It is given that,
\begin{align} \frac{d}{dx} f(x) = 4x^3 - \frac{3}{x^4}\end{align}
∴ Anti derivative of
\begin{align} 4x^3 - \frac{3}{x^4} = f(x)\end{align}
∴ \begin{align} f(x)= \int \left(4x^3 - \frac{3}{x^4}\right).dx\end{align}
\begin{align} f(x)= 4\int x^3.dx - 3\int {x^{-4}}.dx\end{align}
\begin{align} f(x)= 4\left(\frac {x^4}{4}\right) - 3\left(\frac {x^{-3}}{-3}\right) + C\end{align}
∴ \begin{align} f(x)= x^4 + \frac{1}{x^3} + C\end{align}
Also, f(2) = 0
∴ \begin{align} f(2) =\left(2\right)^4 + \frac{1}{\left(2\right)^3} + C = 0 \end{align}
=> \begin{align} 16 + \frac{1}{8} + C = 0 \end{align}
=> \begin{align} C = -\left(16 + \frac{1}{8}\right) \end{align}
=> \begin{align} C = \frac{-129}{8} \end{align}
∴ \begin{align} f(x)= x^4 + \frac{1}{x^3} -\frac{129}{8} \end{align}
Hence, the correct answer is A.
Prove that the function f(x) = 5x – 3 is continuous at x = 0, at x = – 3 and at x = 5.
Determine order and degree(if defined) of differential equation \begin{align} \frac{d^4y}{dx^4}\;+\;\sin(y^m)\;=0\end{align}
Represent graphically a displacement of 40 km, 30° east of north.
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Subject to the constraints:x + y ≤ 4, x ≥ 0, y ≥ 0
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Given that E and F are events such that P(E) = 0.6, P(F) = 0.3 and P(E ∩ F) = 0.2, find P (E|F) and P(F|E).
Show that the Signum Function f : R → R, given by
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Let f, g and h be functions from R to R. Show that
(f + g)oh = foh + goh
(f . g)oh = (foh) . (goh)
\begin{align} y = xsinx:xy{'}=y +x\sqrt{x^2 -y^2}(x\neq0\; and\; x>y\; or\; x<-y)\end{align}
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In each of the following cases, state whether the function is one-one, onto or bijective. Justify your answer.
(i) f : R → R defined by f(x) = 3 – 4x
(ii) f : R → R defined by f(x) = 1 + x2