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11th கணிதவியல் வாரத் தேர்வு -4(தொகை நுண்கணிதம்)

MABS Institution

11th கணிதவியல் வாரத் தேர்வு -4(தொகை நுண்கணிதம்)-Aug 2020

Time : 1.00 Hours

Part - A

Multiple Choice Question - Answer All Question

\(\int { \frac { { sec }^{ 2 }x }{ { tan }^{ 2 }x-1 } } dx\) =

\(2log\left| \frac { 1-tanx }{ 1+tanx } \right| +c\)
\(log\left| \frac { 1+tanx }{ 1-tanx } \right| +c\)
\(\frac { 1 }{ 2 } log\left| \frac { tanx+1 }{ tanx-1 } \right| +c\)
\(\frac { 1 }{ 2 } log\left| \frac { tanx-1 }{ tanx+1 } \right| +c\)

\(\\ \int { \frac { { x }^{ 2 }+{ cos }^{ 2 }x }{ { x }^{ 2 }+1 } { cosec }^{ 2 }xdx= } \)

cot x+sin^-1 x+c
−cot x + tan⁻¹ x + c
− tan x + cot⁻¹ x + c
−cot x − tan⁻¹x + c

\(\int { \frac { secx }{ \sqrt { cos2x } } } dx=\)

tan⁻¹(sin x)+c
2sin⁻¹(tan x)+c
tan⁻¹(cos x) + c
sin⁻¹(tan x) + c

\(\int { \frac { \sqrt { tanx } }{ sin2x } } dx=\)

\(\sqrt { tanx } +c\)
\(2\sqrt { tanx } +c\)
\(\frac { 1 }{ 2 } \sqrt { tanx } +c\)
\(\frac { 1 }{ 4 } \sqrt { tanx } +c\)

\(\int { e^{ -4x } } \) cos x dx =

\(\frac { { e }^{ -4x } }{ 17 } \)[4cosx - sinx]+c
\(\frac { { e }^{ -4x } }{ 17 } \)[-4cosx + sinx] +c
\(\frac { { e }^{ -4x } }{ 17 } \)[4cosx + sinx] + c
\(\frac { { e }^{ -4x } }{ 17 } \) [-4cosx - sin x]+c

\(\int { \sqrt { \frac { 1-x }{ 1+x } } dx } \) =

\(\sqrt { 1-{ x }^{ 2 } } \) + sin^-1 x + c
sin^-1 x - \(\sqrt { 1-{ x }^{ 2 } } \) +c
log | x + \(\sqrt { 1-{ x }^{ 2 } } \) | -\(\sqrt { 1-{ x }^{ 2 } } \) + c
\(\sqrt { 1-{ x }^{ 2 } } \) + log | x + \(\sqrt { 1-{ x }^{ 2 } } \) | +c

(x, y) என்ற ஏதேனும் ஒரு புள்ளியில் ஒரு வளைவரையின் சாய்வு \(\frac { { x }^{ 2 }-4 }{ { x }^{ 2 } } \)ஆகும்.
இவ்வளைவரை (2, 7) என்ற புள்ளி வழியாகச் சென்றால், வளைவரையின் சமன்பாடு.

\(y=x+\frac { 4 }{ x } +3\)
\(y=x+\frac { 4 }{ x } +4\)
y=x²+3x+4
y=x²-3x+6

\(\\ \int { { x }^{ 2 }cosx\quad dx= } \)

x² sin x + 2x cos x − 2sin x + c
x² sin x − 2x cos x − 2sin x + c
−x²sin x + 2x cos x + 2sin x + c
−x² sin x − 2x cos x + 2sin x + c

\(\int { { x }^{ 2 }{ e }^{ \frac { x }{ 2 } }dx } \) =

\({ x }^{ 2 }{ e }^{ \frac { x }{ 2 } }-4x{ e }^{ \frac { x }{ 2 } }-8{ e }^{ \frac { x }{ 2 } }+c\)
\(2x^{ 2 }{ e }^{ \frac { x }{ 2 } }-8{ xe }^{ \frac { x }{ 2 } }-16x{ e }^{ \frac { x }{ 2 } }+c\)
\(2x^{ 2 }{ e }^{ \frac { x }{ 2 } }-8{ xe }^{ \frac { x }{ 2 } }+16x{ e }^{ \frac { x }{ 2 } }+c\)
\({ x }^{ 2 }\frac { { e }^{ \frac { x }{ 2 } } }{ 2 } -\frac { { e }^{ \frac { x }{ 2 } } }{ 4 } +\frac { { e }^{ \frac { x }{ 2 } } }{ 8 } +c\)

\(\int { \frac { { e }^{ 6logx }-{ e }^{ 5logx } }{ { e }^{ 4logx }-{ e }^{ 3logx } } } dx=\)

x+c
\(\frac { { x }^{ 3 } }{ 3 } +c\)
\(\frac { 3 }{ { x }^{ 3 } } +c\)
\(\frac { 1 }{ { x }^{ 2 } } +c\)

\(\int { \frac { 3^{ \frac { 1 }{ x } } }{ { x }^{ 2 } } dx } ={ k }{ (3 }^{ \frac { 1 }{ x } })+c\)எனில், k-ன் மதிப்பு

log 3
-log 3
\(-\frac { 1 }{ log3 } \)
\(\frac { 1 }{ log3 } \)

\(\int { \frac { 1 }{ x\sqrt { \left( logx \right) ^{ 2 }-5 } } } dx\) =

log | x + \(\sqrt { { x }^{ 2 }-5 } \) | +c
log | log x + \(\sqrt { logx-5 } \)| +c
log | log x + \(\sqrt { \left( logx \right) ^{ 2 }-5 } |+c\) |
log | log x - \(\sqrt { \left( logx \right) ^{ 2 }-5 } \)| +c

\(\int { \frac { dx }{ { e }^{ x }-1 } } \) =

log |e^x| - log |e^x - 1| +c
log |e^x| + log | e^x -1| +c
log |e^x - 1|- log|e^x| +c
log |ex + 1| -log | e^x | +c

\(\int { { 2 }^{ \\ x+5 } } dx=\)

\(\frac { 3({ 2 }^{ \\ x+5 }) }{ log\quad 2 } +c\)
\(\frac { { 2 }^{ 3x+5 } }{ 2log(3x+5) } +c\)
\(\frac { { 2 }^{ 3x+5 } }{ 2log3 } +c\)
\(\frac { { 2 }^{ 3x+5 } }{ 3log2 } +c\)

\(\int { \frac { { e }^{ x }(1+x) }{ { cos }^{ 2 }(x{ e }^{ x }) } } dx\)=

cos(xe^x)+c
sec(xe^x)+c
tan(xe^x)+c
cos(xe^x)+c

\(\int { \frac { e^{ x }({ x }^{ 2 }{ tan }^{ -1 }x+{ tan }^{ -1 }x+1) }{ { x }^{ 2 }+1 } dx= } \)

\({ e }^{ x }{ tan }^{ -1 }(x+1)+c\)
\({ tan }^{ -1 }{ (e }^{ x })+c\)
\({ e }^{ x }\frac { { ({ tan }^{ -1 }x) }^{ 2 } }{ 2 } \)
\({ e }^{ x }{ tan }^{ -1 }x+c\)

\(\int { \frac { { sin }^{ 8 }x-{ cos }^{ 8 }x }{ 1-2{ sin }^{ 2 }x{ cos }^{ 2 }x } dx= } \quad \)

\(\frac { 1 }{ 2 } sin2x+c\)
\(-\frac { 1 }{ 2 } sin2x+c\)
\(\frac { 1 }{ 2 } cos2x+c\)
\(-\frac { 1 }{ 2 } cos2x+c\)

\(\int { { e }^{ -7x } } \) sin 5xdx =

\(\frac { { e }^{ -7x } }{ 74 } \) [-7sin 5x - 5cos 5x] + c
\(\frac { { e }^{ -7x } }{ 74 } \) [7sin 5x + 5 cos 5x]+c
\(\frac { { e }^{ -7x } }{ 74 } \) [7 sin 5x - 5 cos5x] + c
\(\frac { { e }^{ -7x } }{ 74 } \) [-7 sin 5x + 5 cos 5x] + c

\(\int { sin } \sqrt { x } dx\) =

2 (- \(\sqrt { x } \) cos \(\sqrt { x } \) + sin \(\sqrt { x } \) ) + c
2 (-\(\sqrt { x } \) cos \(\sqrt { x } \) - sin \(\sqrt { x } \)) +c
2 (-\(\sqrt { x } \) sin \(\sqrt { x } \) - cos \(\sqrt { x } \) ) + c
2 (-\(\sqrt { x } \) sin \(\sqrt { x } \) + cos \(\sqrt { x } \) ) +c

\(\int { { tan }^{ -1 }\left( \sqrt { \frac { 1-cos2x }{ 1+cos2x } } \right) } \)dx=

x²+ c
2x²+c
\(\frac { { x }^{ 2 } }{ 2 } +c\)
-\(\frac { { x }^{ 2 } }{ 2 } +c\)

\(\int { { sin }^{ 3 }xdx= } \)

\(\frac { -3 }{ 4 } cosx-\frac { cos3x }{ 12 } +c\)
\(\frac { 3 }{ 4 } cosx-\frac { cos3x }{ 12 } +c\)
\(\frac { -3 }{ 4 } cosx-\frac { cos3x }{ 12 } +\)c
\(\frac { -3 }{ 4 } cosx-\frac { sin3x }{ 12 } +c\)

\(\int { \frac { x+2 }{ \sqrt { { x }^{ 2 }-1 } } dx } \) =

\(\sqrt { { x }^{ 2 }-1 } \) - 2 log | x + \(\sqrt { { x }^{ 2 }-1 } \) | +c
sin^-1 - 2log | x + \(\sqrt { { x }^{ 2 }-1 } \) | +c
2 log | x + \(\sqrt { { x }^{ 2 }-1 } \) | - sin^-1 x +c
\(\sqrt { { x }^{ 2 }-1 } \) + 2log | x + \(\sqrt { { x }^{ 2 }-1 } \) | + c

\(\int { f(x)dx=g(x)+c } \) எனில், \(\int { f(x){ g }^{ ' }(x)dx } \)என்பது

\(\int { f(x))^{ 2 }dx } \)
\(\int { f(x)g(x) } \)
\(\int { { f }^{ ' }(x)g(x) } \)
\(\int { (g({ x)) }^{ 2 }(x)dx } \)

\(\int { { e }^{ \sqrt { x } } } \) dx =

\(2\sqrt { x } \left( 1-{ e }^{ \sqrt { x } } \right) +c\)
\(2{ e }^{ \sqrt { x } }\left( 1-\sqrt { x } \right) +c\)
\(2\sqrt { x } \left( { e }^{ \sqrt { x } }-1 \right) +c\)
\(2{ e }^{ { e }^{ \sqrt { x } } }\left( \sqrt { x-1 } \right) +c\)

\(\int { { f }^{ ' }(x){ e }^{ { x }^{ 3 } }dx } =(x-1){ e }^{ { x }^{ 3 } }+c\) எனில், f(x) என்பது

\(2{ x }^{ 3 }-\frac { { x }^{ 2 } }{ 2 } +x+c\)
\(\frac { { x }^{ 2 } }{ 2 } +3{ x }^{ 2 }+4x+c\)
\({ x }^{ 3 }+4{ x }^{ 2 }+6x+c\)
\(\frac { { 2x }^{ 3 } }{ 3 } -{ x }^{ 2 }+x+c\)

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