Applications of Integrals Practice Questions - DPP for JEE

 Page 2


= 2 × Area OBCO = 2x
[Since y = 2 –x
2
 is the upper curve and y = x is the lower curve]
2. (d) Given 
Differentiate with respect to b
3. (d) Point of intersection of 
and  are (0,  0) and 
Given  
?   
So, the parabola is 
Page 3


= 2 × Area OBCO = 2x
[Since y = 2 –x
2
 is the upper curve and y = x is the lower curve]
2. (d) Given 
Differentiate with respect to b
3. (d) Point of intersection of 
and  are (0,  0) and 
Given  
?   
So, the parabola is 
Area enclosed by  is
  
4. (b) Point of intersection of y = sin x and y = cos x are 
Since, sin x = cos x on the interval 
?  Area of one such region 
 sq. unit.
5. (b) Area of ?AOB = 
= 
Area of region AOB
Page 4


= 2 × Area OBCO = 2x
[Since y = 2 –x
2
 is the upper curve and y = x is the lower curve]
2. (d) Given 
Differentiate with respect to b
3. (d) Point of intersection of 
and  are (0,  0) and 
Given  
?   
So, the parabola is 
Area enclosed by  is
  
4. (b) Point of intersection of y = sin x and y = cos x are 
Since, sin x = cos x on the interval 
?  Area of one such region 
 sq. unit.
5. (b) Area of ?AOB = 
= 
Area of region AOB
= 
=  = 
? ratio of areas = 
6. (c) f(x) = x
2
 + bx - b;  f '(x) = 2x + b ? f '(1) = b + 2
Equation of tangent :   y - 1 = (b + 2) (x - 1)
Putting x = 0  ? y = 1 - b - 2 = -b - 1 > 0  ? b < -1
Putting y = 0  ? x - 1 = - ? x = + 1
=  > 0 ? b < -2 or b > -1
Combining, the two conditions  = b < -2
Now, |-b -1|  = 2;  (b + 1)
2
 = 4|b + 2| = -4b - 8
? (b + 3)
2
 = 0     ? b = -3 follows the condition b < -2
7. (b) We have y
2 
= 4a(x + a) ...(i), a parabola with vertex (– a, 0)
and y
2
 = 4b (x – a) ...(ii), a parabola with vertex (a, 0)
Solving (i) and (ii), we get 
Page 5


= 2 × Area OBCO = 2x
[Since y = 2 –x
2
 is the upper curve and y = x is the lower curve]
2. (d) Given 
Differentiate with respect to b
3. (d) Point of intersection of 
and  are (0,  0) and 
Given  
?   
So, the parabola is 
Area enclosed by  is
  
4. (b) Point of intersection of y = sin x and y = cos x are 
Since, sin x = cos x on the interval 
?  Area of one such region 
 sq. unit.
5. (b) Area of ?AOB = 
= 
Area of region AOB
= 
=  = 
? ratio of areas = 
6. (c) f(x) = x
2
 + bx - b;  f '(x) = 2x + b ? f '(1) = b + 2
Equation of tangent :   y - 1 = (b + 2) (x - 1)
Putting x = 0  ? y = 1 - b - 2 = -b - 1 > 0  ? b < -1
Putting y = 0  ? x - 1 = - ? x = + 1
=  > 0 ? b < -2 or b > -1
Combining, the two conditions  = b < -2
Now, |-b -1|  = 2;  (b + 1)
2
 = 4|b + 2| = -4b - 8
? (b + 3)
2
 = 0     ? b = -3 follows the condition b < -2
7. (b) We have y
2 
= 4a(x + a) ...(i), a parabola with vertex (– a, 0)
and y
2
 = 4b (x – a) ...(ii), a parabola with vertex (a, 0)
Solving (i) and (ii), we get 
= 
=  sq. units