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Differentiation question

2023 · 24 Jan · Shift 2 · Q35
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  5. /2023 · 24 Jan · Shift 2 · Q35

Differentiation question

2023 · 24 Jan · Shift 2 · Q35

JEE MainMathematicsDifferentiationMCQ+4 / −1
If f(x)=x3−x2f′(1)+xf′′(2)−f′′′(3),x∈Rf(x) = {x^3} - {x^2}f'(1) + xf''(2) - f'''(3),x \in \mathbb{R}f(x)=x3−x2f′(1)+xf′′(2)−f′′′(3),x∈R, then
  1. A
    2f(0)−f(1)+f(3)=f(2)2f(0) - f(1) + f(3) = f(2)2f(0)−f(1)+f(3)=f(2)
  2. B
    f(1)+f(2)+f(3)=f(0)f(1) + f(2) + f(3) = f(0)f(1)+f(2)+f(3)=f(0)
  3. C
    f(3)−f(2)=f(1)f(3) - f(2) = f(1)f(3)−f(2)=f(1)
  4. D
    3f(1)+f(2)=f(3)3f(1) + f(2) = f(3)3f(1)+f(2)=f(3)
View written solutionFree

Correct answer: A

  1. We are given f(x)=x3−x2f′(1)+xf′′(2)−f′′′(3).f(x)=x^3-x^2f'(1)+x f''(2)-f'''(3).f(x)=x3−x2f′(1)+xf′′(2)−f′′′(3).

Let a=f′(1),b=f′′(2),c=f′′′(3).a=f'(1),\quad b=f''(2),\quad c=f'''(3).a=f′(1),b=f′′(2),c=f′′′(3). Then f(x)=x3−ax2+bx−c.f(x)=x^3-ax^2+bx-c.f(x)=x3−ax2+bx−c. So fff is a cubic polynomial.

  1. Differentiate: f′(x)=3x2−2ax+b,f'(x)=3x^2-2ax+b,f′(x)=3x2−2ax+b, f′′(x)=6x−2a,f''(x)=6x-2a,f′′(x)=6x−2a, f′′′(x)=6.f'''(x)=6.f′′′(x)=6.

Now use the self-referential conditions.

  • Since a=f′(1)a=f'(1)a=f′(1), a=3−2a+b  ⟹  3a=3+b  ⟹  b=3a−3.a=3-2a+b \implies 3a=3+b \implies b=3a-3.a=3−2a+b⟹3a=3+b⟹b=3a−3.

  • Since b=f′′(2)b=f''(2)b=f′′(2), b=12−2a.b=12-2a.b=12−2a.

Equating the two expressions for bbb: 3a−3=12−2a3a-3=12-2a3a−3=12−2a 5a=155a=155a=15 a=3.a=3.a=3. Hence b=12−2(3)=6.b=12-2(3)=6.b=12−2(3)=6.

  • Since c=f′′′(3)c=f'''(3)c=f′′′(3) and f′′′(x)=6f'''(x)=6f′′′(x)=6, c=6.c=6.c=6.

Therefore, f(x)=x3−3x2+6x−6.f(x)=x^3-3x^2+6x-6.f(x)=x3−3x2+6x−6.

  1. Compute the required values: f(0)=−6,f(0)=-6,f(0)=−6, f(1)=1−3+6−6=−2,f(1)=1-3+6-6=-2,f(1)=1−3+6−6=−2, f(2)=8−12+12−6=2,f(2)=8-12+12-6=2,f(2)=8−12+12−6=2, f(3)=27−27+18−6=12.f(3)=27-27+18-6=12.f(3)=27−27+18−6=12.

  2. Check each option.

A: 2f(0)−f(1)+f(3)=2(−6)−(−2)+12=−12+2+12=2=f(2).2f(0)-f(1)+f(3)=2(-6)-(-2)+12=-12+2+12=2=f(2).2f(0)−f(1)+f(3)=2(−6)−(−2)+12=−12+2+12=2=f(2). So A is true.

B: f(1)+f(2)+f(3)=−2+2+12=12≠−6=f(0).f(1)+f(2)+f(3)=-2+2+12=12\neq -6=f(0).f(1)+f(2)+f(3)=−2+2+12=12=−6=f(0). So B is false.

C: f(3)−f(2)=12−2=10≠−2=f(1).f(3)-f(2)=12-2=10\neq -2=f(1).f(3)−f(2)=12−2=10=−2=f(1). So C is false.

D: 3f(1)+f(2)=3(−2)+2=−4≠12=f(3).3f(1)+f(2)=3(-2)+2=-4\neq 12=f(3).3f(1)+f(2)=3(−2)+2=−4=12=f(3). So D is false.

  1. Hence the single correct option is A.\boxed{A}.A​.
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