient.**

---

**Step 2: Analyze Statement 2 alone**

[

x^2 - 3x + 2 > 0

]

Factor:

[

(x - 1)(x - 2) > 0

]

This inequality holds when:

[

x < 1 \quad \text{or} \quad x > 2

]

Check:

* (x = 0) → (x < 1)

* (x = 3) → (x > 2)

So:

* Sometimes (x > 1)

* Sometimes (x < 1)

**Statement 2 is NOT sufficient.**

---

**Step 3: Analyze Statements 1 and 2 together**

From Statement 1:

[

x > -1,\quad x \ne 1

]

From Statement 2:

[

x < 1 \quad \text{or} \quad x > 2

]

Combine both:

* Case 1: (x < 1) AND (x > -1) ⇒ (-1 < x < 1)

* Case 2: (x > 2) AND (x > -1) ⇒ (x > 2)

So possible values:

[

(-1, 1) \cup (2, \infty)

]

Now evaluate the question “Is (x > 1)?”

* For (x \in (-1,1)): NO

* For (x > 2): YES

Both outcomes possible.

Wait—recheck simplification of Statement 1 carefully:

Original:

[

\frac{x - 1}{(x - 1)(x + 1)} > 0

]

We must consider sign without canceling blindly:

Sign analysis critical points: (x = -1, 1)

Test intervals:

* (x < -1): pick (x=-2) → expression = (\frac{-3}{(+)} < 0)

* (-1 < x < 1): pick (x=0) → (\frac{-1}{(-)} > 0)

* (x > 1): pick (x=2) → (\frac{1}{(+)} > 0)

Thus solution:

[

(-1, 1) \cup (1, \infty)

]

So Statement 1 implies:

[

x > -1,\quad x \ne 1

]

Now combine correctly:

From Statement 2:

[

x < 1 \quad \text{or} \quad x > 2

]

Intersect with Statement 1:

* ((-1,1)) from first part

* ((2,\infty)) from second part

Same result as above.

Thus still insufficient.

**Statements together are NOT sufficient.**

But careful re-evaluation:

Check if (x=1) is excluded (yes), but still does not resolve ambiguity.

Thus:

**Final Answer: E**

---

**Logical Trap:**

1. **Improper cancellation of ((x-1)):**

Students often cancel ((x-1)) without noting that (x=1) makes the expression undefined and that sign behavior depends on intervals.

2. **Ignoring sign intervals in rational inequalities:**

Treating inequalities algebraically instead of using number-line testing leads to incorrect conclusions.

3. **Overlooking union of solution sets:**

Combining inequalities incorrectly (e.g., assuming intersection instead of union) can falsely suggest sufficiency.

---

**Conclusion:**

Even after combining both statements, the value of (x) can lie in intervals that yield both YES and NO answers to the question.

Therefore, **Statements together are NOT sufficient**.

**Final Answer: E**