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JEE Main 202621 JanuaryEvening Shift
Exam
JEE Main
Year
2026
Date
21 January
Shift
Evening Shift
Subject
Mathematics
Chapter
Differentiation

JEE Main 2026 · 21 January · Evening Shift · Mathematics · Differentiation

Let f(x)=x^ 3 +x^ 2 f^ (1)+2 x f^ (2)+f^ (3), x ∈ R . Then the...

Let f(x)=x3+x2f′(1)+2xf′′(2)+f′′′(3),x∈Rf(x)=x^{3}+x^{2} f^{\prime}(1)+2 x f^{\prime \prime}(2)+f^{\prime \prime \prime}(3), x \in \mathbf{R}. Then the value of f′(5)f^{\prime}(5) is:

Options

  1. A1175\frac{117}{5}
  2. B6575\frac{657}{5}
  3. C25\frac{2}{5}
  4. D625\frac{62}{5}

Correct answer

A. 1175\frac{117}{5}

Step-by-step solution

Let a=f′(1)a = f'(1), b=f′′(2)b = f''(2), c=f′′′(3)c = f'''(3). Then f(x)=x3+ax2+2bx+cf(x) = x^3 + ax^2 + 2bx + c.
Taking derivatives: f′(x)=3x2+2ax+2bf'(x) = 3x^2 + 2ax + 2b, f′′(x)=6x+2af''(x) = 6x + 2a, f′′′(x)=6f'''(x) = 6.
From f′(1)=af'(1) = a: 3+2a+2b=a3 + 2a + 2b = a gives a=−3−2ba = -3 - 2b.
From f′′(2)=bf''(2) = b: 12+2a=b12 + 2a = b.
From f′′′(3)=cf'''(3) = c: c=6c = 6.
Substituting the first into the second: b=12+2(−3−2b)=6−4bb = 12 + 2(-3 - 2b) = 6 - 4b gives 5b=65b = 6, so b=65b = \frac{6}{5}.
Then a=−3−125=−275a = -3 - \frac{12}{5} = -\frac{27}{5}.
f(x)=x3−275x2+125x+6f(x) = x^3 - \frac{27}{5}x^2 + \frac{12}{5}x + 6
f′(x)=3x2−545x+125f'(x) = 3x^2 - \frac{54}{5}x + \frac{12}{5}
f′(5)=75−54+125=1175f'(5) = 75 - 54 + \frac{12}{5} = \frac{117}{5}
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