Basics & LogarithmseasyFree

Basics & Logarithms — JEE Maths practice question

JEE Maths question with a full step-by-step solution.

Question
Let P(x)=x6+ax5+bx4+cx3+dx2+ex+fP(x) = x^6 + ax^5 + bx^4 + cx^3 + dx^2 + ex + f be a polynomial such that P(1)=1P(1)= 1, P(2)=2P(2) = 2, P(3)=3P(3) = 3, P(4)=4P(4) = 4, P(5)=5P(5) = 5 and P(6)=6P(6) = 6. Then the value of P(7)P(7) is equal to:
A616616
B727727correct
C729729
D721721
Solution
Step 1: Construct an auxiliary polynomial Define Q(x)=P(x)−xQ(x) = P(x) - x. Then Q(1)=Q(2)=⋯=Q(6)=0Q(1) = Q(2) = \cdots = Q(6) = 0, so 1,2,3,4,5,61, 2, 3, 4, 5, 6 are roots of Q(x)Q(x). Step 2: Determine Q(x)Q(x) Since P(x)P(x) is monic of degree 6, Q(x)Q(x) is also monic of degree 6 with exactly these six roots:
Q(x)=(x−1)(x−2)(x−3)(x−4)(x−5)(x−6)Q(x) = (x-1)(x-2)(x-3)(x-4)(x-5)(x-6)
 Therefore P(x)=(x−1)(x−2)(x−3)(x−4)(x−5)(x−6)+xP(x) = (x-1)(x-2)(x-3)(x-4)(x-5)(x-6) + x. Step 3: Evaluate at x=7x = 7
P(7)=(6)(5)(4)(3)(2)(1)+7=720+7=727P(7) = (6)(5)(4)(3)(2)(1) + 7 = 720 + 7 = 727
 Answer: (2)
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