MCQ Bank
$${\text{The}}\,\,{\text{equation,}}\,\,\,{r^2} = 4\cos 2\theta ,\,\,{\text{represents}}\,\,{\text{a}}\,\,\_\_\_\_\_\_\_\_\_.$$
- A) $${\text{spiral}}$$
- B) $${\text{lemniscate}}$$
- C) $${\text{rose }}\,\,{\text{curve}}$$
- D) $${\text{cardioids}}$$
If r is a vector-valued function in 2-space or 3- space, then $${\frac{d}{{dt}}\left[ {\int {r(t)dt} } \right]}$$
- A) $${r(t)}$$
- B) $$d\int {r(t)}$$
- C) None of above
- D) $${\int {r(t)dt} }$$
Geometrically, the graph of a continuous vector-valued function is a (an) _________.
- A) unbroken curve
- B) unbroken straight line
- C) broken curve
- D) broken straight line
${\text{While changing Cartesian Integral into Polar integral, we change }}x{\text{ by - - - - - - - - - and }}y{\text{ by - - - - - - - - - - - - }}{\text{.}}$
- A) $Co\sec \,\,\theta ,\,\,\,\,\,\,Sec\,\,\theta$
- B) $Sin\,\,\theta ,\,\,\,\,\,\,Cos\,\,\theta$
- C) $Sec\,\,\theta ,\,\,\,\,\,\,Co\sec \,\,\theta$
- D) $Cos\,\,\theta ,\,\,\,\,\,Sin\,\,\theta$
$$\begin{gathered} {\text{The }}\,\,{\text{orientation}}\,\,{\text{ of }}\,\,{\text{the }}\,\,{\text{rose}}\,\,{\text{ relative}}\,\,{\text{ to}}\,\,{\text{ the}}\,\,{\text{ polar }}\,\,{\text{axis }}\,\,{\text{depends}}\,\,{\text{ on}}\,\,{\text{ the }}\,\,{\text{sign}}\,\,{\text{ of }}\,\,{\text{the }}\,\, \hfill \\ {\text{constant }}\,\,a{\text{ }}\,\,{\text{and }}\,\,{\text{whether}}\,\,{\text{ _________ }}\,\,{\text{appears }}\,\,{\text{in }}\,\,{\text{the}}\,\,{\text{ equation}}{\text{.}} \hfill \\\ \end{gathered}$$
- A) $$(b)\,\,\,\cos \theta$$
- B) $$(c)\,\,\,\tan \theta$$
- C) $$(a)\,\,\,\sin \theta$$
- D) $${\text{(d)}}\,\,\,{\text{Both}}\,\,{\text{(a)}}\,\,{\text{or}}\,\,{\text{(b)}}{\text{.}}$$
$$\eqalign{ & {\text{In 3D - space the parametric equations }}x = {\text{ }}x(t),y{\text{ }} = {\text{ }}y(t),{\text{ }}z = z(t){\text{ can be expressed in single vector }} \cr & {\text{equation as}} \cr}$$
- A) $$\vec r(t) = x(t)i + y(t)j + z(t)k$$
- B) $$\vec r(t) = x(t)j + y(t)i + z(t)k$$
- C) $$\vec r(t) = x(t) - y(t) - z(t)$$
- D) $$\vec r(t) = x(t) + y(t) + z(t)$$
$\begin{gathered} {\text{Let G be the rectangular box defined by the inequalities }}a \leqslant x \leqslant b,\,\,\,c \leqslant y \leqslant d,\,\,\,\,\,e \leqslant z \leqslant f. \hfill \ {\text{If }}f\,\,{\text{is continuous on G, then}}\,\,\int\limits_a^b {\int\limits_c^d {\int\limits_e^f {f(x,\,\,y,\,\,z)} } } \,dz\,\,dy\,\,dx = \,\,\, - - - - - - - - \hfill \\ \end{gathered}$
- A) $\int\limits_e^f {\int\limits_a^b {\int\limits_c^d {f(x,\,\,y,\,\,z)} } } \,dy\,\,dx\,\,dz$
- B) ${\text{All}}\,{\text{three}}\,{\text{options}}\,{\text{are}}\,{\text{true}}{\text{.}}$
- C) $\int\limits_c^d {\int\limits_a^b {\int\limits_e^f {f(x,\,\,y,\,\,z)} } } \,dz\,\,dx\,\,dy\,$
- D) $\int\limits_a^b {\int\limits_e^f {\int\limits_c^d {f(x,\,\,y,\,\,z)} } } \,dy\,\,dz\,\,dx$
\[{\text{The}}\,\,{\text{curve}},\,\,r = \theta \,\,(\theta \geqslant 0),\,\,{\text{is}}\,\,{\text{an}}\,\,{\text{equation}}\,\,{\text{of}}\,\,\_\_\_\_\_\_\_\_\_.\]
- A) \[{\text{cardioids}}\]
- B) \[{\text{spiral}}\,\,{\text{with}}\,\,a = 1\]
- C) \[{\text{roses}}\]
- D) \[{\text{limacons}}\]
$${\text{For a vector valued function }}\vec r(t){\text{ }} = {\text{ }}\sqrt {2t} {\text{ }}i + (t + 1){\text{ }}j{\text{ then lenght of }}\vec r(t){\text{ }}$$
- A) $$\left\| {r(t)} \right\| = \sqrt {4t + {{(t - 1)}^2}}$$
- B) $$\left\| {r(t)} \right\| = \sqrt {4{t^2} + {{(t - 1)}^2}}$$
- C) $$\left\| {r(t)} \right\| = \sqrt {2{t^2} + {{(t - 1)}^2}}$$
- D) $$\left\| {r(t)} \right\| = \sqrt {2t + {{(t - 1)}^2}}$$
$$\begin{gathered} {\text{If}}\,\,\vec r(t) = x(t)\hat i + y(t)\hat j\,\,{\text{is}}\,\,{\text{a}}\,\,{\text{vector - valued}}\,\,{\text{function}}\,\,{\text{in}}\,\,{\text{2 - space,}}\,\,{\text{and}}\,\,{\text{if}}\,\,x(t)\,\,{\text{and}}\,\,y(t)\,\,{\text{are}}\,\,{\text{differentiable,}}\,\, \hfill \\ {\text{then}}\,\,\,\frac{d}{{dt}}\,\left[ {\vec r(t)} \right] = \_\_\_\_\_\_\_\_\_. \hfill \\\ \end{gathered}$$
- A) $$x'(t)\hat i + y'(t)\hat j$$
- B) $$x'(t)\hat i + y(t)\hat j$$
- C) $$x(t)\hat i + y'(t)\hat j$$
- D) $$x'(t)\hat i + y'(t)\hat j + z'(t)\hat k$$
\[{\text{For}}\,\,{\text{a}}\,\,{\text{function}}\,\,\vec r(t) = x(t)\hat i + y(t)\hat j + z(t)\hat k\,\,{\text{in}}\,\,{\text{3 - space}}\,\,{\text{we}}\,\,{\text{define}}\,\,\mathop {\lim }\limits_{t \to \alpha } \,\vec r(t) = \_\_\_\_\_\_\_\_\_.\]
- A) \[\left( {\mathop {\lim }\limits_{t \to \alpha } x(t)} \right)\hat i + \left( {\mathop {\lim }\limits_{t \to \alpha } y(t)} \right)\hat j + z(t)\hat k\]
- B) \[\left( {\mathop {\lim }\limits_{t \to \alpha } x(t)} \right)\hat i + \left( {\mathop {\lim }\limits_{t \to \alpha } y(t)} \right)\hat j + \left( {\mathop {\lim }\limits_{t \to \alpha } z(t)} \right)\hat k\]
- C) \[x(t)\hat i + \left( {\mathop {\lim }\limits_{t \to \alpha } y(t)} \right)\hat j + \left( {\mathop {\lim }\limits_{t \to \alpha } z(t)} \right)\hat k\]
- D) \[\left( {\mathop {\lim }\limits_{t \to \alpha } x(t)} \right)\hat i + \left( {\mathop {\lim }\limits_{t \to \alpha } y(t)} \right)\hat j\]
Graph of a vector-valued function can fail to have a tangent vector at a point because
- A) d) Neither a) nor b)
- B) c) Both a) and b)
- C) a) derivative does not exists at that point
- D) b) derivative is zero at that point
$\begin{gathered} {\text{Let }}\bar r(t)\,\, = \,\,2{t^2}\,\hat i\,\, + \,\,3{t^3}\,\,\hat j,\,\,{\text{then}}\,\,\bar r'(1)\,\, = \,\,4\,\hat i\,\, + \,\,9\,\hat j. \hfill \ \hfill \\ \end{gathered} $
- A) $False$
- B) $True$
- C)
- D)
$${\text{The equation }}r\, = \,a(1 + \cos \,\theta ){\text{ represents - - - - - - - - - }}{\text{.}}$$
- A) $${\text{a straight line}}$$
- B) $${\text{rose curve}}$$
- C) $${\text{lemniscate}}$$
- D) $${\text{cardioid}}$$
\[{\text{If}}\,\,\vec r(t) = 3{t^2}\hat i + 2t\,\hat j,\,\,{\text{then}}\,\,\int {\vec r(t)\,dt = \_\_\_\_\_\_\_\_\_.} \]
- A) \[{t^3}\hat i + {t^2}\,\hat j + {C_1}\hat i + {C_2}\hat j\]
- B) \[{t^3} + {t^2}\, + {C_1}\]
- C) \[{t^3}\hat i + {t^2}\,\hat j + {C_2}\hat j\]
- D) \[6t\,\hat i + 2\,\hat j\]
\[{\text{The differential equation, }}{\kern 1pt} {\kern 1pt} dz{\kern 1pt} = {\kern 1pt} 4xy{\kern 1pt} dx{\kern 1pt} {\kern 1pt} + {\kern 1pt} {\kern 1pt} \left( {2{x^2} + 3{y^2}} \right){\kern 1pt} dy,{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\text{is}}\,\,{\kern 1pt} {\text{an exact differential equation}}{\text{.}}\]
- A) \[{\text{True}}\]
- B) \[{\text{False}}\]
- C)
- D)
$$\begin{gathered} {\text{In polar coordinate system, the equation }}r\, = \,-2a\,\sin \,\theta {\text{ represents a circle passes through the origin,}}\, \ {\text{with center on}}\,{\text{ - - - - - - - }}{\text{.}} \\ \end{gathered}$$
- A) $$x - {\text{axis,}}\,\,{\text{right to the origin}}{\text{.}}$$
- B) $$x - {\text{axis,}}\,\,{\text{left to the origin}}{\text{.}}$$
- C) $$y - {\text{axis,}}\,\,{\text{above the origin}}{\text{.}}$$
- D) $$y - {\text{axis,}}\,\,{\text{below}}\,{\text{the origin}}{\text{.}}$$
${\text{While changing Cartesian Integral into Polar integral, we change }}x{\text{ by - - - - - - - - - and }}y{\text{ by - - - - - - - - - - - - }}{\text{.}}$
- A) $Co\sec \,\,\theta ,\,\,\,\,\,\,Sec\,\,\theta$
- B) $Sin\,\,\theta ,\,\,\,\,\,\,Cos\,\,\theta$
- C) $Cos\,\,\theta ,\,\,\,\,\,Sin\,\,\theta$
- D) $Sec\,\,\theta ,\,\,\,\,\,\,Co\sec \,\,\theta$
$${\text{The}}\,\,{\text{vector}}\,\,\vec r\,\,{\text{is}}\,\,{\text{continuous}}\,\,{\text{at}}\,\,{t_0}\,\,{\text{if}}\,\,\_\_\_\_\_\_\_\_\_.$$
- A) $${\text{(d)}}\,\,\,\,\,\,{\text{All}}\,\,{\text{(a),}}\,\,{\text{(b)}}\,\,{\text{and}}\,\,{\text{(c)}}\,{\text{.}}$$
- B) $${\text{(c)}}\,\,\,\,\,\,\mathop {\lim }\limits_{t \to {t_0}} \vec r(t)\,\, = \vec r({t_0})\,.$$
- C) $${\text{(b)}}\,\,\,\,\,\,\mathop {\lim }\limits_{t \to {t_0}} \vec r(t)\,\,{\text{exist}}.$$
- D) $${\text{(a)}}\,\,\,\,\,\,\vec r({t_0})\,\,{\text{is}}\,\,{\text{defined}}.$$
$${\text{If}}\,\,\vec r(t) = 3{t^2}\hat i + 2t\,\hat j,\,\,{\text{then}}\,\,\int {\vec r(t)\,dt = \_\_\_\_\_\_\_\_\_.}$$
- A) $${t^3} + {t^2}\, + {C_1}$$
- B) $${t^3}\hat i + {t^2}\,\hat j + {C_2}\hat j$$
- C) $$6t\,\hat i + 2\,\hat j$$
- D) $${t^3}\hat i + {t^2}\,\hat j + {C_1}\hat i + {C_2}\hat j$$