Let f : [1, \infty \left.\right) \rightarrow_{}^{} \left[\right. 2 , \infty \left.\right) be a differentiable function such that f (1) = 2. If 6 \textrm{ } \int_{1}^{x} f \left(\right. t \left.\right) \textrm{ } dt= 3x f (x) – x3 for all x \geq 1, then the value of f (2) is

Engineering

Mathematics

Linear Differential Equation

Question

Let f : [1, $\infty) \to_{}^{} [2, \infty)$ be a differentiable function such that f (1) = 2. If $6 \int_{1}^{x} f (t) dt$ = 3x f (x) – x³ for all x $\geq$ 1, then the value of f (2) is

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$6 \int_{I}^{x} f (t) dt = 3 xf (x) - x^{3}$

6 f(x) = 3f(x) + 3xf '(x) –3x²

3f(x) = 3x f '(x) – 3x²

^{$3 y = 3 x \frac{dy}{dx} - 3 x^{2}$}

^{$x \frac{dy}{dx} - y = x^{2}$}

^{$\frac{dy}{dx} - \frac{y}{x} = x$}

^{$I . F . = e^{\int \frac{- 1}{x} dx} = e^{- ℓn x} = \frac{1}{x}$}

^{$y \cdot \frac{1}{x} = \int x \cdot \frac{1}{x} dx \Rightarrow \frac{y}{x} = x + c \Rightarrow y = x^{2} + cx$}

$∵$ f(1) = 2 ⇒ c = 1

y = x² + x

f(2) = 4 + 2 = 6

* The most appropriate answer to this question is (6), but because of ambiguity in language, IIT has declared this question as bonus (marks to all students)

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