MCQ Bank
\begin{gathered} {\text{The}} ~ {\text{graph}}~ {\text{of}} ~ \hfill \ {\text{r = (1 + t)}} {\text{i + ( - 2}} {\text{ + }} {\text{3t)}} {\text{j - 4t}} {\text{k}} ~ \hfill \ {\text{is}} ~{\text{the}}~ \hfill \\ \end{gathered}
- A) line that passes through the point (1, 3, -4)
- B) line that passes through the point (2, 1, -4)
- C) line that passes through the point (1, -2, 0)
- D) None of these
If r is a vector-valued function in 2-space or 3- space, then {\frac{d}{{dt}}\left[ {\int {r(t)dt} } \right]}
- A) None of above
- B) {r(t)}
- C) {\int {r(t)dt} }
- D) d\int {r(t)}
{\text{A vector valued function in 2 - D can be expressed as}}
- A) \vec r(t) = x(t) - y(t)
- B) \vec r(t) = x(t) + y(t)
- C) \vec r(t) = x(t)j + y(t)i
- D) \vec r(t) = x(t)i + y(t)j
\eqalign{ & {\text{If }}x'(t){\text{ and }}y'(t){\text{ are continuous for }}a \leqslant t \leqslant b{\text{, then the}} {\text{arc}} {\text{lenght}} {\text{for}} {\text{the given}} \cr & {\text{parametric}} {\text{equations}} x = x(t),y = y(t){\text{ ;}} \left( {a \leqslant t \leqslant b} \right) {\text{is}} \cr}
- A) L = \int\limits_a^b {\sqrt {{{\left( {dx/dt} \right)}^2} + {{\left( {dy/dt} \right)}^2}} } dt
- B) L = \int\limits_a^b {\sqrt {{{\left( {dx/dt} \right)}^2} - {{\left( {dy/dt} \right)}^2}} } dt
- C) L = \int\limits_a^b {\sqrt {{{\left( {dx/dt} \right)}^2} + {{\left( {dy/dt} \right)}^2}} } dx
- D) L = \int\limits_a^b {\sqrt {{{\left( {dx/dt} \right)}^2} + {{\left( {dy/dt} \right)}^2}} } dy
{\text{Green's Theorem states that}} \iint\limits_R {\left( {\frac{{\partial P}}{{\partial x}} - \frac{{\partial Q}}{{\partial y}}} \right)}dxdy
- A) \oint {(Pdx + Qdy)}
- B) - \oint {(Pdx - Qdy)}
- C) - \oint {(Pdx + Qdy)}
- D) \oint {(Pdx - Qdy)}
{\text{The}}\,\,{\text{grad}}\,\,{\text{operator}}\,\,\nabla \,\,{\text{acts}}\,\,{\text{on}}\,\,{\text{a(an)}}\,\,{\text{_________}}\,\,{\text{and}}\,\,{\text{gives}}\,\,{\text{a}}\,\,{\text{vector}}{\text{.}}
- A) {\text{vector}}
- B) {\text{unit}}\,\,{\text{vector}}
- C) {\text{scalar}}
- D) {\text{constant}}
Which integral gives the arc length of the curve r(t) = \frac{1}{3}{t^3}i + tj + {t^2}k over the interval [1,3]
- A) \int\limits_1^3 {\sqrt {1 + t} } dt
- B) \int\limits_1^3 {\sqrt {{{(1 + t)}^2}} } dt
- C) \int\limits_1^3 {\sqrt {{{(1 + {t^2})}^2}} } dt
- D) \int\limits_1^3 {\sqrt {{{(\frac{1}{3} + t)}^2}} } dt
\begin{gathered} {\text{In}}\,\,{\text{3 - space,}}\,\,\vec r(t) = x(t)\hat i + y(t)\hat j + z(t)\hat k,\,\,{\text{is}}\,\,{\text{smooth}}\,\,{\text{function}}\,\,{\text{of}}\,\,t\,\,{\text{if}}\,\,x'(t),\,\,\,y'(t)\,\,{\text{and}}\,\,z'(t)\,\,{\text{are}}\,\, \hfill \\ {\text{_________}}\,\,{\text{and}}\,\,{\text{there}}\,\,{\text{is}}\,\,{\text{no}}\,\,{\text{value}}\,\,{\text{of}}\,\,{\text{t}}\,\,{\text{at}}\,\,{\text{which}}\,\,{\text{all}}\,\,{\text{three}}\,\,{\text{derivatives}}\,\,{\text{are}}\,\,{\text{zero}}{\text{.}} \hfill \\\ \end{gathered}
- A) {\text{continuous}}
- B) {\text{discontinuous}}
- C)
- D)
The integration taken round a closed curve is ------- provided (Pdx+Qdy) is a(n) ------------ differential.
- A) zero, homogeneous
- B) one, exact
- C) one, homogeneous
- D) zero, exact
\eqalign{ & {\text{If }}g{\text{ is a real valued function, then substituting }}t = g\left( u \right){\text{ for this change in parameter from }}r\left( t \right){\text{ to }}g\left( u \right) \cr & g{\text{ will satisfies the following conditions}} \cr & g{\text{ is differentiable}}{\text{.}} \cr & g{\text{ is continuous}}{\text{.}} \cr & g'(u) \ne 0 {\text{for any }}u{\text{ in the domain in }}g{\text{.}} \cr & {\text{the range of }}g{\text{ is the domain of }}r{\text{.}} \cr}
- A) False
- B) True
- C)
- D)
\[{\text{If}}\,\,Pdx + Qdy + Rdw\,\,{\text{is}}\,\,{\text{an}}\,\,{\text{exact}}\,\,{\text{differential}}\,\,{\text{equation}}\,\,{\text{then}}\,\,\_\_\_\_\_\_\_.\]
- A) \[{\text{(d)}}\,\,\,\,\,{\text{All}}\,\,{\text{(a),}}\,\,{\text{(b)}}\,\,{\text{and}}\,\,{\text{(c)}}{\text{.}}\]
- B) \[{\text{(a)}}\,\,\,\,\,\frac{{\partial P}}{{\partial y}} = \frac{{\partial Q}}{{\partial x}}\]
- C) \[{\text{(b)}}\,\,\,\,\,\frac{{\partial P}}{{\partial w}} = \frac{{\partial R}}{{\partial x}}\]
- D) \[{\text{(c)}}\,\,\,\,\,\frac{{\partial R}}{{\partial y}} = \frac{{\partial Q}}{{\partial w}}\]
$$\eqalign{ & {\text{The arc length of the portation of the circular helix where }}(dx/dt) = - \sin t, (dy/dt) = \cos t \cr & {\text{and}} (dz/dt) = 1 {\text{and }}0 \leqslant t \leqslant \pi {\text{, then the arc lenght is}} \cr} $$
- A) $$L = \int\limits_0^\pi {\sqrt 2 } dy$$
- B) $$L = \int\limits_0^\pi { - \sqrt 2 } dt$$
- C) $$L = \int\limits_0^\pi {\sqrt 2 } dt$$
- D) $$L = \int\limits_0^\pi {\sqrt 2 } dx$$
{\text{ }}Equation{\text{ }}of{\text{ }}the{\text{ }}tangent{\text{ }}line{\text{ }}of{\text{ }}vector{\text{ }}valued{\text{ }}function~ r(t) ~ at~ r({t_0}) ~is
- A) r = r({t_0}) + r'({t_0})
- B) r = r({t_0}) + t
- C) r = r({t_0}) + tr'({t_0})
- D) r = r({t_0})
{\text{The}}\,\,{\text{div}}\,\,{\text{operator}}\,\,\nabla \,\,{\text{acts}}\,\,{\text{on}}\,\,{\text{a(an)}}\,\,{\text{_________}}\,\,{\text{and}}\,\,{\text{gives}}\,\,{\text{a}}\,\,{\text{scalar}}{\text{.}}
- A) {\text{scalar}}
- B) {\text{vector}}
- C) {\text{constant}}
- D) {\text{unit}}\,\,{\text{vector}}
\[{\text{The}}\,\,{\text{div}}\,\,{\text{operator}}\,\,\nabla \,\,{\text{acts}}\,\,{\text{on}}\,\,{\text{a(an)}}\,\,{\text{_________}}\,\,{\text{and}}\,\,{\text{gives}}\,\,{\text{a}}\,\,{\text{scalar}}{\text{.}}\]
- A) \[{\text{unit}}\,\,{\text{vector}}\]
- B) \[{\text{scalar}}\]
- C) \[{\text{constant}}\]
- D) \[{\text{vector}}\]
$$\eqalign{ & {\text{If }}x'(t){\text{ and }}y'(t){\text{ are continuous for }}a \leqslant t \leqslant b{\text{, then the}} {\text{arc}} {\text{lenght}} {\text{for}} {\text{the given}} \cr & {\text{parametric}} {\text{equations}} x = x(t),y = y(t){\text{ ;}} \left( {a \leqslant t \leqslant b} \right) {\text{is}} \cr} $$
- A) $$L = \int\limits_a^b {\sqrt {{{\left( {dx/dt} \right)}^2} + {{\left( {dy/dt} \right)}^2}} } dy$$
- B) $$L = \int\limits_a^b {\sqrt {{{\left( {dx/dt} \right)}^2} + {{\left( {dy/dt} \right)}^2}} } dx$$
- C) $$L = \int\limits_a^b {\sqrt {{{\left( {dx/dt} \right)}^2} + {{\left( {dy/dt} \right)}^2}} } dt$$
- D) $$L = \int\limits_a^b {\sqrt {{{\left( {dx/dt} \right)}^2} - {{\left( {dy/dt} \right)}^2}} } dt$$
$$\eqalign{ & {\text{In 3D - space the parametric equations }}x = {\text{ }}x(t),y{\text{ }} = {\text{ }}y(t),{\text{ }}z = z(t){\text{ can be expressed in single vector }} \cr & {\text{equation as}} \cr} $$
- A) $$\vec r(t) = x(t) - y(t) - z(t)$$
- B) $$\vec r(t) = x(t)j + y(t)i + z(t)k$$
- C) $$\vec r(t) = x(t) + y(t) + z(t)$$
- D) $$\vec r(t) = x(t)i + y(t)j + z(t)k$$
\[\begin{array}{l} If ~x'(t) ~and~ y'(t)~ are ~continuous, ~then~ the ~curve~ given~ by ~the ~parametric ~equation\x = x(t) , y = y(t) ~has ~arc~ length \end{array}\]
- A) $$L = \int\limits_a^b {\sqrt {\frac{{dx}}{{dt}} + \frac{{dy}}{{dt}} } } dt$$
- B) $$L = \int\limits_a^b {\sqrt {{{\left( {\frac{{dx}}{{dt}}} \right)}^2} + {{\left( {\frac{{dy}}{{dt}}} \right)}^2} } } dt$$
- C) None of these
- D) \[L = \int\limits_a^b {\sqrt {{{\left( x \right)}^2} + {{\left( y \right)}^2} } } dxdy\]
\[\begin{gathered} {\text{Consider}}\,\,{\text{the}}\,\,{\text{two}}\,\,{\text{functions,}}\,\,P(x,y)\,\,{\text{and}}\,\,Q(x,y)\,\,{\text{have}}\,\,{\text{continuous}}\,\,{\text{partial}}\,\,{\text{derivatives}}\,\,{\text{in}}\,\,{\text{a}}\,\,{\text{certain}}\,\, \hfill \\ {\text{domain}}\,\,D\,\,{\text{(say)}}{\text{.}}\,\,{\text{The}}\,\,{\text{differential}}\,\,{\text{equation,}}\,\,P(x,y)dx + Q(x,y)dy = 0,\,\,{\text{is}}\,\,{\text{an}}\,\,{\text{exact}}\,\,{\text{differential}}\,\,{\text{equation}} \hfill \\ {\text{if}}\,\,{\text{and}}\,\,{\text{only}}\,\,{\text{if}}\,\,\_\_\_\_\_\_\_\_. \hfill \\\ \end{gathered} \]
- A) \[\frac{{\partial P}}{{\partial y}} \ne \frac{{\partial Q}}{{\partial x}}\]
- B) \[\frac{{\partial Q}}{{\partial x}}\]
- C) \[\frac{{\partial P}}{{\partial y}}\]
- D) \[\frac{{\partial P}}{{\partial y}} = \frac{{\partial Q}}{{\partial x}}\]
\eqalign{ & {\text{Wallis sine formula when n is even}} \cr & \int\limits_0^{\frac{\pi }{2}} {Co{s^4}x} dx = \cr}
- A) \frac{4}{5} \cdot \frac{2}{3}
- B) \frac{4}{3} \cdot \frac{2}{1} \cdot \frac{\pi }{2}
- C) \frac{3}{4} \cdot \frac{1}{2}
- D) \frac{3}{4} \cdot \frac{1}{2} \cdot \frac{\pi }{2}