Question

The number of integers that satisfy the equality \((x^2-5x+7)^{x+1} = 1\) is

• A. 2
• B. 3
• C. 5
• D. 4

Answer 1

The Correct Option is B

The correct answer is (B): \(3\)

\((x^2-5x+7)^{x+1} = 1\)

We know, for \(a^b=1\), if

\(-a =-1\) then \(b\) is even.

\(-a = 1 \) then \(b\) is any number 

\(-a>0\) then \(b=0\)

Case 1: \(x+1=0 ⇒ x = -1\)

Case 2: \(x^2-5x+7 = 1 ⇒ x^2-5x+6=0 ⇒ x = 2 \;or \;3\)

Case 3: \(x^2-5x+7 = -1 ⇒ x^2-5x+8=0\)

but \( x\) is not an integer 

\(∴\) The number of integers satisfies the equation is \(3\)