IPMAT 2024 - Quantitative Aptitude Question 2
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IPMAT 2024 - Quantitative Aptitude
For real solution there are three cases possible:
Case I: (any integer)0 = 1
∴ x2 − 5x + 6 = 0
(x − 2)(x − 3) = 0
x = 2 or 3
Case II: (1)any integer = 1
∴ x2 − 15x + 55 = 1
x2 − 15x + 54 = 0
(x − 9)(x − 6) = 0
x = 6 or 9
Case III: (− 1)any even integer = 1
∴ x2 − 15x + 55 = − 1
x2 − 15x + 56 = 0
(x − 7)(x − 8) = 0
x =7 or 8
But we need to check whether the exponent x2 − 5x + 6 is even or not for x = 7 & 8.
Putting x = 7 in x2 − 5x + 6 we get 20,which is even.
Putting x = 8 in x2 − 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)