Engineering Mathematics - Calculus
Exam Duration: 45 Mins Total Questions : 30
The value of the quantity P, where P=\(\int _{ 0 }^{ 1 }{ { xe }^{ x }dx } \quad is\)eqial to
- (a)
0
- (b)
1
- (c)
e
- (d)
\(\frac { 1 }{ e } \)
If S =\(\int _{ 1 }^{ \infty }{ { x }^{ -3 } } dx\quad then,\quad S\quad has\quad the\quad value\)
- (a)
\(-\frac { 1 }{ 3 } \)
- (b)
\(\frac { 1 }{ 4 } \)
- (c)
\(\frac { 1 }{2 } \)
- (d)
1
At t=0, the function f(t)=\(\frac { sin\quad t }{ t } has\)
- (a)
a minimum
- (b)
a discontinuity
- (c)
a point of inflection
- (d)
a maximum
\(The\quad \underset { x\rightarrow 0 }{ lim } \frac { sin\frac { 2 }{ 3 } x }{ x } is\)
- (a)
\(\frac { 2 }{ 3 } \)
- (b)
1
- (c)
\(\frac { 1 }{ 4 } \)
- (d)
\(\frac { 1 }{2 } \)
What is the value of the definite integral \(\int _{ 0 }^{ a }{ \frac { \sqrt { x } }{ \sqrt { x } +\sqrt { a+x } } } dx?\)
- (a)
0
- (b)
\(\frac { a }{ 2 } \)
- (c)
a
- (d)
2a
Evaluate \(I=\int _{ -\infty }^{ \infty }{ \frac { dx }{ 1+{ x }^{ 2 } } } \)
- (a)
\(\pi \)
- (b)
\(\frac { \pi }{ 2 } \)
- (c)
\(2\pi \)
- (d)
\(\frac { \pi }{ 4 } \)
\(The\quad value\quad of\quad \underset { x\rightarrow 8 }{ lim } \frac { { x }^{ 1/3 }-2 }{ x-8 } is\)
- (a)
\(\frac { 1 }{ 16 } \)
- (b)
\(\frac { 1 }{ 12 } \)
- (c)
\(\frac { 1 }{ 8 } \)
- (d)
\(\frac { 1 }{ 4 } \)
\(\int _{ -a }^{ a }{ ({ sin }^{ 6 }x+{ sin }^{ 7 }x)dx\quad is\quad equal\quad to\quad } \)
- (a)
\(2\int _{ 0 }^{ a }{ { sin }^{ 6 } } x\quad dx\)
- (b)
\(2\int _{ 0 }^{ a }{ { sin }^{ 7 } } x\quad dx\)
- (c)
\(2\int _{ 0 }^{ a }{ { sin }^{ 6 } } x+{ sin }^{ 7 }x)\quad dx\)
- (d)
Zero
\(\underset { x\rightarrow 0 }{ lim } \frac { { sin }^{ 2 }x }{ x } is\quad equal\quad to\quad \)
- (a)
0
- (b)
\(\infty \)
- (c)
1
- (d)
-1
\(\underset { x\rightarrow 1 }{ lim } \frac { { x }^{ 2 }-1 }{ x-1 } is\)
- (a)
\(\infty \)
- (b)
0
- (c)
2
- (d)
1
Which of the following integrals is unbounded?
- (a)
\(\int _{ p }^{ \pi /4 }{ tanx\quad dx } \)
- (b)
\(\int _{ 0 }^{ \infty }{ \frac { 1 }{ { x }^{ 2 }+1 } } dx\)
- (c)
\(\int _{ 0 }^{ \infty }{ { xe }^{ -x } } dx\)
- (d)
\(\int _{ 0 }^{ 1 }{ \frac { 1 }{ 1-x } dx } \)
\(\underset { x\rightarrow 0 }{ lim } \frac { { e }^{ x }-\left( 1+x+\frac { { x }^{ 2 } }{ 2 } \right) }{ { x }^{ 3 } } \) is equal to
- (a)
0
- (b)
\(\frac { 1 }{ 6 } \)
- (c)
\(\frac { 1 }{3 } \)
- (d)
1
Assuming i=\(\sqrt { -1 } \)and t is a real number \(\int _{ 0 }^{ \pi }{ { e }^{ it } } dt\) is
- (a)
\(\frac { \sqrt { 3 } }{ 2 } +\frac { i }{ 2 } \)
- (b)
\(\frac { \sqrt { 3 } }{ 2 } -\frac { i }{ 2 } \)
- (c)
\(\frac { 1 }{ 2 } +i\frac { \sqrt { 3 } }{ 2 } +\frac { 1 }{ 2 } +i\left( 1-\frac { \sqrt { 3 } }{ 2 } \right) \)
- (d)
0
If f(x) \(\frac { 2{ x }^{ 2 }-7x+3 }{ 5{ x }-12x-9 } ,then\quad \underset { x\rightarrow 3 }{ lim } f(x)will\quad be\)
- (a)
\(-\frac { 1 }{ 3 } \)
- (b)
\(\frac { 5 }{ 18 } \)
- (c)
0
- (d)
\(\frac { 2 }{ 5 } \)
The volume of an object expressed in spherical coordinates is given by V=\(\int _{ 0 }^{ 2\pi }{ \int _{ 0 }^{ \pi /3 }{ \int _{ 0 }^{ 1 }{ { r }^{ 2 }sin\phi dr\phi d\phi d\theta } } } \)
- (a)
\(\frac { \pi }{ 3 } \)
- (b)
\(\frac { \pi }{ 6 } \)
- (c)
\(\frac { 2\pi }{ 3 } \)
- (d)
\(\frac { \pi }{4 } \)
\(\int _{ 0 }^{ \pi /2 }{ \int _{ 0 }^{ \pi /2 }{ sin\quad (x+y)dx\quad dy } } is\)
- (a)
0
- (b)
\(\pi \)
- (c)
\(\frac { \pi }{ 2 } \)
- (d)
2
The value of the Integral I=\(\int _{ 0 }^{ \pi /2 }{ { x }^{ 2 } } sin\quad xdx\quad is\)
- (a)
\(\frac { x+2 }{ 2 } \)
- (b)
\(\frac { 2 }{ \pi -2 } \)
- (c)
\(\pi -2\)
- (d)
\(\pi +2\)
The value of \(\int _{ 0 }^{ 1 }{ \left| 5x-3 \right| } dx\quad is\)
- (a)
\(-\frac { 1 }{ 2 } \)
- (b)
\(\frac { 13 }{ 10 } \)
- (c)
\(\frac { 1 }{ 2 } \)
- (d)
\(\frac { 23 }{ 10 } \)
The value of \(\underset { x\rightarrow 0 }{ lim } \frac { cot\quad \theta }{ \left( \frac { \pi }{ 2 } -0 \right) } is\)
- (a)
1
- (b)
2
- (c)
0
- (d)
\(\frac { 1 }{ 2 } \)
The integral\(\frac { 1 }{ 2\pi } \int _{ 0 }^{ 2\pi }{ sin(t-\tau )cos\tau } d\tau \quad equals\)
- (a)
sin t cos t
- (b)
0
- (c)
\(\frac { 1 }{ 2 } \)cos t
- (d)
\(\frac { 1 }{ 2 } \)sin t
Consider function f(x)=(x2-4)2 where X is real number. Then, the function has.
- (a)
Only one minima
- (b)
only two minima
- (c)
three minima
- (d)
three maxima
A cubic polynomial with real coefficients
- (a)
cab possibly have no extrema and no zero crossing.
- (b)
may have up to three extreme and up to two zero crossings
- (c)
cannot have more than two extreme and more than three zero crossings
- (d)
will always have an equal number of extreme and zero crossing.
For the function f(x) = x2e-x, the maximum occurs when x is equal to
- (a)
2
- (b)
1
- (c)
0
- (d)
-1
\(\int { \frac { dx }{ 1+3{ sin }^{ 2 }x } } \)is equal to
- (a)
\(2{ tan }^{ -1 }\left( \frac { 1 }{ 2 } tanx \right) \)
- (b)
\(\frac { 1 }{ 2 } { tan }^{ -1 }(2tanx)\)
- (c)
\(2{ tan }^{ -1 }(tanx)\)
- (d)
\(\frac { 1 }{ 2 } { tan }^{ -1 }(tanx)\)
\(\int { \frac { sinx+cosx }{ \sqrt { 1+sin2x } } } \)dx is equal to
- (a)
tanx
- (b)
cosx
- (c)
sinx
- (d)
x
\(\int { \frac { dx }{ sinx+cosx } } \)is equal to
- (a)
\(\frac { 1 }{ \sqrt { 2 } } log\quad tan\left( \frac { x }{ 4 } +\frac { \pi }{ 4 } \right) \)
- (b)
\(\frac { 1 }{ \sqrt { 2 } } log\quad tan\left( \frac { x }{ 2 } +\frac { \pi }{ 8 } \right) \)
- (c)
\(\frac { 1 }{ \sqrt { 2 } } log\quad tan\left( \frac { x }{ 2 } +\frac { \pi }{ 6 } \right) \)
- (d)
\(\frac { 1 }{ \sqrt { 2 } } log\quad tan\left( x+\frac { \pi }{ 4 } \right) \)
The area in first equation under curve \(y=\frac { 1 }{ { x }^{ 2 }+6x+10 } \)is
- (a)
\(\frac { \pi }{ 2 } \)
- (b)
\(\frac { \pi }{ 4 } -2{ tan }^{ -1 }3\)
- (c)
\(\frac { \pi }{ 2 } -{ tan }^{ -1 }3\)
- (d)
\(\frac { \pi }{ 2 } +{ tan }^{ -1 }3\)
\(\underset { x->\infty }{ lim } \left( \sqrt { { x }^{ 2 }+1 } -\sqrt { x+1 } \right) \)equals
- (a)
0
- (b)
\(\infty \)
- (c)
1
- (d)
e
The value of \(l=\int _{ 0 }^{ a }{ \int _{ \frac { { x }^{ 2 } }{ a } }^{ 2a-z }{ xy\quad dx\quad dy } } \)is
- (a)
\(\frac { 3 }{ 8 } { a }^{ 4 }\)
- (b)
\(\frac { 5 }{ 8 } { a }^{ 4 }\)
- (c)
\(\frac { 7 }{ 8 } { a }^{ 4 }\)
- (d)
\({ a }^{ 4 }\)
Let \(f(x)=x(x+3){ e }^{ -\frac { x }{ 2 } },-3\le x\le 0.\quad Let\quad c\quad \leftrightarrow ]-3,0[\) such that f' (c)=0. t5hen, the value of c is
- (a)
\(-\frac { 1 }{ 2 } \)
- (b)
-2
- (c)
-3
- (d)
3
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