IPMAT Indore 2024 - QA (SA) Q2 Explanation

Explanation:

(6) (^2 −15𝑥+55)^𝑥2−5𝑥+6 = 1

For real solution there are three cases possible:

Case I: (any integer)^0 = 1

=x^2 −5x+6=0

= (x−2)(x−3)=0

= x=2 or 3

Case II: (1) any integer = 1

= x^2 −15𝑥+55=1

= x^2−15x+54=0

=(x−9)(x−6)=0

x=6 𝑜𝑟 9

Case III: (−1) any even integer = 1

=x^2−15x+55=−1

=x^2−15x+56=0

= (x−7)(x−8)=0

= x=7 𝑜𝑟 8

But we need to check whether the exponent x^2 −5x +6 is even or not for  x = 7 & 8.

Putting x = 7 in x^2−5x+6 we get 20, which is even.

Putting x = 8 in x^2−5x +6 we get 30, which is again even.

Thus both 7 and 8 are valid solutions,as well.

Total the total number of real solutions for the given equation is 6 i.e (2, 3, 6, 9, 7, 8) Ans.