MCQ Bank
$${\text{The integral }}\int {{{\sec }^2}(2{x^2})\,.4x\,dx\,} {\text{will be equal to ?}}$$
- A) $$\sec (2{x^2}).\tan (2{x^2}) + c$$
- B) $$\tan (2x) + c$$
- C) $${\text{se}}{{\text{c}}^2}(2x) + c$$
- D) $$\tan (2{x^2}) + c$$
Sum the first three terms of the series, whose general term is 5ki
Where first term=k1=10,
Second term= k2=14
Third term=k3= -2
The correct choice is …………
Note: 1, 2, 3 and i with k are in subscript
- A) 101
- B) 011
- C) 110
- D) 111
The polynomial function f(x)=6x^2-30x+36 has the critical point over the real line is ………
- A) 5/2
- B) 5
- C) 2/5
- D) 2
What is the estimated area under f(x) = 10-x2 from x = 0 to x = 3 with left end points for n = 3?
- A) 25
- B) 15
- C) 30
- D) 21
\[{\text{The integral }}\int {\sqrt {4x - 3} \,dx\,} {\text{will be equal to ?}}\]
- A) \[\frac{{{{(4x + 3)}^{\frac{3}{2}}}}}{6} + c\]
- B) \[{\text{None of these}}\]
- C) \[\frac{{{{(4x - 3)}^{\frac{3}{2}}}}}{3} + c\]
- D) \[\frac{{{{(4x - 3)}^{\frac{3}{2}}}}}{6} + c\]
$$\begin{align} & \text{A function }f\text{ is said to have a relative minimum at }{{x}_{0}}\text{, if}\,\text{ }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ }\,\text{for all }x\text{ } \\ & \text{in some open interval containing }{{x}_{0}}. \\ \end{align}$$
- A) $$f\left( {{x}_{0}} \right)\ge \text{ }f\left( x \right)$$
- B) $$f\left( {{x}_{0}} \right)=f\left( x \right)$$
- C) $$f\left( {{x}_{0}} \right)>\text{ }f\left( x \right)$$
- D) $$f\left( {{x}_{0}} \right)\le \text{ }f\left( x \right)$$
\[\begin{align} & \text{A function }f\text{ is said to have a }\,\text{ }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ }\!\!\_\!\!\text{ }\,\text{at }{{x}_{0}}\text{, if }f({{x}_{0}})\ge f(x)\text{ for all }x\text{ } \\ & \text{in some open interval containing }{{x}_{0}}. \\ \end{align}\]
- A) Relative minimum
- B) Relative maximum
- C)
- D)
If [-8,8] is subdivided into ‘16’ equally spaced subintervals, then the MIDDLE point of 8th
sub-interval will be--------.
- A) 1.5
- B) 2.5
- C) 0.5
- D) -0.5
\[{\text{The integral }}\int {{{\sec }^2}(2{x^2})\,.4x\,dx\,} {\text{will be equal to ?}}\]
- A) \[\tan (2{x^2}) + c\]
- B) \[\tan (2x) + c\]
- C) \[{\text{se}}{{\text{c}}^2}(2x) + c\]
- D) \[\sec (2{x^2}).\tan (2{x^2}) + c\]
\[\text{If }{f}'(x)={{x}^{2}}-1.\text{ Then the critical points of the function }f\text{ are}\]
- A) 1, -1
- B) 1, 2
- C) 0, 2
- D) 0, 1
$$\text{If we say a function }f\text{ have a relative extremum at a point }{{x}_{0}},\text{ then it means that }f\text{ has }\!\!~\!\!\text{ }\_\_\_\_\_\_\_\_\text{ at }{{x}_{0}}.$$
- A) Either a relative maximum or relative minimum
- B) Relative minimum
- C) None of these
- D) Relative maximum
To get better approximation to actual area under a continuous curve over a closed interval, we have to increase ………
- A) Size of the interval
- B) Number of subintervals
- C) Total area
- D) Width of the subintervals
Right end point ,left end point, and midpoint evaluation all converges to same result as number of subintervals tends to +ive infinity.
- A) False
- B) True
- C)
- D)
Which of the following will be left end points if the interval [-2,2] is divided into 4 equal subintervals.
- A) -1,0,1,2
- B) None of these
- C) -2,-1,1,2
- D) -2,-1,0,1
\[{\text{The integral }}\int {\frac{x}{{1 + {x^2}}}\,dx\,} {\text{will be equal to ?}}\]
- A) \[\frac{{\ln \left( {1 + {x^2}} \right)}}{2} + c\]
- B) \[2\ln (1 + {x^2}) + c\]
- C) \[{\text{ln(1 + }}{{\text{x}}^2}) + c\]
- D) \[\frac{{\ln \left( {1 - {x^2}} \right)}}{2} + c\]
\[{\text{The integral }}\int {\cot (2x)\,dx\,} {\text{will be equal to ?}}\]
- A) \[\frac{1}{2}\ln \left| {\sin (2x)} \right| + c\]
- B) \[\frac{1}{2}\ln \left| {\sec (2x)} \right| + c\]
- C) \[{\text{ln}}\left| {\sin (2x)} \right| + c\]
- D) \[\ln \left| {{\text{sec}}(2x)} \right| + c\]
$${\text{The integral }}\int {{{\left( {{x^3} + 1} \right)}^{10}}\,.3{x^2}\,dx\,} {\text{will be equal to ?}}$$
- A) $$- \frac{{{{\left( {{x^3} + 1} \right)}^{11}}}}{{11}} + c$$
- B) $$\frac{{{{\left( {{x^3} - 1} \right)}^{11}}}}{{11}} + c$$
- C) $${\text{None of these}}$$
- D) $$\frac{{{{\left( {{x^3} + 1} \right)}^{11}}}}{{11}} + c$$
\[{\text{The integral }}\int {\sec (x).\tan (x)\,dx\,} {\text{will be equal to ?}}\]
- A) \[\cos ec(x) + c\]
- B) \[\sec (x) + c\]
- C) \[{\text{None of these}}\]
- D) \[ - \ln \left| {\cos (x)} \right| + c\]
\[{\text{The integral }}\int {\cos e{c^2}(3{x^2})\,.6x\,dx\,} {\text{will be equal to ?}}\]
- A) \[ - \cot (3{x^2}) + c\]
- B) \[\sec (3{x^2}) + c\]
- C) \[{\text{cot(3}}{{\text{x}}^2}) + c,\]
- D) \[{\text{None of these}}\]
If integral of ‘f(x)’ from [1,2] = 5 ,and integral of ‘f(x)’ from [3,2] = 4 , than integral of ‘f(x)’ from [1,3] is ……….
- A) 1
- B) 2
- C) 9
- D) 3