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JEE Main
Exam
JEE Main
Subject
Mathematics
Chapter
Differentiation

JEE Main · Mathematics · Differentiation

Let P(x) be a polynomial of degree 3 such that it satisfies the...

Let P(x)P(x) be a polynomial of degree 33 such that it satisfies the identity P(x)+P′(x)+P′′(x)=x3+5x2+13x+11P(x) + P'(x) + P''(x) = x^3 + 5x^2 + 13x + 11 for all real xx. The value of P(2)P(2) is equal to:

Options

  1. A36
  2. B26
  3. C65
  4. D30

Correct answer

B. 26

Step-by-step solution

Let the cubic polynomial be P(x)=ax3+bx2+cx+dP(x) = ax^3 + bx^2 + cx + d.

The derivatives are:
P′(x)=3ax2+2bx+cP'(x) = 3ax^2 + 2bx + c
P′′(x)=6ax+2bP''(x) = 6ax + 2b

Adding these gives:
P(x)+P′(x)+P′′(x)=ax3+(b+3a)x2+(c+2b+6a)x+(d+c+2b)P(x) + P'(x) + P''(x) = ax^3 + (b + 3a)x^2 + (c + 2b + 6a)x + (d + c + 2b)

We are given that this sum equals x3+5x2+13x+11x^3 + 5x^2 + 13x + 11. Equating the corresponding coefficients:

Coefficient of x3x^3: a=1a = 1

Coefficient of x2x^2: b+3a=5⇒b+3(1)=5⇒b=2b + 3a = 5 \Rightarrow b + 3(1) = 5 \Rightarrow b = 2

Coefficient of xx: c+2b+6a=13⇒c+2(2)+6(1)=13⇒c+10=13⇒c=3c + 2b + 6a = 13 \Rightarrow c + 2(2) + 6(1) = 13 \Rightarrow c + 10 = 13 \Rightarrow c = 3

Constant term: d+c+2b=11⇒d+3+2(2)=11⇒d+7=11⇒d=4d + c + 2b = 11 \Rightarrow d + 3 + 2(2) = 11 \Rightarrow d + 7 = 11 \Rightarrow d = 4

Thus, the polynomial is P(x)=x3+2x2+3x+4P(x) = x^3 + 2x^2 + 3x + 4.

We need to find P(2)P(2):
P(2)=(2)3+2(2)2+3(2)+4=8+8+6+4=26P(2) = (2)^3 + 2(2)^2 + 3(2) + 4 = 8 + 8 + 6 + 4 = 26.

Answer: 26
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