The area bounded by y=e^(x) , y-axis and the line y=e is -Turito

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The area bounded by $y equals e to the power of x end exponent$ , Y-axis and the line y=e is

2
e
$\frac{e}{2}$
1

The correct answer is: 1

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maths-

If $32 Tan to the power of 8 theta equals 2 Cos squared space alpha minus 3 cos space alpha text and end text 3 cos space 2 theta equals 1$ then the general value of 'α ' is

maths-General

General

maths-

The parabolas $y to the power of 2 end exponent equals 4 x comma x to the power of 2 end exponent equals 4 y$ divide the square region bounded by the lines x=4, y=4 and the co-ordinate axes. If $S subscript 1 end subscript comma S subscript 2 end subscript comma S subscript 3 end subscript$ are respectively the areas of these parts numbered from top to bottom then $S subscript 1 end subscript colon S subscript 2 end subscript colon S subscript 3 end subscript$ is

maths-General

General

physics-

The figure shows a double slit experiment P and Q are the slits. The path lengths PX and QX are $n lambda$ and $left parenthesis n plus 2 right parenthesis lambda$ respectively, where n is a whole number and $lambda$ is the wavelength. Taking the central fringe as zero, what is formed at X

For brightness, path difference

equals n lambda equals 2 lambda

So second is bright.

The figure shows a double slit experiment P and Q are the slits. The path lengths PX and QX are $n lambda$ and $left parenthesis n plus 2 right parenthesis lambda$ respectively, where n is a whole number and $lambda$ is the wavelength. Taking the central fringe as zero, what is formed at X

physics-General

For brightness, path difference

equals n lambda equals 2 lambda

So second is bright.

General

maths-

If $gamma Sin space theta equals 3 comma gamma equals 4 left parenthesis 1 plus Sin space theta right parenthesis comma 0 less or equal than theta less or equal than 2 pi$ then $theta$ = ---

maths-General

General

physics-

A particle starts from rest. Its acceleration $open parentheses a close parentheses blank v e r s u s$ time $open parentheses t close parentheses$ is as shown in the figure. The maximum speed of the particle will be

Area under acceleration-time graph gives the change in velocity. Hence,

v subscript m a x end subscript equals fraction numerator 1 over denominator 2 end fraction cross times 10 cross times 11 equals 55 blank m s to the power of negative 1 end exponent

Therefore, the correct option is [b].

A particle starts from rest. Its acceleration $open parentheses a close parentheses blank v e r s u s$ time $open parentheses t close parentheses$ is as shown in the figure. The maximum speed of the particle will be

physics-General

Area under acceleration-time graph gives the change in velocity. Hence,

v subscript m a x end subscript equals fraction numerator 1 over denominator 2 end fraction cross times 10 cross times 11 equals 55 blank m s to the power of negative 1 end exponent

Therefore, the correct option is [b].

General

physics-

In figure, one car at rest and velocity of the light from head light is $c$ , tehn velocity of light from head light for the moving car at velocity $v$ , would be

Since

c greater than greater than v

(negligible)

In figure, one car at rest and velocity of the light from head light is $c$ , tehn velocity of light from head light for the moving car at velocity $v$ , would be

physics-General

Since

c greater than greater than v

(negligible)

General

maths-

If α and β are two different solutions lying between $fraction numerator negative pi over denominator 2 end fraction$ and of the equation $2 Tan space theta plus Sec space theta equals 2$ then Tan α + Tan β is

maths-General

General

physics-

A particle of mass $m$ is initially situated at the point $P$ inside a hemispherical surface of radius $r$ as shown in figure. A horizontal acceleration of magnitude $a subscript 0 end subscript$ is suddenly produced on the particle in the horizontal direction. If gravitational acceleration is neglected, the time taken by particle to touch the sphere again is

Let the particle touches the sphere t the point

A.

Let

P A equals 1

therefore P B equals fraction numerator l over denominator 2 end fraction

increment O P B comma cos invisible function application alpha equals fraction numerator P B over denominator r end fraction

therefore P B equals r cos invisible function application a

fraction numerator l over denominator 2 end fraction equals r cos invisible function application a

therefore l equals 2 r cos invisible function application alpha

B u t blank l equals fraction numerator 1 over denominator 2 end fraction a subscript 0 end subscript t to the power of 2 end exponent

therefore blank t equals square root of open parentheses fraction numerator 2 l over denominator a subscript 0 end subscript end fraction close parentheses end root equals square root of open parentheses fraction numerator 2 cross times 2 r cos invisible function application a over denominator a subscript 0 end subscript end fraction close parentheses end root equals square root of open parentheses fraction numerator 4 blank r cos invisible function application a over denominator a subscript 0 end subscript end fraction close parentheses end root

A particle of mass $m$ is initially situated at the point $P$ inside a hemispherical surface of radius $r$ as shown in figure. A horizontal acceleration of magnitude $a subscript 0 end subscript$ is suddenly produced on the particle in the horizontal direction. If gravitational acceleration is neglected, the time taken by particle to touch the sphere again is

physics-General

Let the particle touches the sphere t the point

A.

Let

P A equals 1

therefore P B equals fraction numerator l over denominator 2 end fraction

increment O P B comma cos invisible function application alpha equals fraction numerator P B over denominator r end fraction

therefore P B equals r cos invisible function application a

fraction numerator l over denominator 2 end fraction equals r cos invisible function application a

therefore l equals 2 r cos invisible function application alpha

B u t blank l equals fraction numerator 1 over denominator 2 end fraction a subscript 0 end subscript t to the power of 2 end exponent

therefore blank t equals square root of open parentheses fraction numerator 2 l over denominator a subscript 0 end subscript end fraction close parentheses end root equals square root of open parentheses fraction numerator 2 cross times 2 r cos invisible function application a over denominator a subscript 0 end subscript end fraction close parentheses end root equals square root of open parentheses fraction numerator 4 blank r cos invisible function application a over denominator a subscript 0 end subscript end fraction close parentheses end root

General

physics-

A cyclist starts from the centre $O$ of a circular park of radius 1 km, reaches the edge $P$ of the park, then cycles along the circumference and returns to the point $O$ as shown in figure. If the round trip takes 10 min, the net displacement and average speed of the cyclist (in metre and kilometer per hour) are

Since, the initial position of cyclist coincides with final position, so his net displacement is zero.

A v e r a g e blank s p e e d equals fraction numerator t o t a l blank d i s t a n c e blank t r a v e l l e d over denominator t o t a l blank t i m e blank t a k e n end fraction

equals fraction numerator O P plus P Q plus Q O over denominator 10 end fraction k m blank m i n to the power of negative 1 end exponent

equals fraction numerator 1 plus fraction numerator pi over denominator 2 end fraction cross times 1 plus 1 over denominator 10 end fraction k m blank m i n to the power of negative 1 end exponent

equals fraction numerator pi plus 4 over denominator 20 end fraction cross times 60 blank k m h to the power of negative 1 end exponent equals 21.4 blank k m h to the power of negative 1 end exponent

A cyclist starts from the centre $O$ of a circular park of radius 1 km, reaches the edge $P$ of the park, then cycles along the circumference and returns to the point $O$ as shown in figure. If the round trip takes 10 min, the net displacement and average speed of the cyclist (in metre and kilometer per hour) are

physics-General

Since, the initial position of cyclist coincides with final position, so his net displacement is zero.

A v e r a g e blank s p e e d equals fraction numerator t o t a l blank d i s t a n c e blank t r a v e l l e d over denominator t o t a l blank t i m e blank t a k e n end fraction

equals fraction numerator O P plus P Q plus Q O over denominator 10 end fraction k m blank m i n to the power of negative 1 end exponent

equals fraction numerator 1 plus fraction numerator pi over denominator 2 end fraction cross times 1 plus 1 over denominator 10 end fraction k m blank m i n to the power of negative 1 end exponent

equals fraction numerator pi plus 4 over denominator 20 end fraction cross times 60 blank k m h to the power of negative 1 end exponent equals 21.4 blank k m h to the power of negative 1 end exponent

General

physics-

A the instant a motor bike starts from rest in a given direction, a car overtakes the motor bike, both moving in the same direction. The speed-time graphs for motor bike and car are represented by $O A B$ and $C D$ respectively Then

Distance travelled by motor bike at

t equals 18

s subscript b i k e end subscript equals s subscript 1 end subscript equals fraction numerator 1 over denominator 2 end fraction

(18)(60)=540 m
Distance travelled by car at

t equals 18

s subscript c a r end subscript equals s subscript 2 end subscript

=(18)(60)=720 m
Therefore, separation between them at

t equals 18

s is 180m. Let, separation between them decreases to zero at time

t

beyond 18s.
Hence,

s subscript b i k e end subscript equals 540 plus 60 t

and

s subscript c a r end subscript equals 720 plus 40 t

s subscript c a r end subscript minus s subscript b i k e end subscript equals 0

rightwards double arrow blank 720 plus 40 t equals 540 plus 60 t

rightwards double arrow blank t equals 9

s beyond 18s or
Hence,

t equals open parentheses 18 plus 9 close parentheses

s=27s from start and distant travelled by both is

s subscript b i k e end subscript

s subscript c a r end subscript equals 1080

A the instant a motor bike starts from rest in a given direction, a car overtakes the motor bike, both moving in the same direction. The speed-time graphs for motor bike and car are represented by $O A B$ and $C D$ respectively Then

physics-General

Distance travelled by motor bike at

t equals 18

s subscript b i k e end subscript equals s subscript 1 end subscript equals fraction numerator 1 over denominator 2 end fraction

(18)(60)=540 m
Distance travelled by car at

t equals 18

s subscript c a r end subscript equals s subscript 2 end subscript

=(18)(60)=720 m
Therefore, separation between them at

t equals 18

s is 180m. Let, separation between them decreases to zero at time

t

beyond 18s.
Hence,

s subscript b i k e end subscript equals 540 plus 60 t

and

s subscript c a r end subscript equals 720 plus 40 t

s subscript c a r end subscript minus s subscript b i k e end subscript equals 0

rightwards double arrow blank 720 plus 40 t equals 540 plus 60 t

rightwards double arrow blank t equals 9

s beyond 18s or
Hence,

t equals open parentheses 18 plus 9 close parentheses

s=27s from start and distant travelled by both is

s subscript b i k e end subscript

s subscript c a r end subscript equals 1080

General

physics-

Assertion : Owls can move freely during night.
Reason : They have large number of rods on their retina.

Owls can move freely during night, because they have large number of cones on their retina which help them to see in night.

Assertion : Owls can move freely during night.
Reason : They have large number of rods on their retina.

physics-General

Owls can move freely during night, because they have large number of cones on their retina which help them to see in night.

General

physics-

A particle shows distance-time curve as given in this figure. The maximum instantaneous velocity of the particle is around the point

Instantaneous velocity is given by the slope of the curve at that instant

v equals fraction numerator d s over denominator d t end fraction equals tan invisible function application theta

from the figure it is clear that slope of the curve is maximum at point ‘

C

’

A particle shows distance-time curve as given in this figure. The maximum instantaneous velocity of the particle is around the point

physics-General

Instantaneous velocity is given by the slope of the curve at that instant

v equals fraction numerator d s over denominator d t end fraction equals tan invisible function application theta

from the figure it is clear that slope of the curve is maximum at point ‘

C

’

General

Maths-

The area bounded by y=3x and $y equals x to the power of 2 end exponent$ is

Maths-General

General

Maths-

The area bounded by $y equals x squared plus 2 comma$ X- axis, x=1 and x=2 is

Maths-General

General

maths-

If $Cos space 2 theta times Cos space 3 theta times cos space theta equals 1 fourth text for end text 0 less than theta less than pi$ then $theta$ =

maths-General

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Question

The area bounded by $y equals e to the power of x end exponent$ , Y-axis and the line y=e is

2
e
$\frac{e}{2}$
1

Answer

The correct answer is: 1

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Related Questions to study

If $32 Tan to the power of 8 theta equals 2 Cos squared space alpha minus 3 cos space alpha text and end text 3 cos space 2 theta equals 1$ then the general value of 'α ' is

If $32 Tan to the power of 8 theta equals 2 Cos squared space alpha minus 3 cos space alpha text and end text 3 cos space 2 theta equals 1$ then the general value of 'α ' is

The figure shows a double slit experiment P and Q are the slits. The path lengths PX and QX are $n lambda$ and $left parenthesis n plus 2 right parenthesis lambda$ respectively, where n is a whole number and $lambda$ is the wavelength. Taking the central fringe as zero, what is formed at X

The figure shows a double slit experiment P and Q are the slits. The path lengths PX and QX are $n lambda$ and $left parenthesis n plus 2 right parenthesis lambda$ respectively, where n is a whole number and $lambda$ is the wavelength. Taking the central fringe as zero, what is formed at X

If $gamma Sin space theta equals 3 comma gamma equals 4 left parenthesis 1 plus Sin space theta right parenthesis comma 0 less or equal than theta less or equal than 2 pi$ then $theta$ = ---

If $gamma Sin space theta equals 3 comma gamma equals 4 left parenthesis 1 plus Sin space theta right parenthesis comma 0 less or equal than theta less or equal than 2 pi$ then $theta$ = ---

A particle starts from rest. Its acceleration $open parentheses a close parentheses blank v e r s u s$ time $open parentheses t close parentheses$ is as shown in the figure. The maximum speed of the particle will be

A particle starts from rest. Its acceleration $open parentheses a close parentheses blank v e r s u s$ time $open parentheses t close parentheses$ is as shown in the figure. The maximum speed of the particle will be

In figure, one car at rest and velocity of the light from head light is $c$ , tehn velocity of light from head light for the moving car at velocity $v$ , would be

In figure, one car at rest and velocity of the light from head light is $c$ , tehn velocity of light from head light for the moving car at velocity $v$ , would be

If α and β are two different solutions lying between $fraction numerator negative pi over denominator 2 end fraction$ and of the equation $2 Tan space theta plus Sec space theta equals 2$ then Tan α + Tan β is

If α and β are two different solutions lying between $fraction numerator negative pi over denominator 2 end fraction$ and of the equation $2 Tan space theta plus Sec space theta equals 2$ then Tan α + Tan β is

A the instant a motor bike starts from rest in a given direction, a car overtakes the motor bike, both moving in the same direction. The speed-time graphs for motor bike and car are represented by $O A B$ and $C D$ respectively Then

A the instant a motor bike starts from rest in a given direction, a car overtakes the motor bike, both moving in the same direction. The speed-time graphs for motor bike and car are represented by $O A B$ and $C D$ respectively Then

Assertion : Owls can move freely during night.
Reason : They have large number of rods on their retina.

Assertion : Owls can move freely during night.
Reason : They have large number of rods on their retina.

A particle shows distance-time curve as given in this figure. The maximum instantaneous velocity of the particle is around the point

A particle shows distance-time curve as given in this figure. The maximum instantaneous velocity of the particle is around the point

The area bounded by y=3x and $y equals x to the power of 2 end exponent$ is

The area bounded by y=3x and $y equals x to the power of 2 end exponent$ is

The area bounded by $y equals x squared plus 2 comma$ X- axis, x=1 and x=2 is

The area bounded by $y equals x squared plus 2 comma$ X- axis, x=1 and x=2 is

If $Cos space 2 theta times Cos space 3 theta times cos space theta equals 1 fourth text for end text 0 less than theta less than pi$ then $theta$ =

If $Cos space 2 theta times Cos space 3 theta times cos space theta equals 1 fourth text for end text 0 less than theta less than pi$ then $theta$ =

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Question

The area bounded by , Y-axis and the line y=e is

2 e 1

Answer

The correct answer is: 1

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Related Questions to study

If then the general value of 'α ' is

If then the general value of 'α ' is

The parabolas divide the square region bounded by the lines x=4, y=4 and the co-ordinate axes. If are respectively the areas of these parts numbered from top to bottom then is

The parabolas divide the square region bounded by the lines x=4, y=4 and the co-ordinate axes. If are respectively the areas of these parts numbered from top to bottom then is

The figure shows a double slit experiment P and Q are the slits. The path lengths PX and QX are and respectively, where n is a whole number and is the wavelength. Taking the central fringe as zero, what is formed at X

The figure shows a double slit experiment P and Q are the slits. The path lengths PX and QX are and respectively, where n is a whole number and is the wavelength. Taking the central fringe as zero, what is formed at X

If then = ---

If then = ---

A particle starts from rest. Its acceleration time is as shown in the figure. The maximum speed of the particle will be

A particle starts from rest. Its acceleration time is as shown in the figure. The maximum speed of the particle will be

In figure, one car at rest and velocity of the light from head light is , tehn velocity of light from head light for the moving car at velocity , would be

In figure, one car at rest and velocity of the light from head light is , tehn velocity of light from head light for the moving car at velocity , would be

If α and β are two different solutions lying between and of the equation then Tan α + Tan β is

If α and β are two different solutions lying between and of the equation then Tan α + Tan β is

A the instant a motor bike starts from rest in a given direction, a car overtakes the motor bike, both moving in the same direction. The speed-time graphs for motor bike and car are represented by and respectively Then

A the instant a motor bike starts from rest in a given direction, a car overtakes the motor bike, both moving in the same direction. The speed-time graphs for motor bike and car are represented by and respectively Then

Assertion : Owls can move freely during night.Reason : They have large number of rods on their retina.

Assertion : Owls can move freely during night.Reason : They have large number of rods on their retina.

A particle shows distance-time curve as given in this figure. The maximum instantaneous velocity of the particle is around the point

A particle shows distance-time curve as given in this figure. The maximum instantaneous velocity of the particle is around the point

The area bounded by y=3x and is

The area bounded by y=3x and is

The area bounded by X- axis, x=1 and x=2 is

The area bounded by X- axis, x=1 and x=2 is

If then =

If then =

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The area bounded by $y equals e to the power of x end exponent$ , Y-axis and the line y=e is

2
e
$\frac{e}{2}$
1

If $32 Tan to the power of 8 theta equals 2 Cos squared space alpha minus 3 cos space alpha text and end text 3 cos space 2 theta equals 1$ then the general value of 'α ' is

If $32 Tan to the power of 8 theta equals 2 Cos squared space alpha minus 3 cos space alpha text and end text 3 cos space 2 theta equals 1$ then the general value of 'α ' is

The figure shows a double slit experiment P and Q are the slits. The path lengths PX and QX are $n lambda$ and $left parenthesis n plus 2 right parenthesis lambda$ respectively, where n is a whole number and $lambda$ is the wavelength. Taking the central fringe as zero, what is formed at X

The figure shows a double slit experiment P and Q are the slits. The path lengths PX and QX are $n lambda$ and $left parenthesis n plus 2 right parenthesis lambda$ respectively, where n is a whole number and $lambda$ is the wavelength. Taking the central fringe as zero, what is formed at X

If $gamma Sin space theta equals 3 comma gamma equals 4 left parenthesis 1 plus Sin space theta right parenthesis comma 0 less or equal than theta less or equal than 2 pi$ then $theta$ = ---

If $gamma Sin space theta equals 3 comma gamma equals 4 left parenthesis 1 plus Sin space theta right parenthesis comma 0 less or equal than theta less or equal than 2 pi$ then $theta$ = ---

A particle starts from rest. Its acceleration $open parentheses a close parentheses blank v e r s u s$ time $open parentheses t close parentheses$ is as shown in the figure. The maximum speed of the particle will be

A particle starts from rest. Its acceleration $open parentheses a close parentheses blank v e r s u s$ time $open parentheses t close parentheses$ is as shown in the figure. The maximum speed of the particle will be

In figure, one car at rest and velocity of the light from head light is $c$ , tehn velocity of light from head light for the moving car at velocity $v$ , would be

In figure, one car at rest and velocity of the light from head light is $c$ , tehn velocity of light from head light for the moving car at velocity $v$ , would be

If α and β are two different solutions lying between $fraction numerator negative pi over denominator 2 end fraction$ and of the equation $2 Tan space theta plus Sec space theta equals 2$ then Tan α + Tan β is

If α and β are two different solutions lying between $fraction numerator negative pi over denominator 2 end fraction$ and of the equation $2 Tan space theta plus Sec space theta equals 2$ then Tan α + Tan β is

A the instant a motor bike starts from rest in a given direction, a car overtakes the motor bike, both moving in the same direction. The speed-time graphs for motor bike and car are represented by $O A B$ and $C D$ respectively Then

A the instant a motor bike starts from rest in a given direction, a car overtakes the motor bike, both moving in the same direction. The speed-time graphs for motor bike and car are represented by $O A B$ and $C D$ respectively Then

Assertion : Owls can move freely during night.
Reason : They have large number of rods on their retina.

Assertion : Owls can move freely during night.
Reason : They have large number of rods on their retina.

The area bounded by y=3x and $y equals x to the power of 2 end exponent$ is

The area bounded by y=3x and $y equals x to the power of 2 end exponent$ is

The area bounded by $y equals x squared plus 2 comma$ X- axis, x=1 and x=2 is

The area bounded by $y equals x squared plus 2 comma$ X- axis, x=1 and x=2 is

If $Cos space 2 theta times Cos space 3 theta times cos space theta equals 1 fourth text for end text 0 less than theta less than pi$ then $theta$ =

If $Cos space 2 theta times Cos space 3 theta times cos space theta equals 1 fourth text for end text 0 less than theta less than pi$ then $theta$ =