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.
If a line makes angles 90°, 135°, 45° with x, y and z-axes respectively, find its direction cosines.
Maximise Z = 3x + 4y
Subject to the constraints:x + y ≤ 4, x ≥ 0, y ≥ 0
Find the area of the region bounded by the curve y2 = x and the lines x = 1, x = 4 and the x-axis.
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).
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
Find the direction cosines of a line which makes equal angles with the coordinate axes.
Maximise Z = 3x + 4y
Subject to the constraints:x + y ≤ 4, x ≥ 0, y ≥ 0
Let f, g and h be functions from R to R. Show that
(f + g)oh = foh + goh
(f . g)oh = (foh) . (goh)
Determine order and degree(if defined) of differential equation \begin{align} \frac{d^4y}{dx^4}\;+\;\sin(y^m)\;=0\end{align}
Show that f : [–1, 1] → R, given by is one-one. Find the inverse of the function f : [–1, 1] → Range f.
(Hint: For y ∈ Range f, y =, for some x in [ - 1, 1], i.e.,
)
\begin{align} y= \sqrt{1+x^2} : y^{'}=\frac{xy}{1+x^2}\end{align}