\((x^2-7x+11)^{(x^2-13x+42)}=1\)
We know if \(a^b = 1\)
\(⇒ a = 1\) and \(b\) is any number
or \(a = -1\) and \(b\) is even
\(a > 0\) and \(b\) is \(0\)
case 1: \(x^2-13x+42 = 0 ⇒ x = 6,7\)
case 2: \(x^2-7x+11 = 1 ⇒ x^2-7x+10 = 0 ⇒ x = 2 \;or\; 5\)
case 3: \(x^2-7x+11 = -1 ⇒ x^2-7x+12 = 0\)\(\ ⇒ x=4 \;or \ 3\)
Thus, the number of solutions are \(6\).
So, the correct option is (A): \(6\)
\((x2-7x+11)^{(x^{2}-13x+42)}= 1\)
Only possible thing is x2-13x+42 = 0
Or x2-7x+11 = 1
Solving these two, x2-13x+42 = 0
(x – 6) (x – 7) = 0
x = 6 or x = 7
x2-7x+10 = 0
(x – 2) (x – 5) = 0
x = 2 or x = 5
At this moment we have 4 values possible, but there is one more way we can arrive this
(-1)Even = 1
So, x2-7x+11 = -1
(x – 3) (x – 4) = 0
x = 3 or x = 4
Thus, 4 + 2 = 6 values
So, the correct option is (A): \(6\)