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)
\(\frac { 1 }{ e } \)
\(P=\int _{ 0 }^{ 1 }{ { xe }^{ x }dx } \)
\(=[x\int { { e }^{ x } } dx{ ] }_{ 0 }^{ 1 }-\int _{ 0 }^{ 1 }{ \left\{ \frac { d }{ dx } (x)\int { { e }^{ x } } dx \right\} } dx\\ =[x.{ e }^{ x }{ ] }_{ 0 }^{ 1 }-\int _{ 0 }^{ 1 }{ { e }^{ x } } dx\\ =(1.e-0)-[{ e }^{ x }{ ] }_{ 0 }^{ 1 }\\ =e-e+{ e }^{ 0 }=1\)
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
\(S=\int _{ 0 }^{ \infty }{ { x }^{ -3 } } dx\\ =\left[ \frac { { x }^{ -2 } }{ -2 } \right] =-\frac { 1 }{ 2 } { \left[ \frac { 1 }{ { x }^{ 2 } } \right] }_{ 1 }^{ \infty }\\ \quad \quad \quad =-\frac { 1 }{ 2 } \left[ \frac { 1 }{ \infty } -\frac { 1 }{ 1 } \right] =-\frac { 1 }{ 2 } (0-1)\\ \quad \quad \quad =\frac { 1 }{ 2 } \)
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)
\(\frac { 1 }{2 } \)
\(\underset { x\rightarrow 0 }{ lim } \frac { sin\left[ \frac { 2 }{ 3 } x \right] }{ x } =\underset { x\rightarrow 0 }{ lim } \frac { sin\left[ \frac { 2 }{ 3 } x \right] }{ \left[ \frac { 2 }{ 3 } x \right] } \times \frac { 2 }{ 3 } \left[ \therefore \quad \underset { x\rightarrow 0 }{ lim } \frac { sinx }{ x } =1 \right] \\ \quad \quad \quad \quad \quad =1.\left( \frac { 2 }{ 3 } \right) =\frac { 2 }{ 3 } \)
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
\(I=\int _{ 0 }^{ a }{ \frac { \sqrt { x } }{ \sqrt { x } +\sqrt { a+x } } } dx\\ By\quad property\int _{ 0 }^{ a }{ f(x } )=\int _{ 0 }^{ a }{ f(a-x)dx } \\ \quad \quad \quad \quad \quad \quad \quad I=\int _{ 0 }^{ a }{ \frac { \sqrt { x } }{ \sqrt { a-x } +\sqrt { x } } } dx\\ Adding\quad both\quad intergrals,\quad we\quad get\quad \\ \quad \quad \quad \quad \quad \quad \quad I=\frac { a }{ 2 } \quad \quad \quad \quad \quad \)
Evaluate \(I=\int _{ -\infty }^{ \infty }{ \frac { dx }{ 1+{ x }^{ 2 } } } \)
- (a)
\(\pi \)
- (b)
\(\frac { \pi }{ 2 } \)
- (c)
\(2\pi \)
- (d)
\(\frac { \pi }{ 4 } \)
\(I=\int _{ -\infty }^{ \infty }{ \frac { dx }{ 1+{ x }^{ 2 } } dx } which\quad is\quad an\quad even\quad function\\ I=2\int _{ 0 }^{ \infty }{ \frac { dx }{ 1+{ x }^{ 2 } } } \\ \left\{ I=\int _{ -a }^{ a }{ f(x)dx=2 } \int _{ 0 }^{ a }{ f(x)dx } \right\} \\ I=2{ [{ tan }^{ -1 }x] }_{ 0 }^{ \infty }\\ =2[{ tan }^{ -1 }\infty { -tan }^{ -1 }0]\\ I=2\left( \frac { \pi }{ 2 } -0 \right) ,\\ I=\pi \)
\(The\quad value\quad of\quad \underset { x\rightarrow 8 }{ lim } \frac { { x }^{ 1/3 }-2 }{ x-8 } is\)
- (a)
\(\frac { 1 }{ 16 } \)
- (b)
\(\frac { 1 }{ 12 } \)
- (c)
\(\frac { 1 }{ 8 } \)
- (d)
\(\frac { 1 }{ 4 } \)
\(Given.\quad \underset { x\rightarrow 8 }{ lim } \frac { { x }^{ 1/3 }-2 }{ x-8 } \quad \quad \quad \left[ \frac { 0 }{ 0 } From \right] \\ applyL'Hospital\quad rule\\ =\underset { x\rightarrow 8 }{ lim } \frac { \frac { 1 }{ 3 } { x }^{ -2/3 }-0 }{ 1-0 } =\underset { x\rightarrow 8 }{ lim } \frac { 1 }{ { 3x }^{ 2/3 } } \\ =\frac { 1 }{ 3\times 4 } =\frac { 1 }{ 12 } \)
\(\int _{ -a }^{ a }{ ({ sin }^{ 6 }x+{ sin }^{ 7 }x)dx\quad is\quad equal\quad to\quad } \)
- (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
\(\int _{ -a }^{ a }{ ({ sin }^{ 6 }x-{ sin }^{ 7 }x)dx } \\ \Rightarrow \int _{ -a }^{ a }{ { sin }^{ 6 }x\quad dx } -\int _{ -a }^{ a }{ { sin }^{ 7 }x\quad dx } \\ \Rightarrow \int _{ -a }^{ a }{ 0\quad dx+2 } \int _{ 0 }^{ a }{ { sin }^{ 6 }x } dx\\ Here,\quad { sin }^{ 6 }x\quad is\quad even\quad function,\quad while\quad { sin }^{ 7 }x\quad is\quad an\quad odd\quad fuction\\ \because \int _{ -a }^{ a }{ \left\{ 0,\quad \quad \quad \quad f(x)\quad is\quad odd\\ 2\int _{ 0 }^{ a }{ f(x)\quad dx,\quad f(x)is\quad even } \right\} } \)
\(\underset { x\rightarrow 0 }{ lim } \frac { { sin }^{ 2 }x }{ x } is\quad equal\quad to\quad \)
- (a)
0
- (b)
\(\infty \)
- (c)
1
- (d)
-1
\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\underset { x\rightarrow 0 }{ lim } \frac { { sin }^{ 2 }x }{ x } =\underset { x\rightarrow 0 }{ lim } \left( \frac { sinx }{ x } \right) .sinx\\ \quad \quad \quad =1.sin0\\ \quad \quad \quad =1\times 0=0\)
\(\underset { x\rightarrow 1 }{ lim } \frac { { x }^{ 2 }-1 }{ x-1 } is\)
- (a)
\(\infty \)
- (b)
0
- (c)
2
- (d)
1
\(\underset { x\rightarrow 1 }{ lim } \frac { { x }^{ 2 }-1 }{ x-1 } \\ \underset { x\rightarrow 1 }{ lim } \frac { (x-1)(x+1) }{ (x-1) } =\underset { x\rightarrow 1 }{ lim } (x+1)=2\)
\(\underset { x\rightarrow 0 }{ lim } \frac { { e }^{ x }-\left( 1+x+\frac { { x }^{ 2 } }{ 2 } \right) }{ { x }^{ 3 } } \) is equal to
- (a)
0
- (b)
\(\frac { 1 }{ 6 } \)
- (c)
\(\frac { 1 }{3 } \)
- (d)
1
\(\underset { x\rightarrow 0 }{ lim } \frac { { e }^{ x }-\left( 1+x+\frac { { x }^{ 2 } }{ 2 } \right) }{ { x }^{ 3 } } \quad \quad \left[ \frac { 0 }{ 0 } by\quad L'Hospital\quad rule \right] \\ =\underset { x\rightarrow 0 }{ lim } \frac { { e }^{ x }-(1+x) }{ 3{ x }^{ 2 } } \quad \quad \quad \quad \quad \quad \left[ \frac { 0 }{ 0 } From \right] \\ =\underset { x\rightarrow 0 }{ lim } \frac { { e }^{ x }-1 }{ 6x } \quad \quad \quad \quad \quad \quad \quad \quad \quad \left[ \frac { 0 }{ 0 } From \right] \\ =\underset { x\rightarrow 0 }{ lim } \frac { { e }^{ x } }{ 6 } =\frac { { e }^{ x } }{ 6 } =\frac { 1 }{ 6 } \)
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
\(\int _{ 0 }^{ \pi }{ { e }^{ it } } dt={ { \left[ \frac { { e }^{ it } }{ i } \right] } }_{ 0 }^{ \pi /3 }\\ \quad \quad \quad \quad \quad =\left[ \frac { cos\frac { \pi }{ 3 } -isin\frac { \pi }{ 3 } }{ i } \right] -\left[ \frac { cos\quad 0-isin0 }{ i } \right] \\ \quad \quad \quad \quad \quad =\frac { -\frac { 1 }{ 2 } i+{ i }^{ 2 }\frac { \sqrt { 3 } }{ 2 } }{ { i }^{ 2 } } =-\left[ -\frac { i }{ 2 } -\frac { \sqrt { 3 } }{ 2 } \right] \\ \quad \quad \quad \quad \quad =\left[ \frac { \sqrt { 3 } }{ 2 } +\frac { 1 }{ 2 } \right] \)
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)
\(\frac { 2 }{ 5 } \)
\(\underset { x\rightarrow 3 }{ lim } =\frac { 2{ x }^{ 2 }-7x+3 }{ 5{ x }^{ 2 }-12x-9 } \quad \quad \quad \left[ \frac { 0 }{ 0 } From \right] \\ \quad \quad \quad =\underset { x\rightarrow 3 }{ lim } \frac { 4x-7 }{ 10x-12 } \frac { 5 }{ 18 } [By\quad L'Hospital\quad rule]\)
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)
\(\frac { \pi }{4 } \)
\(V=\int _{ 0 }^{ 2\pi }{ \int _{ 0 }^{ \pi /3 }{ \int _{ 0 }^{ 1 }{ { r }^{ 2 }sin\phi dr\phi d\phi d\theta } } } \\ \quad =\int _{ 0 }^{ 2\pi }{ \int _{ 0 }^{ \pi /3 }{ { \left[ \frac { { r }^{ 3 } }{ 3 } \right] }_{ 0 }^{ 1 } } } sin\phi d\phi d\theta \\ \quad =\int _{ 0 }^{ 2\pi }{ \int _{ 0 }^{ \pi /3 }{ \frac { 1 }{ 3 } } } sin\quad \phi d\phi \quad d\theta \\ \quad =\frac { 1 }{ 3 } \int _{ 0 }^{ 2\pi }{ { [-cos\phi ] }_{ 0 }^{ \pi /3 } } d\theta \\ \quad =\frac { 1 }{ 3 } \int _{ 0 }^{ 2\pi }{ \left[ \frac { 1 }{ 2 } \right] } d\theta \\ \quad =\frac { 1 }{ 6 } \times 2\pi =\frac { \pi }{ 3 } \)
\(\int _{ 0 }^{ \pi /2 }{ \int _{ 0 }^{ \pi /2 }{ sin\quad (x+y)dx\quad dy } } is\)
- (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)
\(\pi +2\)
Integrating by parts,
\(I=\int _{ 0 }^{ \pi /2 }{ { x }^{ 2 } } sinxdx\\ \quad ={ \left[ -{ x }^{ 2 }cosx \right] }_{ 0 }^{ \pi /2 }+\int _{ 0 }^{ \pi /2 }{ { 2x } } cosx\quad dx\\ \quad ={ \left[ -{ x }^{ 2 }cosx \right] }_{ 0 }^{ \pi /2 }+{ \left[ 2xsinx \right] }_{ 0 }^{ \pi /2 }-2\int _{ 0 }^{ \pi /2 }{ sinx\quad dx } \\ \quad ={ \left[ -{ x }^{ 2 }cosx+2x\quad sin\quad x+2cosx \right] }_{ 0 }^{ \pi /2 }\\ \quad =\pi -2\)
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)
\(\frac { 23 }{ 10 } \)
\(\int _{ 0 }^{ 1 }{ \left| 5x-3 \right| } dx\quad \\ =\int _{ 0 }^{ 3/5 }{ \left| 5x-3 \right| } dx+\int _{ 3/5 }^{ 1 }{ \left| 5x-3 \right| } dx\\ ={ \left( -\frac { 5 }{ 2 } { x }^{ 2 }+3x \right) }_{ 0 }^{ 3/5 }+{ \left( \frac { 5{ x }^{ 2 } }{ 2 } -3x \right) }_{ 3/5 }^{ 1 }\\ =\left( \frac { 9 }{ 10 } \right) +\left( -\frac { 1 }{ 2 } +\frac { 9 }{ 10 } \right) \\ =\frac { 13 }{ 10 } \)
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)
\(\frac { 1 }{ 2 } \)
\(Putting\frac { \pi }{ 2 } -\theta =y,\\ So\quad that\quad as,\quad \theta \rightarrow \frac { \pi }{ 2 } ,y\rightarrow 0\\ \therefore \underset { \theta \rightarrow \frac { \pi }{ 2 } }{ lim } \frac { cot\theta }{ \left( \frac { \pi }{ 2 } -\theta \right) } =\underset { \theta \rightarrow \frac { \pi }{ 2 } }{ lim } \frac { tan\left( \frac { \pi }{ 2 } -\theta \right) }{ \frac { \pi }{ 2 } -\theta } \\ \underset { y\rightarrow 0 }{ lim } \frac { tany }{ y } =1\)
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
\(\frac { 1 }{ 2\pi } sin(t-\tau )cos\quad \tau \quad d\tau \\ =\frac { 1 }{ 2\pi } \int _{ 0 }^{ 2\pi }{ sint\quad { cos }\tau -cost\quad sin\tau )cos\tau \quad d\tau } \\ =\frac { 1 }{ 2\pi } \int _{ 0 }^{ 2\pi }{ sin\quad t{ cos }^{ 2 }\tau d\tau - } \frac { 1 }{ 2\pi } \int _{ 0 }^{ 2\pi }{ cost\quad sin\tau \quad cos\tau \quad d\tau } \\ =\frac { 1 }{ 4\pi } sint\int _{ 0 }^{ 2\pi }{ (1+cos2\tau } )d\tau -\frac { 1 }{ 4\pi } cost\int _{ 0 }^{ 2\pi }{ sin2\tau \quad d\tau } \\ =\frac { 1 }{ 4\pi } sint{ \left[ \tau +\frac { sin2\tau }{ 2 } \right] }_{ 0 }^{ 2\pi }-\frac { 1 }{ 4\pi } cost{ \left[ -\frac { 1 }{ 2 } cos2\tau \right] }_{ 0 }^{ 2\pi }\\ =\frac { 1 }{ 4\pi } sint\left[ 2\pi +\frac { 1 }{ 2 } .0-0 \right] -\frac { 1 }{ 4\pi } cost\left[ -\frac { 1 }{ 2 } .1+\frac { 1 }{ 2 } .1 \right] \\ =\frac { 1 }{ 4\pi } sint(2\pi )-\frac { 1 }{ 4\pi } cost\left( -\frac { 1 }{ 2 } +\frac { 1 }{ 2 } \right) \\ =\frac { 1 }{ 2 } sint\)
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
\( f(x)=\left( { x }^{ 2 }-4 \right) ^{ 2 }\\ { f }^{ ' }(x)=2\left( x^{ 2 }-4 \right) \times 2x=0\\ x=0,x=2\quad and\quad x=-2\quad are\quad the\quad stationary\quad points.\\ { f }^{ '' }(x)=4[x(2x)+({ x }^{ 2 }-4)]\\ \quad \quad \quad \quad =4[2{ x }^{ 2 }+{ x }^{ 2 }+4]=4[{ 3x }^{ 2 }-4]\\ \quad \quad \quad \quad =12{ x }^{ 2 }-16\\ { f }^{ '' }(x)=-16<\quad ,\quad maximum\quad at\quad x=0\\ f(2)=(12){ 2 }^{ 2 }-16\\ \quad \quad \quad =31<0,\quad minimum\quad at\quad x=2\\ { f }^{ ' }(-2)=12({ -2 })^{ 2 }-16\\ \quad \quad \quad =32>0,minimum\quad at\quad x=-2\\ There\quad is\quad only\quad one\quad maxima\quad and\quad only\quad one\quad minima\quad for\quad \)
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.
A cubic polynomial with real coefficients cannot have more than two extrema and more than three zero crossings.
For the function f(x) = x2e-x, the maximum occurs when x is equal to
- (a)
2
- (b)
1
- (c)
0
- (d)
-1
\(f(x)={ x }^{ 2 }{ e }^{ -x }\\ { f }^{ ' }(x)={ x }^{ 2 }({ -e }^{ -x })+{ e }^{ -x }\times 2x\\ \quad \quad \quad \quad ={ e }^{ -x }(2x-{ x }^{ 2 })\\ Putting\quad { f }^{ ' }(x)=0,\\ { e }^{ -x }(2x-{ x }^{ 2 })=0\\ { e }^{ -x }x(2-x)=0\\ x=0\quad and\quad x=2\quad are\quad the\quad stationary\quad points.\\ { f }^{ '' }(x)={ e }^{ -x }(2-2x)+(2x+{ x }^{ 2 })+({ -e }^{ -x })\\ \quad \quad \quad \quad ={ e }^{ -x }[2-2x-(2x-{ x }^{ 2 })]\\ \quad \quad \quad \quad ={ e }^{ -x }({ x }^{ 2 }-4x+2)\\ At\quad x=0,\quad { f }^{ ' }(0)={ e }^{ -0 }(0-0+2)=2\\ Since,\quad { f }^{ '' }(0)=2\quad is\quad >0,\quad at\quad x=0,\quad we\quad have\quad a\quad minima.\\ Now,\quad atx=2\\ { f }^{ '' }(2)={ e }^{ -2 }({ 2 }^{ 2 }-4\times 2+2)\\ \quad \quad \quad \quad =-2{ e }^{ -2 }<0\\ \therefore At\quad x=2,\quad we\quad have\quad a\quad maxima.\)
\(\int { \frac { dx }{ 1+3{ sin }^{ 2 }x } } \)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)
\(\frac { 1 }{ 2 } { tan }^{ -1 }(tanx)\)
\(Let\quad I=\int { \frac { dx }{ 1+3{ sin }^{ 2 }x } } \\ \quad \quad \quad \quad =\int { \frac { { cosec }^{ 2 }x }{ { cosec }^{ 2 }x+3 } dx } \\ \quad \quad \quad \quad =\int { \frac { { cosec }^{ 2 }x }{ \left( 1+{ cot }^{ 2 }x \right) +3 } dx } \\ putting\quad cot\quad x=t\quad \Rightarrow \quad { -cosec }^{ 2 }x\quad dx\quad =\quad dt\\ \quad \quad \quad I=\int { \frac { -dt }{ 4+{ t }^{ 2 } } =\frac { 1 }{ 2 } { cot }^{ -1 }\frac { t }{ 2 } } \\ \quad \quad \quad \quad =\frac { 1 }{ 2 } { cot }^{ -1 }\left( \frac { cotx }{ 2 } \right) \quad \quad \quad \left( \because { cot }^{ -1 }x={ tan }^{ -1 }\frac { 1 }{ x } \right) \\ \quad \quad \quad \quad =\frac { 1 }{ 2 } { tan }^{ -1 }(2tanx)\)
\(\int { \frac { sinx+cosx }{ \sqrt { 1+sin2x } } } \)dx is equal to
- (a)
tanx
- (b)
cosx
- (c)
sinx
- (d)
x
\(\int { \frac { sinx+cosx }{ \sqrt { 1+sin\quad 2x } } } \\ \quad \quad \quad \quad =\int { \frac { sinx+cosx }{ \sqrt { { sin }^{ 2 }x+{ cos }^{ 2 }x+2sinxcosx } } } \\ \quad \quad \quad \quad =\int { \frac { sinx+cosx }{ { \sqrt { { \left( sinx+cosx \right) }^{ 2 } } } } } dx=\int { dx } =x\)
\(\int { \frac { dx }{ sinx+cosx } } \)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)
\(\frac { 1 }{ \sqrt { 2 } } log\quad tan\left( x+\frac { \pi }{ 4 } \right) \)
\(\int { \frac { dx }{ sinx+cosx } =\frac { 1 }{ \sqrt { 2 } } \int { \frac { dx }{ sinxcos\frac { \pi }{ 4 } +cosxsin\frac { \pi }{ 4 } } } } \\ =\frac { 1 }{ \sqrt { 2 } } \int { \frac { dx }{ sin\left( x+\frac { \pi }{ 4 } \right) } } \\ =\frac { 1 }{ \sqrt { 2 } } \int { cosec\left( x+\frac { \pi }{ 4 } \right) dx } \\ =\frac { 1 }{ \sqrt { 2 } } \left[ -log\quad cot\frac { 1 }{ 2 } \left( x+\frac { \pi }{ 4 } \right) \right] \\ =\frac { 1 }{ \sqrt { 2 } } log\quad tan\left( \frac { x }{ 2 } +\frac { \pi }{ 8 } \right) \)
\(tan\left( \frac { \pi }{ 4 } +x \right) \) when expanded in Taylor's series gives
- (a)
\(1+\frac { { x }^{ 2 } }{ 2! } +\frac { { x }^{ 4 } }{ 4! } +...\)
- (b)
\(1+2x+2{ x }^{ 2 }+\frac { 8 }{ 3 } { x }^{ 3 }+...\)
- (c)
\(1+x+{ x }^{ 2 }+\frac { 4 }{ 3 } { x }^{ 3 }+...\)
- (d)
None of the above
\(Let\quad f(x)=tan\quad x\\ Then,\quad f\left( \frac { \pi }{ 4 } +x \right) =f\left( \frac { \pi }{ 4 } \right) +x{ f }^{ ' }\left( \frac { \pi }{ 4 } \right) +\frac { { x }^{ 2 } }{ 2! } { f }^{ " }\left( \frac { \pi }{ 4 } \right) ....\\ { f }^{ ' }(x)={ sec }^{ 2 }x,{ f }^{ " }(x)=2{ sec }^{ 2 }x\quad tan\quad x\\ { f }^{ "' }(x)=2{ sec }^{ 4 }x+4{ sec }^{ 2 }x{ tan }^{ 2 }x+......\\ Now,\quad f\left( \frac { \pi }{ 4 } \right) =1,\quad { f }^{ ' }\left( \frac { \pi }{ 4 } \right) =2,{ f }^{ " }\left( \frac { \pi }{ 4 } \right) =4......\\ Thus,\quad tan\left( \frac { \pi }{ 4 } +x \right) =1+2x+\frac { { x }^{ 2 } }{ 2 } .4+\frac { { x }^{ 3 } }{ 6 } .16+.....\\ \quad \quad \quad \quad =1+2x+2{ x }^{ 2 }+\frac { 8 }{ 3 } { x }^{ 3 }+.......\)
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)
\(\frac { \pi }{ 2 } +{ tan }^{ -1 }3\)
\(A=\int _{ 0 }^{ \infty }{ \frac { dx }{ { x }^{ 2 }+6x+10 } } \\ \quad =\int _{ 0 }^{ \infty }{ \frac { dx }{ { \left( { x }+3 \right) }^{ 2 }+1 } } \\ \quad ={ \left[ { tan }^{ -1 }(x+3) \right] }_{ 0 }^{ \infty }\\ \quad =\frac { \pi }{ 2 } -{ tan }^{ -1 }3\)
\(\underset { x->\infty }{ lim } \left( \sqrt { { x }^{ 2 }+1 } -\sqrt { x+1 } \right) \)equals
- (a)
0
- (b)
\(\infty \)
- (c)
1
- (d)
e
\(\underset { x->\infty }{ lim } \left( \sqrt { { x }^{ 2 }+1 } -\sqrt { x+1 } \right) \\ =\underset { x->\infty }{ lim } \left( \sqrt { { x }^{ 2 }+1 } -\sqrt { x+1 } \right) \left( \frac { \sqrt { { x }^{ 2 }+1 } +\sqrt { x+1 } }{ \sqrt { { x }^{ 2 }+1 } +\sqrt { x+1 } } \right) \\ =\underset { x->\infty }{ lim } \frac { \left( { x }^{ 2 }+1 \right) -\left( x+1 \right) }{ \sqrt { { x }^{ 2 }+1 } +\sqrt { x+1 } } \\ =\underset { x->\infty }{ lim } \frac { x(x-1) }{ x\left[ \sqrt { 1+\frac { 1 }{ { x }^{ 2 } } +\frac { 1 }{ \sqrt { x } } \sqrt { 1+\frac { 1 }{ x } } } \right] } \\ =\underset { x->\infty }{ lim } \frac { (x-1) }{ x\left[ \sqrt { 1+\frac { 1 }{ { x }^{ 2 } } +\frac { 1 }{ \sqrt { x } } \sqrt { 1+\frac { 1 }{ x } } } \right] } =\frac { \infty }{ 0+1 } =\infty \)
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)
\({ a }^{ 4 }\)
\(\int _{ 0 }^{ a }{ dx } \int _{ \frac { { x }^{ 2 } }{ a } }^{ 2a-x }{ xy } dy=\frac { 1 }{ 2 } \int _{ 0 }^{ a }{ x } dx\left[ \left( 2a-x \right) ^{ 2 }-\frac { { x }^{ 4 } }{ { a }^{ 2 } } \right] \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad =\frac { 1 }{ 2 } \int _{ 0 }^{ 2 }{ \left[ 4{ a }^{ 2 }x-4a{ x }^{ 2 }+{ x }^{ 3 }-\frac { { x }^{ 5 } }{ { a }^{ 2 } } \right] dx } \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad =\frac { 1 }{ 2 } \left[ 2{ a }^{ 4 }-\frac { 4 }{ 3 } { a }^{ 4 }+\frac { { a }^{ 4 } }{ 4 } -\frac { { a }^{ 4 } }{ 6 } \right] \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad =\frac { 3{ a }^{ 4 } }{ 8 } \)
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
Since a polynomial functions as well as an exponential function is continuous and the product of two continuous functions is continuous, so f(x) is continuous in ]-3,0[.
\({ f }^{ ' }(x)=(2x+3){ e }^{ -\frac { x }{ 2 } }\\ \quad \quad \quad \quad =\frac { -1 }{ 2 } { e }^{ -\frac { x }{ 2 } }({ x }^{ 2 }+3x)\\ \quad \quad \quad \quad ={ e }^{ -\frac { x }{ 2 } }\left[ \frac { x+6-{ x }^{ 2 } }{ 2 } \right] \)
Which clearly exists for all x <-> ]-3,0[, f'(x) is differentiable in ]-3,0[.
Also, f(-3)=f(0)=0
\(By\quad Rolle's\quad theroem\quad c\leftrightarrow ]-3,0[such\quad that\\ \quad \quad \quad \quad f'(c)=0\\ { e }^{ -c/2 }\left[ \frac { c+6-{ c }^{ 2 } }{ 2 } \right] =0\\ c+6-{ c }^{ 2 }=0\\ { c }^{ 2 }-c-6=0\\ c=-2,c=3\\ c=-2\leftrightarrow ]-3,0[\)