Discussion

Explanation:

Given, (x² - $10{)}^{x^2-3x-10}$

For this to be equal to 1

Case 1: x² - 10 = 1

⇒ x = $\sqrt{11}$
Rejected as x is not an integer.

Case 2: x² – 3x – 10 = 0
⇒ (x – 5)(x + 2) = 0
⇒ x = 5 or -2 i.e., 2 integral values of x.

Case 3: x² – 10 = -1 and x² – 3x – 10 = even
If x² – 10 = -1 ⇒ x = ±3
For both x = +3 and -3 x² – 3x – 10 is even, hence, 2 integral values of x.

⇒ Total 4 integral values of x are possible.

Hence, 4.

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