Continuity & Differentiability Practice Questions - DPP for JEE

 Page 2


 f(0) =0 and f ' (0) = 3
? f(x) = 3x + c, Q f(0) = 0 ? c = 0
? f(x) = 3x
2. (b)
? yy' + x = 0
....(i)
 
3. (c) Since f(x) is continuous at x = 0 ?  
Take any point x = a, then at x = a
  
? f(x) is continuous at x = a. Since x = a is any arbitrary point,
therefore f(x) is continuous for all x.
4. (a) Let   ...... (i)
Page 3


 f(0) =0 and f ' (0) = 3
? f(x) = 3x + c, Q f(0) = 0 ? c = 0
? f(x) = 3x
2. (b)
? yy' + x = 0
....(i)
 
3. (c) Since f(x) is continuous at x = 0 ?  
Take any point x = a, then at x = a
  
? f(x) is continuous at x = a. Since x = a is any arbitrary point,
therefore f(x) is continuous for all x.
4. (a) Let   ...... (i)
and ...... (ii)
In equation (i) put, x = tan
   = tan
–1
 (tan 2 )
 u = 2    ...... (a)
In equation (ii), put  x = tan
? v =   ........ (b)
From equations (a) and (b),
 Required differential coefficient will be 1.
5. (c) In the definition of the function,  then f(x) will be undefined
in x > 0.
Q  f(x) is continuous at x = 0,    ? LHL = RHL = f(0)
?
?
?
? a + 2 
? a + 2   ? 
Page 4


 f(0) =0 and f ' (0) = 3
? f(x) = 3x + c, Q f(0) = 0 ? c = 0
? f(x) = 3x
2. (b)
? yy' + x = 0
....(i)
 
3. (c) Since f(x) is continuous at x = 0 ?  
Take any point x = a, then at x = a
  
? f(x) is continuous at x = a. Since x = a is any arbitrary point,
therefore f(x) is continuous for all x.
4. (a) Let   ...... (i)
and ...... (ii)
In equation (i) put, x = tan
   = tan
–1
 (tan 2 )
 u = 2    ...... (a)
In equation (ii), put  x = tan
? v =   ........ (b)
From equations (a) and (b),
 Required differential coefficient will be 1.
5. (c) In the definition of the function,  then f(x) will be undefined
in x > 0.
Q  f(x) is continuous at x = 0,    ? LHL = RHL = f(0)
?
?
?
? a + 2 
? a + 2   ? 
6. (d) f(x) = [x]
2
 – [x
2
] = (–1)
2
 – (0)
2
 = 0, – 1< x < 0  
= 0 – 0 = 0,  0 = x < 1 and = 1 – 1 = 0, 1 = x < 
and = 1 – 3 = –2,  = x <
? From above it is clear that the function is discontinuous at 
 except at x = 1.
7. (d) For f (x) to be continuous at x = 0, we should have
  f (x) = f (0) = 12(log 4)
3
 f (x) = 
= (log 4)
3
 · 1 · p · 
= 3p (log 4)
3
 · Hence p = 4.
8. (c)
? 
?  ?  ? 
9. (d) We have, 
Page 5


 f(0) =0 and f ' (0) = 3
? f(x) = 3x + c, Q f(0) = 0 ? c = 0
? f(x) = 3x
2. (b)
? yy' + x = 0
....(i)
 
3. (c) Since f(x) is continuous at x = 0 ?  
Take any point x = a, then at x = a
  
? f(x) is continuous at x = a. Since x = a is any arbitrary point,
therefore f(x) is continuous for all x.
4. (a) Let   ...... (i)
and ...... (ii)
In equation (i) put, x = tan
   = tan
–1
 (tan 2 )
 u = 2    ...... (a)
In equation (ii), put  x = tan
? v =   ........ (b)
From equations (a) and (b),
 Required differential coefficient will be 1.
5. (c) In the definition of the function,  then f(x) will be undefined
in x > 0.
Q  f(x) is continuous at x = 0,    ? LHL = RHL = f(0)
?
?
?
? a + 2 
? a + 2   ? 
6. (d) f(x) = [x]
2
 – [x
2
] = (–1)
2
 – (0)
2
 = 0, – 1< x < 0  
= 0 – 0 = 0,  0 = x < 1 and = 1 – 1 = 0, 1 = x < 
and = 1 – 3 = –2,  = x <
? From above it is clear that the function is discontinuous at 
 except at x = 1.
7. (d) For f (x) to be continuous at x = 0, we should have
  f (x) = f (0) = 12(log 4)
3
 f (x) = 
= (log 4)
3
 · 1 · p · 
= 3p (log 4)
3
 · Hence p = 4.
8. (c)
? 
?  ?  ? 
9. (d) We have, 
Since        ?   f(x) is differentiable at x = 0
Since every differentiable function is continuous, therefore, f(x) is continuous
at x = 0.
10. (a)
Hence  exists if .
If g(x) = a ? 0 (constant) then
Thus  doesn’t exist in this case.
? exists in case of (b), (c) and (d) each.
11. (d) f (x)   =  max. {x, x
3
}
= 
? f ' (x) = 
Clearly f is not differentiable at  – 1, 0 and 1.
12. (b) f is  continuous at , if .
Now, 
?