Calculus - Study Mode
[#86] For a position vector [{
m{r}} = {
m{xhat i}} + {
m{yhat j}} + {
m{zhat k}}] xa0 xa0the norm of the vector can be defined as $$left| {overrightarrow { ext{r}} }
ight| = sqrt {{{ ext{x}}^2} + {{ ext{y}}^2} + {{ ext{z}}^2}} .$$ xa0 xa0 Given a function $$phi = ln left| {overrightarrow { ext{r}} }
ight|,$$ xa0 its gradient $$
abla phi $$ xa0is
Correct Answer
(C) $$frac{{overrightarrow { ext{r}} }}{{overrightarrow { ext{r}} cdot overrightarrow { ext{r}} }}$$
[#87] According to the Mean Value Theorem, for a continuous function f(x) in the interval [a, b], there exists a value $$xi $$ in this interval such that $$intlimits_{ ext{a}}^{ ext{b}} {{ ext{f}}left( { ext{x}}
ight){ ext{dx}} = } $$
Correct Answer
(A) $${ ext{f}}left( xi
ight)left( {{ ext{b}} - { ext{a}}}
ight)$$
[#88] Let $$
abla cdot left( {{ ext{f}}overrightarrow { ext{v}} }
ight) = {{ ext{x}}^2}{ ext{y}} + {{ ext{y}}^2}{ ext{z}} + {{ ext{z}}^2}{ ext{x}},$$ xa0 xa0 xa0where f and v are scalar and vector fields respectively. If $$overrightarrow { ext{v}} = { ext{y}}overrightarrow { ext{i}} + { ext{z}}overrightarrow { ext{j}} + { ext{x}}overrightarrow { ext{k}} ,$$ xa0 xa0 then $$overrightarrow { ext{v}} cdot
abla { ext{f}}$$ xa0 is
Correct Answer
(A) x 2 y + y 2 z + z 2 x
[#89] The polynomial p(x) = x 5 + x + 2 has
Correct Answer
(C) 1 real and 4 complex roots
[#90] A function y = 5x 2 + 10x is defined over an open interval x = (1, 2). At least at one point in this interval, $$frac{{{ ext{dy}}}}{{{ ext{dx}}}}$$xa0is exactly
Correct Answer
(B) 25