Quizrr Chapter-Wise Test for JEE Advanced - 2025

By: C I P H Ξ R

Q.1

Area bounded by y = π [sin2 x

2
​ + cos x2 ] and y = ∣∣sin x − 12 ∣∣ , x = 0, x = π where [.] represents the
​ ​ ​ ​ ​

greatest integer function is.

(A) 1 (7π + 12(1 − 3))

6
​ ​

(B) 7π + 12(1 + 3)
​

(C) π2 ​ +1− 3 ​

(D) π2 ​ +1+ 3 ​

Q.2

Area bounded by the curve y = max{sin x, cos x} and x-axis, between the lines x = π/4 and x = 2π is
equal to

(A) (4 2−1) ​

2
​

sq. units (
(B) (4 2 − 1) sq.units ​

(C) (4 2−1) sq. units ​

2 ​

(D) None of these

Q.3

Area enclosed by the figure described by the equation x4 + 1 = 2x2 + y 2 is

(A) 3
(B) 83 ​

(C) 73 ​

(D) 10
3
​

Q.4

Area bounded by the curves y = ∣x∣ − 2 and y = 1 − ∣x − 1∣ is equal to -

(A) 4 sq. units

(B) 6 sq. units

(C) 2 sq. units
(D) 8 sq. units
@IITJEE_Advanced
Q.5

ln x
The area bounded by the curves y = ex ln x and y = ex
is
​

5
(A) 4e ​ + 4e
​

5
(B) 4e ​ − 4e
​

4
(C) 4e ​ − 5e
​

4
(D) 4e ​ + 5e
​

Q.6

The area between the curve y = 2x4 − x2 , the x - axis and the ordinates of the two minimum-points on the curve
is
7
(A) 12 ​

7
(B) 30 ​

7
(C) 60 ​

7
(D) 120 ​

Q.7

A rea bounded by the curves y = sin x, tangent drawn to it at x = 0 and the line x = π/2, is equal to
2
(A) π −4 sq. units
2
​

2
(B) π −8 sq. units
8
​

2
(C) π −2 sq.units
4 ​

2
(D) π −2 sq. units
2
​

Q.8

The area between the curves y = x and y = x is

​

(A) 1/3
(B) 1/6
(C) 2/3
(D) 1

Q.9

Consider the curve whose equation is y = 1 + 4x − x2 .

(A) A rea between the curve and the x-axis is less than 10 5 sq. units
​

(B) A rea between the curve, the y -axis and the line through the origin and that highest point of the curve is 73 sq.
​

units.
@IITJEE_Advanced
(C) If the line y = mx bisects the area between the curve, y = 0, x = 0 and x = 23 , then m =
​
13
16
​

(D) The curve is symmetrical about the line x = 2.

Q.10

For which of the following values of m, is the area of the region bounded by the curve y = x − x2 and the line
y = x − x2 and the line y = mx equals 9/2 ?
(A) -4

(B) -2

(C) 2
(D) 4

Q.11

The area of the region bounded between the curves ∣y∣ − ∣ sin x∣ ≥ 0 and x2 + y 2 − π 2 ≤ 0 is π 3 − A. Find
the values greater than A.

(A) 9

(B) 10
(C) 11
(D) 12

Q.12

Passage
≡ [f(y)]2/3 + [f(x)]1/3 = 0 and C2 ≡ [f(y)]2/3 + [f(x)]2/3 = 12, satisfying the relation
Two curves C1 ​ ​

(x − y)f(x + y) − (x + y)f(x − y) = 4xy (x2 − y2 ).

Question:
The area bounded by C1 and C2 is - ​ ​

(A) 2π − 3 3 sq. units ​

(B) 2π + 3 sq. units

​

(C) π+ 6 sq. units

​

(D) 2 3 − π sq. units

​

Q.13

Passage
Two curves C1 ≡ [f (y)]2/3 + [f (x)]1/3 = 0 and C2 ≡ [f (y)]2/3 + [f (x)]2/3 = 12, satisfying the
​ ​

relation (x − y)f (x + y) − (x + y)f (x − y) = 4xy (x2 − y 2 ).

Question:
The area bounded by the curve C2 and ∣x∣ + ∣y∣ ​ = 12 is - ​

(A) 12π − 24 sq. units @IITJEE_Advanced

(B) 6− 12 sq. units
​

(C) 2 12 − 6 sq. units

​

(D) None of these

Q.14

Passage
Two curves C1 ≡ [f (y)]2/3 + [f (x)]1/3 = 0 and C2 ≡ [f (y)]2/3 + [f (x)]2/3 = 12, satisfying the
​ ​

relation (x − y)f (x + y) − (x + y)f (x − y) = 4xy (x2 − y 2 ).

Question:
The area bounded by C1 and x + y ​ + 2 = 0 is -
(A) 5/2 sq. units
(B) 7/2 sq. units
(C) 9/2 sq. units
(D) None of these

Q.15

Passage
−1 −1
Consider the function defined implicity by the equation y2 − 2yesin x + x2 − 1 + [x] + e2 sin x
= 0,
(where [x] denotes the greatest integer function).
Question:
The area of the region bounded by the curve and the line x = −1 is -

(A) π + 1 sq. units

(B) π − 1 sq. units
π
(C) 2 ​ + 1 sq. units
(D) π2 ​ − 1 sq. units

Q.16

Passage
−1 −1
Consider the function defined implicity by the equation y 2 − 2yesin x + x2 − 1 + [x] + e2 sin x = 0,
(where [x] denotes the greatest integer function).
Question:
Line x = 0 divides the region mention above in two parts. The ratio of area of the left-hand side of line to that of
right-hand side of line is -

(A) 1+π :π
(B) 2−π :π
(C) 1:1
(D) π+2:π
@IITJEE_Advanced
Q.17

Passage
−1 −1
Consider the function defined implicity by the equation y 2 − 2yesin x
+ x2 − 1 + [x] + e2 sin x
= 0,
(where [x] denotes the greatest integer function).
Question:
1
The area of the region of curve and the line x = 0 and x = 2
​is −
3 π
(A)
4
​

​
+ 6 sq. units
​

3 π
(B)
2
​

​ + 6 sq. units
​

3 π
(C)
4
​

​ − 6
​sq. units
3 π
(D)
2
​

​ − 6
​sq. units

Q.18

Match the column -

(A) A-p ; B-q ; C-q ; D-r

(B) A-p ; B-q ; C-r ; D-q

(C) A-p ; B-r ; C-r ; D-q

(D) A-q ; B-s ; C-s ; D-r

Q.19

Consider the collection of all curves of the form y= a − bx2 that passes through the point (2, 1) where a and b
are positive real numbers. If the minimum area of the region bounded by y = a − bx2 and the x-axis is A, then ​

find the value of A

8
. ​

Q.20
@IITJEE_Advanced
 = 1 − x2 at x = α, where 0 < α < 1, meets the axes at P and Q. If α varies and
If the tangent to the curve y
the minimum value of the area of the triangle OPQ is k times the area bounded by the axes and the part of the curve
0 < x < 1, then find 3k . ​

Answers & Solutions

Q.1 Answer:
—
Solution:

1 ≤ sin2 x
2
​ + cos ( x2 ) < 2. ⇒ [sin2
​ ​
x
2
​ + cos x2 ] = 1
​ ​

∴ required area is

= π − 2 (∫ (sin x − ) dx)
π/6 π/2
1 1
( − sin x) dx + ∫
2 2
​ ​ ​ ​

0 π/6
​ ​

7π
= + 2(1 − 3)
6
​ ​

Q.2 Answer:
(4 2−1) ​

2
​

sq. units (
Solution:

Bold lines represents the graph of y = max{sin x, cos x}. Required area,

@IITJEE_Advanced
 π 5π/4 3π/2 2π (4 2−1)
Δ = ∫π/4 sin xdx − ∫π sin xdx − ∫5π/4 cos xdx + ∫3π/2 cos xdx = sq.units
​

2
​ ​ ​ ​ ​

Q.3 Answer:
8
3
​

Solution:
2
y 2 = ( x2 − 1 )
So y = x2 − 1 or y = 1 − x2
​ ​

1
(1 − x ) dx = 4 [x − ]
1 2 x3 8
So required area = 4 ∫0 ​

3
​ ​ = 3
​

Q.4 Answer:
4 sq. units
Solution:

Bounded figure ABCD is a rectangle.

AB = 1+1=​ 2, BC =
​ 4 + 4 = 2 2 Thus, bounded area = ( 2)(2 2) = 4 sq. units.
​ ​ ​ ​

@IITJEE_Advanced
Q.5 Answer:
e 5
4
​ − 4e
​

Solution:
ln x
The curves meet at x given by ex ln x = ex
​

1
⇒x= ​

e
1
ln x
A=∫ ​ ( − ex ln x) dx
​

1
e
​
ex
1 1
1∣ x2 x2 ∣
(ln x) − e ( ln x − )
​ ​

2
=
2e ∣ 1 2 4 ∣1
​ ​ ​ ​ ​ ​ ​

e e
​ ​

1 e 3 e 5
=− + − = −
2e 4 4e 4 4e
​ ​ ​ ​ ​

Q.6 Answer:
7
120
​

Solution:

y = 2x4 − x2 , y ′ = 8x3 − 2x = 2x (4x2 − 1)

y ′ = 0 ⇒ x = 0, ± 12 ​

y is maximum at x = 0
and minimum at x = ± 12 ​

2 [ x3 x]
1 1 3
2 4 2 4 2 5 2 1 1 7
​

A = ∫− 1 (x − 2x ) dx = 2 ∫0 (x − 2x ) dx =
2 2
− = 2 [ 24 − ] =
​ ​

5 80 120
​ ​ ​ ​ ​ ​ ​ ​

2
​

Q.7 Answer:
π 2 −8
8 sq. units
​

Solution:

@IITJEE_Advanced
 The tangent drawn to y = sin x at x = 0 is the line y = x. Clearly the line y = x lies above the graph
π
of y = sin x∀x ∈ (0, 2
).​

π/2
( x2 + cos x)
2
π/2 π 2 −8
Thus required area Δ = ∫0 (x ​
− sin x)dx = ​ ​
= 8 sq. units.
​

Q.8 Answer:
1/6
Solution:

The points of intersection of curves are x = 0 and x = 1

1
[ 2x3 ]
1 3/2
x2
∴ required area = ∫0 ( ​ x − x)dx =
​ ​ − 2
​ ​ = 2
3
​ − 1
2
​ = 1
6
​

Q.9 Answer:
A rea between the curve and the x-axis is less than 10 5 sq. units
​

Solution:

(x − 2)2 = −(y − 5) which is a parabola with vertex at (2, 5) and inverted U type.
This meets the x-axis at (2 − 5, 0) and (2 + 5, 0) ​ ​

Area < area of rectangle with CB and DA as sides = (2 5) × 5 = 10 5 ​ ​

The line OA is y = 25 x ​

2
{(1 + 4x − x2 ) − ( 25 x)} dx = [x + 43 x2 − 3 ]
2 x3 7
Area required = ∫0 ​ ​ ​ ​ ​ = 3 ​

@IITJEE_Advanced
 3/2 1 3/2
∫0 ​ mxdx = 2
​ ∫0 ​ (1 + 4x − x2 ) dx

⇒ m2 ( 94 ) = 21 [ 32 + 2 ( 94 ) − 13 ( 27
​ ​ ​ ​

8 )] ⇒ m = ​ ​ ​
39
18 ​

∴ C is not true
D is evidently true.

Q.10 Answer:
-2
Solution:

Case I: When m =0

In this case y = x − x2 ...(i)

and y = 0 ...(ii)

are two given curves, y > 0 is total region above x-axis.

Therefore, area between y = x − x2 and y = 0 is area between y = x − x2 and above the x-axis

@IITJEE_Advanced
 1 1
x2 x3
∴ A = ∫ (x − x ) dx = [ − ] 2
2 3 0
​ ​ ​ ​

0
1 1 1 9
​ ​

= − = = 
2 3 6 2
​ ​ ​ ​

Hence, no solution exists.

Case II: When m <0

In this case area between

y = x − x2 and y = mx is

Case II: When m <0

In this case area between y = x − x2 and y = mx is

1
Put tan x2 ​
=t⇒ 2
​
sec2 x2 dx = dt
​

@IITJEE_Advanced
 tan π
4tdt
= ∫0 8
​

(1+t2 ) 1−t2
​ ​

as tan π8 = ​ 2−1
​

2−1 4tdt
So, ∫0 .
​

(1+t2 ) 1−t2
​ ​

OABCO and points of intersection are (0, 0) and {1 − m, m(1 − m)}

∴ Area of curve OABCO

1−m
= ∫0 ​ [x − x2 − mx] dx

1−m
x2 x3
= [(1 − m) − ]
2 3 0
​ ​ ​

1 1 1
= (1 − m)3 − (1 − m)3 = (1 − m)3
2 3 6
​ ​ ​

1 9
​

∴ (1 − m)3 = (given) ​

6 2
​ ​

⇒ (1 − m)3 = 27
⇒ 1−m=3
⇒ m = −2

Case III: When m >0

In this case y = mx and y = x − x2 intersect in (0, 0) and {(1 − m), m(1 − m)} as shown in Fig.

Area of shaded region

@IITJEE_Advanced
 0
=∫ ​ (x − x2 − mx) dx
1−m
x2 x3
⌊= (1 − m) − ⌋
2 3 1−m
​ ​ ​

1 1
= − (1 − m)(1 − m)2 + (1 − m)3
2 3
​ ​

1
= − (1 − m)3
​ ​

6
​

9 1
⇒ = − (1 − m)3 (given)
2 6
​

⇒ (1 − m)3 = −27
⇒ (1 − m) = −3
⇒ m = 3 + 1 = 4.

Therefore, (b) and (d) are the answers.

Q.11 Answer:
9
Solution:
π
Required area = π (π 2 ) − 4 ∫0 sin xdx = π 3 − 8
​

Q.12 Answer:
2π + 3 sq. units
​

Solution:

(i) Given : (x − y)f (x + y) − (x + y)f (x − y) = 4xy (x2 − y 2 )

= (x2 − y 2 ) [(x + y)2 − (x − y)2 ]
= (x − y)(x + y)3 − (x + y)(x − y)3
⇒ f(x + y) = (x + y)3 ⇒ f(x) = x3 , f(y) = y3
Now equation of given curves are @IITJEE_Advanced
 y2 + x = 0
x2 + y2 = 12
​ ​

∣ −3 ∣ ∣ 0 ∣
A = 2 [ ∫−2 3 12 − x2 dx + ∫−3 −xdx ]
∣−3 ∣ ∣−π ∣
​ ​ ​ ​ ​ ​ ​ ​

2
I1 = 2 ∫−2 3 12 − x dx = 2 ∫−π
​ ​ ​ ​

= 12 [∫−π/2 (1 + cos 2θ)dθ] = 12

​

−π/3
​

= 12 [ π6 −
​

4
3
] ​

​ = 2π − 3 3 ​

Solving eq. (1) and (2), we get x = −3, y = ± 3 ​

0
0 2[(−x)3/2 ]
= 2 ∫−3 −xdx = −3
= − 43 [0 − 33/2 ] = 4 3
​

The area bounded by curves I2 ​ ​ ​

−3/2
​ ​ ​

@IITJEE_Advanced
A = 2π − 3 3 + 4 3 = 2π +
​ ​ 3 sq. units
​

(ii) The required area is = area of circle − area of square

= 12π − 24 sq. units
2
(iii) The required area = ∫−1 (−y 2 − (−y − 2)dy
​

@IITJEE_Advanced
 2
y2 y3
= [ + 2y − ]
2 3 −1
​ ​ ​

4 8 1 1 9
= [ + 4 − − ( − 2 + )] = sq. units
​ ​

2 3 2 3 2
​ ​ ​ ​ ​

Q.13 Answer:
12π − 24 sq. units
Solution:

(i) Given : (x − y)f (x + y) − (x + y)f (x − y) = 4xy (x2 − y 2 ) =

(x2 − y 2 ) [(x + y)2 − (x − y)2 ]
= (x − y)(x + y)3 − (x + y)(x − y)3
⇒ f(x + y) = (x + y)3 ⇒ f(x) = x3 , f(y) = y3
Now equation of given curves are
y2 + x = 0

@IITJEE_Advanced
x2 + y2 = 12 ............. (2)

Solving eq. (1) and (2), we get x = −3, y = ± 3 ​

The area bounded by curves

∣ −3 ∣ ∣ 0 ∣
A = 2 [ ∫−2 3 12 − x2 dx + ∫−3 −xdx ]
∣ ∣ ∣ ∣
​ ​ ​ ​ ​ ​ ​ ​

−3 −π/3 2
I1 = 2 ∫−2 3 12 − x dx = 2 ∫−π/2 12 cos θdθ
​
2
​
​ ​ ​

= 12 [∫−π/2 (1 + cos 2θ)dθ] = 12 [θ + sin θ −π/3

= 12 [− π3 − + π2 ]
−π/3 3
​

​ ]
2 −π/2
​ ​ ​

4
​

​ ​

= 12 [ π6 −
​

4 ]
3 ​

​
= 2π − 3 3 ​

@IITJEE_Advanced
 0
0 2[(−x)3/2 ]
I2 = 2 ∫−3−xdx = = − 43 [0 − 33/2 ] = 4 3
−3
​

−3/2
​ ​ ​ ​ ​ ​

A = 2π − 3 3 + 4 3 = 2π + 3 sq. units
​ ​ ​

(ii) The required area is = area of circle − area of square = 12π − 24 sq. units
2
(iii) The required area = ∫−1 ​ (−y 2 − (−y − 2)dy

@IITJEE_Advanced
 2
= [ y2 + 2y − ]
2
y3
3
​ ​ ​

−1 ​

= [ 24 ​ +4− 8
3
​ − ( 12 −
​ 2+ 1
3
​)] = 9
2
​ sq. units

Q.14 Answer:
9/2 sq. units
Solution:

(i) Given : (x − y)f (x + y) − (x + y)f (x − y) = 4xy (x2 − y 2 ) =

(x2 − y 2 ) [(x + y)2 − (x − y)2 ]
= (x − y)(x + y)3 − (x + y)(x − y)3
⇒ f(x + y) = (x + y)3 ⇒ f(x) = x3 , f(y) = y3
Now equation of given curves are
y2 + x = 0

@IITJEE_Advanced
x2 + y2 = 12 ............. (2)

Solving eq. (1) and (2), we get x = −3, y = ± 3 ​

The area bounded by curves

∣ −3 ∣ ∣ 0 ∣
A = 2 [ ∫−2 3 12 − x2 dx + ∫−3 −xdx ]
∣ ∣ ∣ ∣
​ ​ ​ ​ ​ ​ ​ ​

−3 −π/3 2
I1 = 2 ∫−2 3 12 − x dx = 2 ∫−π/2 12 cos θdθ
​
2
​
​ ​ ​

= 12 [∫−π/2 (1 + cos 2θ)dθ] = 12 [θ + sin θ −π/3

= 12 [− π3 − + π2 ]
−π/3 3
​

​ ]
2 −π/2
​ ​ ​

4
​

​ ​

= 12 [ π6 −
​

4 ]
3 ​

​
= 2π − 3 3 ​

@IITJEE_Advanced
 0
0 2[(−x)3/2 ]
I2 = 2 ∫−3−xdx = = − 43 [0 − 33/2 ] = 4 3
−3
​

−3/2
​ ​ ​ ​ ​ ​

A = 2π − 3 3 + 4 3 = 2π + 3 sq. units
​ ​ ​

(ii) The required area is = area of circle − area of square = 12π − 24 sq. units
2
(iii) The required area = ∫−1 ​ (−y 2 − (−y − 2)dy

@IITJEE_Advanced
 2
= [ y2 + 2y − ]
2
y3
3
​ ​ ​

−1 ​

= [ 24 ​ +4− 8
3
​ − ( 12 −
​ 2+ 1
3
​)] = 9
2
​ sq. units

Q.15 Answer:
π
2
​ + 1 sq. units
Solution:

(i) For −1 ≤x<0

2
(y − esin ) = 2 − x2
−1
x

−1
y = esin x
± 2 − x2 ​

0 0
A=∫
​

​ (e sin−1 x
+ 2− x2 ) dx
​ − (e sin−1 x
− 2− x2 ) dx
​ = 2∫ ​ 2 − x2 dx
​
​

−1 −1
0 0
= 2( x ) = [1 + 2 (0 − (− ))] = + 1 sq. units
1 ∣ 2 x ∣ π π
2 − x2 + sin−1
2 2 2 ∣−1 4 2
​ ​ ​ ​ ​ ​ ​ ​ ​

∣−1 ​

@IITJEE_Advanced
 For 0 ≤ x < 1, y = sin−1 x ± 1 − x2 ​

1 ∣1
= 2[ ] = 0 + sin−1 (1) = sq. units
x ∣ 1
1
−1 x π
A = 2∫ 1− x2 dx 1− x2 + sin
2 ∣0 2 1 ∣0 2
​ ​ ​ ​ ​ ​ ​ ​ ​ ​

Total area = ( π2 + 1) + ​
π
2
​ =π+1
1 − x2 dx = 2 [ x2 1 − x2 ∣∣0 − sin−1 x∣∣0 ] =
π
+1 1/2 1/2 1 1/2
(ii) Ratio = 2
= π+2
(iii) A = 2 ∫0
​

π
π 2
​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​

2 ​

3 π
4
​

​ + 6
​sq. units

Q.16 Answer:
π+2:π
Solution:

(i) For −1 ≤x<0

2
(y − esin ) = 2 − x2
−1
x
−1
y = esin x ± 2 − x2 ​

A = ∫−1 (esin x + 2 − x2 ) dx − (esin x − 2 − x2 ) dx = 2 ∫−1

0 −1 −1 0
​ ​ ​ ​ 2 − x2 dx ​
​

0
= 2 ( 12 x 2 − x2 ∣∣−1 + x ∣
)
0 2
​ ​ ​ ​

2
​ sin−1 2 ∣−1
​

​
​ ​
= [1 + 2 (0 − (− π4 ))] = ​
π
2
​
+ 1 sq. units
For 0 ≤ x < 1, y = sin−1 x ±
1 − x2 ​

A = 2 ∫0 1 − x2 dx = 2 [ x2 1 − x2 ∣∣0 + sin−1 x1 ∣∣0 ] = 0 + sin−1 (1) =

1 1 1 1 π
​ ​ ​ ​ ​ ​

2
​ ​ ​ ​

2
​sq. units
Total area = ( π2 + 1) + π2 = π + 1 ​ ​

π
+1
(ii) Ratio = 2 π = π+2
​

π
​ ​

2 ​

1 − x2 dx = 2 [ x2 1 − x2 ∣∣0 − sin−1 x∣∣0 ] =

1/2 1/2 1 1/2 3
(iii) A = 2 ∫0 ​ ​ ​ ​ ​ ​

2
​ ​ ​

4
​

​ + π
6
sq. units
​

Q.17 Answer:
3
4
​

​ + π
6
sq. units
​

Solution:

(i) For −1 ≤x<0

2
(y − esin ) = 2 − x2
−1
x
−1
y = esin x ± 2 − x2 ​

A = ∫−1 (esin x + 2 − x2 ) dx − (esin x − 2 − x2 ) dx = 2 ∫−1

0 −1 −1 0
​ ​ ​ 2 − x2 dx ​
​

0
= 2 ( 12 x 2 − x2 ∣∣−1 + x ∣
)
0 2
​ ​ ​ ​

2
​ sin−1 2 ∣−1
​

​
​ ​ = [1 + 2 (0 − (− π4 ))] = ​
π
2
​ + 1 sq. units

For 0 ≤ x < 1, y = sin−1 x ± 1 − x2 ​

1 − x2 dx = 2 [ x2 sin−1 x1 ∣∣0 ] = 0 + sin−1 (1) =@IITJEE_Advanced

1 1 1
A = 2 ∫0 ​ ​ ​
1 − x2 ∣∣0 + ​ ​ ​
1
2
​ ​ ​ ​
π
2 sq. units
​
 Total area = ( π2 + 1) + ​
π
2
​ =π+1
π
+1 π+2
(ii) Ratio = 2 π = π
​

​ ​

2 ​

∣1/2
1 − x 2 dx = 2 [ − sin x ] =
1/2
x ∣ 1/2 1 3 π
(iii) A = 2 ∫ 1 − x2 −1
+ sq. units
​

2 ∣0 2 4 6
​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​

0 ∣0

Q.18 Answer:
A-p ; B-q ; C-r ; D-q
Solution:

(A) – p, (B) – q, (C) – r, (D) – q

(A) x∈ (−8, 8), y = 2 x ∈ (−8 2, −8] ∪ [8, 8 2), y = 3 ​ ​

Required area = 2 × 3 − 21 × 2 × 2 = 6 − 2 = 4 sq. units ​

e e
(B) The required area = ∫1 ​ ydx = ∫1 log xdx
​

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[x log x − x]e1 = (e log e − e) − (0 − 1) = e − e + 1 = 1
​

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(C)

The required area,

π/2
=2 ∫ ​ xdy where y = sin−1 x i.e. x
0
​

π/2 ​

= 2∫ ​ sin ydy = 2 [cos y [0π/2 = 2

0

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 1
2 ∫0 ydx, where y = x = 2 ∫0 xdx = 2 [ x2 ] = 1
1 1 2
(D) The required area = ​ ​ ​ ​

Q.19 Answer:
—
Solution:

y = a − bx2
when x = 2, y = 1
1=a−4b⇒a=1+4b
a/b
(a − bx2 ) dx = 2 [ax − bx3
] = 2 [a 3 b b]
a/b
​

a ba a
A = 2 ∫0 −
​
​ ​

3 ​ ​

b
​ ​ ​

0 ​

3 [ b ] 3 [ b ]
2 3a a a a 2 2a a 4 (1+4 b) 1+4 b
=
​

−
​

= ​

= 3
​

b b
​ ​ ​ ​ ​ ​ ​

​ ​ ​ ​

Since area (A) has to be minimum

dA
db
​ = 0.
As a result we get,

⇒ 12 b = 1 + 4 b ⇒ 8 b = 1 ⇒ b = 81 , a = ​
3
2
​ ∴ Amin = ​
4
3
​ ( 32 ) ​
3
2
​ ​ ⋅2 2=
​
4 3 2
​

2 ​
​

A
∴ Minimum area = 48 ⇒ a = 48 ⇒
​

8 ​ =6

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Q.20 Answer:
—
Solution:

A1 = area under the curve and axes

​

1
= ∫0 (1 − x2 ) dx = [x − ]
1 x3 2
​

3
​ ​ = 3
​ ……… ..(1)
0
Now, y = 1 − x2 ∴ y′ = −2x ∴ y′ (α, 1 − α2 ) = −2α Equation of tangent to the curve y =
1 − x2

⇒ (y − (1 − α2 )) = −2α(x − α)
⇒ 2αx + y = α2 + 1
α2 + 1
∴ P≡( , 0) ; Q ≡ (0, α2 + 1)
2α
​

​ ​

2
(α2 +1)
A = Area of triangle POQ = 21 (OP)(OQ) = 14 α ​ ​ ​

2
1 α(α +1)2α−(α +1)
2 2
dA 1 3α4 +2α2 −1
∴ dα = 4 ​ ​

α 2 = 4 α2
​ ​ ​

dA 1 1
For maximum/minimum, dα = 0 ⇒ 3α + 2α2 − 1 = 0 4
⇒ α2 = 3
⇒α=± 3
​ ​ ​

2
Since, 0 < α < 1 ⇒ α = 1 , ddαA2 > 0
3
​ ​

2
1 ( 3 +1)
1
Hence, A is minimum, A = 4 = 343
​

1
​ ​ ​

3
​
​

Since A = kA1 (given) ⇒ 4 = k2 ⇒ k = 2

3 3 3
​ ​ ​ ​

​ ​

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