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previous year questions Of Areas Bounded by Regions For NDA

previous year questions Of Areas Bounded by Regions

previous year questions
Q 2116501479

Consider the function
`f(x) = | x - 1| + x^(2)` where `x in R`.

What is the area of the region bounded by X-axis, the
curve `y = f( x)` and the two ordinates `x =1/2` and `x = 1?`
NDA Paper 1 2016
(A)

`5/12` sq unit

(B)

`5/6` sq unit

(C)

`7/6` sq unit

(D)

`2` sq unit

Solution:

Given `f(x) = | x -1 | + x^2`

`:. f(x) = { tt (( x^2 + (x - 1) , text(for x > 1) ), ( x^2 - x + 1 ,text(for x < 1)) )`

`f(x) = | x - 1| + x^(2)`

When `1/2 < x < 1`

`f(x) = x^( 2) - x + 1`

:. Area of required region will be

` int _(1//2)^(1) f(x) dx = int _(1//2)^(1) (x^(2) - x -1)dx = [x^(3)/3 - x^(2)/2 + x]_(1//2)^(1)`

` = [ (1)^(3)/3 - (1)^(2)/2 + (1) ] -[(1/2)^(3)/3 - (1/2)^(2)/2 +1/2]`

` = ( 1/3 - 1/2 +1) -( 1/24 - 1/8 +1/2 )`

` = ((2-3+6)/6 ) - ((1-3 +12)/24)`

` 5/6 - 10/24`

`(20-10)/24 = 10/24 = 5/12` sq unit.
Correct Answer is `=>` (A) `5/12` sq unit
Q 2136601572

Consider the function
`f(x) = | x - 1| + x^(2)` where `x in R`.

What is the area of the region bounded by X-axis, the
curve `y = f(x)` and the two ordinates `x =1` and `x = 3/2?`

NDA Paper 1 2016
(A)

`5/12` sq unit

(B)

`7/12` sq unit

(C)

`2/3` sq unit

(D)

`11/12` sq unit

Solution:

Given `f(x) = | x -1 | + x^2`

`:. f(x) = { tt (( x^2 + (x - 1) , text(for x > 1) ), ( x^2 - x + 1 ,text(for x < 1)) )`

`f(x) =| x- 1| + x^(2)`

When `1 < x < 3/2`

`f(x) = x ^(2) + x- 1`

:. Area of required region will be

` int _(1)^(3//2) f(x) dx = int _(1)^(3//2) (x^(2) + x -1) dx`

` = [x^(3)/3 +x^(2)/2 - x]_(1)^(3//2)`

`= [(3/2)^(3)/3 + (3/2)^(2)/2 - 3/2 ] - [(1)^(3)/3 + (1)^(2)/2 -1]`

`= (27/24 +9/8 -3/2 )- (1/3 +1/2 -1)`

`= ((27+27-36)/24) - ((2+3-6)/6) `

` = 18/24 +1/6 = (18 +4)/24 = 22/24`

` = 11/12 `sq unit
Correct Answer is `=>` (D) `11/12` sq unit
Q 2136878772

Given curves are
`f(x) =x | x |- 1` and `g(x) = { tt (((3x)/2 ,x>0) , (2x, x <= 0))`

What is the area bounded by the curves ?
NDA Paper 1 2016
(A)

`(17)/6` sq units

(B)

`8/3` sq units

(C)

`2` sq units

(D)

`1/3` sq units

Solution:

When `x > 0, f(x) =x^(2) -1, g(x) = (3x)/2`

and when `x < 0, f(x) = - x^(2) - 1, g(.x) = 2x`

` = | int_(0)^(2) [(3x)/2 - (x^(2) -1)] dx | + | int_(-1)^(0) [(-x^(2) -1)-2x]| dx`

` = | int_(0)^(2) ((3x)/2 - x^(2) + 1) dx | + | int_(-1)^(0) (-1) (x^(2) + 2x - 1) | dx`

` = [3/2 .x^(2)/2 - x^(3)/3 +x ]_(0)^(2) + [x^(3)/3 - (2x^2)/2 + x]_(-1)^(0)`

` =(3/2 xx 4/2 - 8/3 +2 ) + [ 0 - (-1/3 +1 -1)]`

` = (3 - 8/3 +2 ) +1/3 `

`= 5 -7/3 = 8/3` sq units
Correct Answer is `=>` (B) `8/3` sq units
Q 2166380275

Consider the curves `y = | x - 1 |` and `| x |= 2`

What is the area of the region bounded by the curves and
x-axis?
NDA Paper 1 2016
(A)

`3` sq units

(B)

`4` sq units

(C)

`5` sq units

(D)

`6` sq units

Solution:

Given curves

`y = | x - 1|` and `| x | = 2`

` => { tt ((x - 1, x >= 1 ),( -x (x-1) , x < 1 text(and) x = pm 2))`

`:.` Required area

` = int _(-2)^(1) - (x- 1) dx + int _(1)^(2) (x -1) dx`

` = int _(-2)^(1) -x) dx + int _(1)^(2) dx`

` = [x - x^( 2)/2]_(-2)^(1) + [ x^( 2)/2 - x]_(1)^(2)`

` = [(1- 1/2) -( -2- 4/2)] + [(4/2 -2 )-(1/2 -1 )]`

`= [1/2 - (-4) ] + [ 0 - (-1/2) ]`

` = (1/2 + 4) + 1/2 = 5` sq units
Correct Answer is `=>` (C) `5` sq units
Q 2751380224

What is the area bounded by the curves `|y| = 1-x^2 ?`
NDA Paper 1 2016
(A)

`4/3` square units

(B)

`8/3` square units

(C)

`4` square units

(D)

`16/3` square units

Solution:

`y = 1-x^2`

`|y| = 1-x^2`

`A = 4 int_(0)^(1) (1-x^2)dx`

` = 4 [ x - x^3/3]_(0)^(1) = 4 (1-1/3) = 8/3`
Correct Answer is `=>` (B) `8/3` square units
Q 1649801713

Consider the line `x = sqrt(3)y`
and the circle `x^2 + y^ 2 = 4`.

What is the area of the region in the first quadrant
enclosed by the `X`-axis, the line `x = sqrt(3)` and the
circle?
NDA Paper 1 2015
(A)

`pi/3 - sqrt(3)/2`

(B)

`pi/2 - sqrt(3)/2`

(C)

`pi/3 - 1/2`

(D)

None of the above

Solution:

Required area is shaded area represented in the
following figure

:. Required area `= int _sqrt(3)^2 sqrt(4 - x^2 ) dx`

` = 1/2 [ x sqrt(4 - x^2 ) + 4 sin^(-1)(x/2)]_sqrt(3)^2`

` = 1/2 [ 4 sin^(-1) (1) - sqrt(3) sqrt(1) - 4 sin^(-1) ( sqrt(3)/2)]`

` = 1/2 [ 4 . pi/2 - sqrt(3) -4 xx pi/3]`

` = pi - (2pi)/3 - sqrt(3)/2 = pi/3 - sqrt(3)/2`
Correct Answer is `=>` (A) `pi/3 - sqrt(3)/2`
Q 1669801715

Consider the line `x = sqrt(3)y`
and the circle `x^2 + y^ 2 = 4`.

What is the area of the region in the first quadrant
enclosed by the X-axis, the line `x = sqrt(3)y` and the
circle?

NDA Paper 1 2015
(A)

`pi/3`

(B)

`pi/6`

(C)

`pi/3 - sqrt(3)/2`

(D)

None of these

Solution:

Required area is the shaded area represented in

the following figure

Let us first find the intersection of line `x = sqrt(3) y` and

circle `x^ 2 + y^2 = 4`

On solving these two, we get

`x = pm - sqrt(3)` and `y = pm 1`

:. Required area = Area of region `OBCO`

= Area of ` Delta OAC +` Area of region `ABCA`

` = 1/2 xx sqrt(3) xx 1 + pi/3 - sqrt(3)/2 = pi/3`
Correct Answer is `=>` (A) `pi/3`
Q 1609801718

Consider the curves `y = sin x` and `y = cos x`.

What is the area of the region bounded by the
above two curves and the lines `x = 0` and `x = pi/4 `?
NDA Paper 1 2015
(A)

`sqrt(2) - 1`

(B)

`sqrt(2) + 1`

(C)

`sqrt(2)`

(D)

`2`

Solution:

Given equation of curves are

`y = sinx ... (i)`

and `y = cosx ... (ii)`

The graph of above curves between `0` to `pi/2` is


Required area = Area of region `OABO`

`= int _0^(pi/4) (cosx-sinx) dx`

`=[sin x + cos x]_0^(pi/4)`

`= 1/sqrt(2) + 1/sqrt(2) - 0 - 1 = sqrt(2) - 1`
Correct Answer is `=>` (A) `sqrt(2) - 1`
Q 1629001811

Consider the curves `y = sin x` and `y = cos x`.


What is the area of the region bounded by the
above two curves and the lines `x = pi/4` and `x = pi/2`?
NDA Paper 1 2015
(A)

`sqrt(2) -1`

(B)

`sqrt(2) +1`

(C)

`2sqrt(2)`

(D)

`2`

Solution:

Given equation of curves are

`y = sinx ... (i)`

and `y = cosx ... (ii)`

The graph of above curves between `0` to `pi/2` is

Required area = Area of region `ACDA`

` = int_(pi/4)^(pi/4) ( sinx - cosx )dx`

`= [- cosx- sinx]_(pi/4)^(pi/2)`

`=- [cosx + sinx]_(pi/4)^(pi/2)`

`= - [0+1-(1/sqrt(3) + 1/sqrt(3) )]= sqrt(2) -1`
Correct Answer is `=>` (A) `sqrt(2) -1`
Q 2241480323

The area bounded by the coordinate axes and the curve
`sqrt(x) + sqrt(y) = 1` is
NDA Paper 1 2015
(A)

`1` sq unit

(B)

`1/2` sq unit

(C)

`1/3` sq unit

(D)

`1/6` sq unit

Solution:

`:.` Required area `= int _0^1 ( 1- sqrt(x))^2 dx`

` = int _0^1 (1+ x-2 sqrt(x)) dx`

`= [ x +x^2/2 -2 . x^(3//2)/(3//2)]_0^1`

` = 1 + 1/2 - 2 xx 2/3 (1)`

` = 1 + 1/2 - 4/3 = 1/6 `sq unit
Correct Answer is `=>` (D) `1/6` sq unit
Q 1659267114

The line `2y = 3x + 12` cuts the
parabola `4 y = 3x^2`

What is the area enclosed by the parabola and the line?
NDA Paper 1 2014
(A)

27 sq units

(B)

36 sq units

(C)

48 sq units

(D)

S4 sq units

Solution:

Area enclosed by the parabola and the line

` = int _(-2)^4 [((3x + 12)/2 - (3x^2)/4 )] dx`

` = 1/2 [ (3x^2)/2 + 12x ]_(-2)^(4) - 3/4 [(x^3)/3] _(-2)^(4)`

` = 1/2 [ {(3(4)^2)/2 + 12(4) } - { (3(-2)^2)/2 + 12(-2) }] - 3/4 [ 4^3/3 - (-2)^3/3]`

` = 1/2 [24 + 48- 6 + 24] - 3/4 [(64 + 8)/3]`

` = 1/2 (90) - 18`

` = 45 - 18 = 27` sq units
Correct Answer is `=>` (A) 27 sq units
Q 1619767610

What is the area of the parabola `y^2 = 4bx` bounded by its
latus rectum?
NDA Paper 1 2014
(A)

`(2b^2) /3` sq units

(B)

`(4b^2) /3` sq units

(C)

` b^2 ` sq units

(D)

`(8b^2) /3` sq units

Solution:

Equation of parabola is `y^2 = 4bx`.

Area of parabola bounded by its latus rectum

` 2. int _0^b sqrt(4bx) dx`

` = 4 sqrt(b) x 2/3 [ x^(3//2)] _0^b = (8 sqrt(b))/3 [ b^(3//2)-0]`

` = (8b^2)/3 ` sq units
Correct Answer is `=>` (D) `(8b^2) /3` sq units
Q 1753223144

Consider an ellipse, `x^2/a^2 + y^2/b^2 = 1`

What is the area included between the ellipse and the
greatest rectangle inscribed in the ellipse?
NDA Paper 1 2014
(A)

`ab (pi - 1)`

(B)

`2ab (pi - 1)`

(C)

`ab (pi - 2)`

(D)

None of these

Solution:

Given ellipse, `x^2/a^2 + y^2/b^2 = 1`

We know that,

Area of the ellipse `x^2/a^2 + y^2/b^2 = 1` is, `Delta ' = pi ab`

`:.` Required area =Area of shaded region

=Area of an ellipse

-Area of greatest rectangle

`= pi ab - 2ab`

` = ab (pi -2)`
Correct Answer is `=>` (C) `ab (pi - 2)`
Q 2379667516

What is the area bounded by the lines `x = 0, y = 0`
and `x + y + 2 = 0?`
NDA Paper 1 2013
(A)

`1/2` sq unit

(B)

`1` sq unit

(C)

`2` sq units

(D)

`4` sq units

Solution:

Given equation of lines `x= 0, y = 0` and `x + y + 2 = 0`


`therefore` Required area = Area of `Delta OAB`


` = 1/2 xx OA xx OB = 1/2xx 2 xx 2 = 2 qs ` units


Alternate Method

`therefore` Required area = `| underset(-2) overset(0) inty dx| = |underset(-2) overset(0) int(-2-x)dx|`

` = |[-2x-x^2/2]_(-2)^(0)| = |[0+2(-2)+(-2)^2/2]|`


` = |-4+2| = 2 qs ` units
Correct Answer is `=>` (C) `2` sq units
Q 2349767613

What is the area of the parabola `x^2 = y` bounded
by the line `y = 1?`
NDA Paper 1 2013
(A)

`1/3` qs units

(B)

`2/3` qs units

(C)

`4/3` qs units

(D)

`2` qs units

Solution:

Given equation of parabola and line are


`x^2 = y` ................(i)


and `y = 1` ................(ii)

On solving Eqs. (i) and (ii), we get


`x^2 = 1 => x = pm1`


`therefore` Required area = `2 xx` Area of OPBO


` = 2 int_0^1 sqrty dy`


`2[(2y^(3/2))/3]_0^1 = 4/3(1-0) = 4/3` sq units

Area along x axis

`= int_(-1)^1 (1 - y) dx = int_(-1)^1 [1 - x^2] dx `

`= 2 int_0^1 [1 - x^2] dx `

`= 2 ([x - x^3/3]_0)^1 = 2 [1-1/3]`

`4/3`
Correct Answer is `=>` (C) `4/3` qs units
Q 2369867715

What is the area bounded by `y = tanx, y = 0` and `x = pi/4 ?`
NDA Paper 1 2013
(A)

`log 2` sq units

(B)

`(log2)/2` sq units

(C)

`2 (log 2)` sq units

(D)

None of these

Solution:

Given equation of curves
`y = tanx` ................(i)

and `y= 0` and `x = pi/4` .........(ii)


`therefore` Requrred area = `int_0^(pi/4) ydx = int_0^(pi/4) tanxdx`


`[log|secx|_0^(pi/4) = log|sec(pi)|-log|sec0|`


`log|sqrt2|-log|1| = log sqrt2-0 = 1/2 log2` sq units
Correct Answer is `=>` (B) `(log2)/2` sq units
Q 2319067810

What is the area of the region enclosed by
`y = 2 |x|` and `y = 4?`
NDA Paper 1 2013
(A)

2 sq units

(B)

4 sq units

(C)

8 sq units

(D)

16 sq units

Solution:

Given curves



`therefore` Required area = area of `Delta OAB = 1/2xxOP xx AB`

` = 1/2xx4xx4 = 8` sq units
Correct Answer is `=>` (C) 8 sq units
Q 2379067816

What is the area of the parabola `y^2 = x` bounded
by its latusrectum?
NDA Paper 1 2013
(A)

`1/12` sq units

(B)

`1/6` sq units

(C)

`1/3` sq units

(D)

None of these

Solution:

Given curve, `y^2= x`

On compare with `y^2 = 4ax`, we get



`4a = 1`


`a = 1/4`

`therefore` Focus of the parabola, `S = (a , 0) = (1/4 , 0)`


`therefore` Required area =`2 * int_0^(1/4) sqrtxdx =2[2/3x^(3/2)]_0^(1/4)`


` = 4/3[(1/2)^3-0]`


` = 4/3 * (1/2)^3 = 4/3xx1/8 = 1/6 ` sq units
Correct Answer is `=>` (B) `1/6` sq units
Q 2319167910

What is the area of the portion of the curve
`y = sin x`, lying between `x = 0, y = 0` and `x = 2pi?`
NDA Paper 1 2012
(A)

1 sq unit

(B)

2 sq units

(C)

4 sq units

(D)

8 sq units

Solution:

`therefore` Required area (OBAB'C)

` = int_0^pisinxdx+int_pi^(2pi)-sinxdx`


` = [-cosx]_0^(pi)+[cosx]_0^(2pi)`


` = (cospi-cos0)+(cos2pi-cospi)`

` = -(-1-1)+(1+1) = 4` sq units
Correct Answer is `=>` (C) 4 sq units
Q 2369167915

What is the area of the region bounded by the lines `y = x , y = 0` and `x = 4 ?`
NDA Paper 1 2012
(A)

4 sq units

(B)

8 sq units

(C)

12 sq units

(D)

16 sq units

Solution:

`therefore` Required area = area `(DeltaOAB) = 1/2xx4xx4 = 8` sq units




Alternate Method
`therefore` Requird area OABO = `int_0^4ydx`


` = int_0^4xdx = [x^2/2]_0^4 = 1/2(4)^2 = 8` sq units
Correct Answer is `=>` (B) 8 sq units
Q 2319178919

The area bounded by the curve `x = f(y)`, the
Y-axis and the two lines `y = a` and `y = b` is equa to
NDA Paper 1 2012
(A)

`int_a^bydx`

(B)

`int_a^by^2dx`

(C)

`int_a^bxdy`

(D)

None of these

Solution:

Requires area ` = int_(y = a)^(y = b)xdy`
Correct Answer is `=>` (C) `int_a^bxdy`
Q 2359180014

What is the i area bounded by the curve
`sqrtx + sqrty = sqrta` where, `(x,y ge 0)` and the coordinate
axes?
NDA Paper 1 2011
(A)

`(5a^2)/6`

(B)

`a^2/3`

(C)

`a^2/2`

(D)

`a^2/6`

Solution:

The given equation of curve


`sqrtx+sqrty = sqrta` (where,`x,y ge 0`)


`=> sqrty = sqrta-sqrtx => (sqrty)^2 = (sqrta-sqrtx)^2`


`y = ( sqrta-sqrtx)^2`


At `x = 0 , sqrty = sqrta => y = a`

At `y = 0 , sqrtx = sqrta => x = a`


So, curve cuts the axes at (a, 0) and (0, a), respectively


`therefore` Required area `= int_(x = 0)^(x = a) ydx = int_0^a (sqrta-sqrtx)^2dx`


` = int_0^a (a+x-2sqrtasqrtx)dx = [ax+x^2/2-4/3 sqrt(ax^(3/2))dx]_0^a`

` = a^2+a^2/2-4/3 sqrta * a^(3/2) = (3a^2)/2-4/3a^2 = (9-8)/6a^2 = a^2/6`
Correct Answer is `=>` (D) `a^2/6`
Q 2309180018

What is the area bounded by the curve `y = cos 3x`,
where `0 - le x le pi/6` and the coordinate axes?
NDA Paper 1 2011
(A)

`1` sq unit

(B)

`1/2` sq unit

(C)

`1/3` sq unit

(D)

`1/4` sq unit

Solution:

Given curve `y = cos 3x`, `0 - le x le pi/6`


`therefore` Required area =` int_(x = 0)^(pi/6)ydx = int_0^(pi/6)cos3xdx`



` = [(sin3x)/3]_0^(pi/6) = 1/3{sin(pi/2)-sin0}`


` = 1/3xx1 = 1/3` sq unit
Correct Answer is `=>` (C) `1/3` sq unit
Q 2319280110

What is the area enclosed by the equatic


`x^2+y^2 = 2?`
NDA Paper 1 2011
(A)

`4pi` sq units

(B)

`2pi` sq units

(C)

`4pi^2`sq units

(D)

`4`sq units

Solution:

Given curve is the equation of circle,

`x^2+y^2 = (sqrt2)^2`


`therefore` Required area = `4xx int_(x = 0)^(x = sqrt2 )ydx`


` = 4int_0^sqrt2 sqrt(2-x^2)dx`


` = 4[x/2sqrt(2-x^2)+2/2sin^(-1) (x/sqrt2)]_0^sqrt2`


` = 4(pi/2-0) = 2pi` sq units


Alternate Method
We know that. Area of cricle = `pi (text(radius))^2`

` = pi(sqrt2)^2 = 2pi` sq units
Correct Answer is `=>` (B) `2pi` sq units
Q 2349280113

What is the area bounded by the curves `y = e^x`,
`y =e^(-x)` and the straight line `x = 1?`
NDA Paper 1 2011
(A)

`(e+1/e)` sq unit

(B)

`(e-1/e)` sq unit

(C)

`(e+1/e -2 )` sq unit

(D)

`(e- 1/e - 2)` sq unit

Solution:

The equation of curves are `y =e^x` and `y= e^(-x)`



`therefore e^x = 1/e^x`


`=> e^(2x) = e^0 => x = 0`


`therefore` Required area = `int_0^1(e^x-e^(-x))dx = [e^x+e^(-x)]_0^1`



` = e+e^(-1)-e^0-e^0`


` = (e+1/e-2)` sq unit
Correct Answer is `=>` (C) `(e+1/e -2 )` sq unit
Q 2369280115

What is the area under the curve `f(x) = xe^x` above
the X-axis and between the lines `x = 0` and `x = 1?`
NDA Paper 1 2010
(A)

`1/3` sq unit

(B)

`1` sq unit

(C)

`3/2` sq unit

(D)

`2` sq unit

Solution:

Required area = `int_0^1se^xdx = [xe^x-inte^xdx]_0^1`



` = [xe^x-e^x]_0^1 = (e-e) -(0-1)`

`= 1` sq units
Correct Answer is `=>` (B) `1` sq unit
Q 2319280119

What is the area bounded by the curve `y = x^2` and the line `y = 16?`
NDA Paper 1 2010
(A)

`32/3` sq units

(B)

`64/3` sq units

(C)

`256/3` sq units

(D)

`128/3` sq units

Solution:

The equations of curves are
`y= x^2`............(i)
and `y = 16`................(ii)

On solving Eqs. (i) and (ii), we get

`x = 4` and `- 4`
So, the points of intersection are (4, 16) and {-4, 16).

`therefore` Required area = `int_(-4)^4(16-x)^2dx = 2int_0^4(16-x^2)dx`

`[because f(x) = 16-x^2` is an even function `=> f(-x) = f(x)]`


` = 2[16x-x^3/3]_0^4 = 2[64-64/3]`


` = (128xx2)/3 = 256/3` qs units
Correct Answer is `=>` (C) `256/3` sq units
Q 2369380215

If `f(x) = 1-x^2/4 , x xi [-2 , 2]` then find the area covered by X - axis
NDA Paper 1 2010
(A)

`8/3` sq units

(B)

`4/3` sq units

(C)

`2/3` sq units

(D)

`1/3` sq units

Solution:

Required area = `int_(-2)^(2)(1-x^2/4)dx`


` = 2(x-x^3/12)_0^2 = 2(2-2^3/12)`


`[ because f(x) = (1-x^2/4)` is an even function `=> f(-x) = f(x)]`


` = 2(1-2/3) = 8/3` sq units
Correct Answer is `=>` (A) `8/3` sq units
Q 2319480310

What is the area enclosed between the curves
`y^2 = 12x` and the lines `x = 0` and `y= 6?`
NDA Paper 1 2010
(A)

2 sq units

(B)

4 sq units

(C)

6 sq units

(D)

8 sq units

Solution:

Equation of curve is
`y^2 = 12x`

At `y = 6, 36 =' 12x => x = 3`



`therefore` Required area = `int_0^3(6-sqrt(12x)) dx`


` = [6x]_0^3-sqrt(12)((2x^(3/2))/3)_0^3`


` = [6xx3]-(sqrt(12) xx2 xx sqrt(27))/3`


` = 18-12 = 6` sq units
Correct Answer is `=>` (C) 6 sq units
Q 2319580410

What is the area bounded by the curve
`y = 4x- x^2- 3` and the X-axis?
NDA Paper 1 2009
(A)

`2/3` sq units

(B)

`4/3` sq units

(C)

`5/3` sq units

(D)

`3/5` sq units

Solution:

Given `y -4x -x^2-3`


`=> -(x^2-4x) = y+3`


`=> -(x^2-4x+4) = y+3-4`


`=> (x-2)^2 = -(y-1)`

Let `X^2 = -y`

Where `X = x-2` and `Y = y-1`

which is a required equation of parabola intercept on X- axis

Put `y = 0`

`x^2-4x+3 = 0`

`=> (x-1)(x-3) = 0`

` x = 1 ,3`
Correct Answer is `=>` (B) `4/3` sq units
Q 2359580414

What is the value of K if the area bounded by the curve `y = sinkx , y = 0 , x = pi/k , x = pi/(3k)` is 3 sq units ?
NDA Paper 1 2009
(A)

`1/2`

(B)

`1`

(C)

`3/2`

(D)

`2`

Solution:

Area `A = int_(pi/(3k))^(pi/k) sinkxdx`



`=> 3 = -((coskx)/k)_(pi/(3k))^(pi/k)`


`=> 3 = -1/k(cos pi - cos (pi/3))`

`=> 3 = -1/k(-1-1/2)`

`=> 3 = 3/(2k)`

`therefore k = 1/2`
Correct Answer is `=>` (A) `1/2`
Q 2329680511

What is the area of the region bounded by the line `3x-5y = 15 , x = 1 , x = 3` and X- axis?
NDA Paper 1 2008
(A)

`36/5` sq units

(B)

`18/5` sq units

(C)

`9/5` sq units

(D)

`3/5` sq units

Solution:

The given equation of line can be rewritten as

`x/5-y/3 = 1` and `y = (3x-15)/5`


`therefore` Required area = `int_1^3ydx`


` = int_1^3((3x-15)/5)dx = 1/5int_1^3(3x-15)dx`


`= 1/5((3x^2)/2-15)_1^3 = 1/5(27/2-45-3/2+15)`

` = 1/5(24/2-30) = 1/5(12-30)`

` = -18/5 = 18/5` sq units (neglective -ve sign)
Correct Answer is `=>` (B) `18/5` sq units
Q 2339780612

What is the area enclosed by the curve

`2x^2+y^2 = 1?`
NDA Paper 1 2007
(A)

`2pi`

(B)

`pi`

(C)

`pi/2``

(D)

`pi/sqrt2`

Solution:

Given curve is `2x^2+y^2 = 1`

This curve can be written as `x^2/(1/2)+y^2/1 = 1` which is an ellipse.

Where `a^2 = 1/2 , a = 1/sqrt2` and `b^2 = 1 , b = 1`



`therefore` area of an ellipse = 4 `xx` Area of circle ABO


` = 4 int_0^(1/sqrt2)ydx = 4 int_0^(1/sqrt2) sqrt(1-2x^2)dx = 4int_0^(1/sqrt2) sqrt(1-sqrt(2x)^2)dx`


Let `t = sqrt2x dt = sqrt2 *dx = 4/sqrt2[int_0^1sqrt(1-t^2)dt]`



` = 4/sqrt2{t/2 sqrt(1-t^2)+1/2sin^(-1)t}_0^1`


` = 4/sqrt2[1/2*0+1/2* sin^(-1)(1)-0-0] = 4/sqrt2*1/2*pi/2 = pi/sqrt2`


Alternate Method
We know that, the area of an ellipse `= pi ab`

` = pi(1/sqrt2)(1) = pi/sqrt2`
Correct Answer is `=>` (D) `pi/sqrt2`
 
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