\begin{align} \int\left(\sqrt x + \frac {1}{\sqrt x}\right).dx \end{align}
\begin{align}= \int x^\frac{1}{2}.dx + \int x^\frac{-1}{2}.dx \end{align}
\begin{align}= \left(\frac {x^\frac{3}{2}}{\frac{3}{2}}\right) + \left(\frac {x^\frac{1}{2}}{\frac{1}{2}}\right) + C\end{align}
\begin{align}= \frac {2}{3} . x^{\frac{3}{2}}+ 2x^\frac{1}{2} + C\end{align}
Hence, the Correct Answer is C.
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
Show that the Modulus Function f : R → R, given by f(x) = |x|, is neither oneone nor onto, where | x | is x, if x is positive or 0 and |x| is – x, if x is negative.
Prove that the Greatest Integer Function f : R → R, given by f(x) = [x], is neither one-one nor onto, where [x] denotes the greatest integer less than or equal to x.
The total revenue in Rupees received from the sale of x units of a product is given by
R (x) = 13x2 + 26x + 15
Find the marginal revenue when x = 7.
Consider f : R → R given by f(x) = 4x + 3. Show that f is invertible. Find the inverse of f.
State with reason whether following functions have inverse
(i) f : {1, 2, 3, 4} → {10} with
f = {(1, 10), (2, 10), (3, 10), (4, 10)}
(ii) g : {5, 6, 7, 8} → {1, 2, 3, 4} with
g = {(5, 4), (6, 3), (7, 4), (8, 2)}
(iii) h : {2, 3, 4, 5} → {7, 9, 11, 13} with
h = {(2, 7), (3, 9), (4, 11), (5, 13)}
The vertices of ΔABC are A (3, 5, −4), B (−1, 1, 2), and C (−5, −5, −2).