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IPMAT Indore 2024 - QA (MCQ) Q25 Explanation
November 24, 2024
Explanation:
(b) (𝑥^2 −5)^4 +(𝑥^2 −7)^4 = 16
(𝑥^2 −6+1)^4 +(𝑥^2 −6−1)^4 = 16
Take 𝑥^2 −6 = 𝑡….(i)
(𝑡 +1)^4 +(𝑡 −1)^4 = 16
(𝑡^4 + 4𝑡^3+ 6𝑡^2 +4𝑡+1)+ (𝑡^4−4𝑡^3+ 6𝑡^2−4𝑡+1) = 16
2(𝑡^4 +6𝑡^2 +1) = 16
t^4 +6𝑡^2 −7 =0
Let t^4 = k^2
k^2 +6𝑘−7 = 0
(𝑘 +7)(𝑘 −1) = 0
k = −7, 𝑘 = 1
k = 𝑡^2 = −7 (not possible, because a perfect square can't be negative)
∴ 𝑡^2 = 1
⇒ 𝑡=±1
Putting t = 1 in the equation (i), we get
x2 − 6 = 1
x2 = 7
x = ±√7
Putting t = -1 in the equation (i), we get
x^2−6 = −1
x^2=5