Applications of Derivatives Practice Questions - DPP for JEE

 Page 2


If  x = 0 then y = 0
If  then 
Therefore, the points are (0, 0) and .
2. (d) f '(x) = 4x – 1/x
f '(x) is monotonic increasing when f '(x) > 0
? 4x – 1/x > 0
?  > 0
? 
But x > 0, 4x
2
 – 1> 0 ? x
2
 > 1/4 ? | x | > 1/2
? x ? (1/2, 8 )
and x < 0, 4x
2
 – 1< 0 ? x
2
 < 1/4 ? | x | < 1/2
? x ? (–1/2, 0 )
? x ? (–1/2, 0) ? (1/2, 8)
3. (a) f (x) = 
Let 
Let  g (t) = t
3
 – 3t – 4t
2
 + 8
g (t) = t
3
 – 4t
2
 – 3t + 8
g' (t) = 3t
2
 – 8t – 3 = (t – 3) (3t + 1)
g' (t) = 0  ? t = 3  (t ? –1/3)
g'' (t) = 6t – 8
Page 3


If  x = 0 then y = 0
If  then 
Therefore, the points are (0, 0) and .
2. (d) f '(x) = 4x – 1/x
f '(x) is monotonic increasing when f '(x) > 0
? 4x – 1/x > 0
?  > 0
? 
But x > 0, 4x
2
 – 1> 0 ? x
2
 > 1/4 ? | x | > 1/2
? x ? (1/2, 8 )
and x < 0, 4x
2
 – 1< 0 ? x
2
 < 1/4 ? | x | < 1/2
? x ? (–1/2, 0 )
? x ? (–1/2, 0) ? (1/2, 8)
3. (a) f (x) = 
Let 
Let  g (t) = t
3
 – 3t – 4t
2
 + 8
g (t) = t
3
 – 4t
2
 – 3t + 8
g' (t) = 3t
2
 – 8t – 3 = (t – 3) (3t + 1)
g' (t) = 0  ? t = 3  (t ? –1/3)
g'' (t) = 6t – 8
g'' (3) = 10 > 0  ?  g (3) is minimum
g (3) = 27 – 9 – 36 + 8 = – 10
4. (c) Given,  6y = x
3
 + 2
On differentiating w.r.t. t, we get
?
? 3x
2 
= 48 ? x
2
 = 16
?
When x = 4, then 6y = (4)
3
 + 2
? 6y = 64 + 2 ?  = 11
When x = – 4, then 6y = (– 4)
3
 + 2
? 6y = – 64 + 2 ? 
Hence, the required points on the curve are (4, 11) and 
5. (c) (a) x
2
y
2
 = 16a
4
 ? xy = 4a
2
y + xy' = 0
 ;  
? (a) is true.
(b)  i.e. x = 2
? (2, 1) is the point on the curve x
2
 = 4y at which
the normal is
y – 1 = – 1 (x – 2) i.e. x + y = 3   ?  (b) is true
(c) y = – 4x
2
, y = e
–x/2
The curves are non-intersecting
?  curves are not orthogonal i.e. (c) is false.
Page 4


If  x = 0 then y = 0
If  then 
Therefore, the points are (0, 0) and .
2. (d) f '(x) = 4x – 1/x
f '(x) is monotonic increasing when f '(x) > 0
? 4x – 1/x > 0
?  > 0
? 
But x > 0, 4x
2
 – 1> 0 ? x
2
 > 1/4 ? | x | > 1/2
? x ? (1/2, 8 )
and x < 0, 4x
2
 – 1< 0 ? x
2
 < 1/4 ? | x | < 1/2
? x ? (–1/2, 0 )
? x ? (–1/2, 0) ? (1/2, 8)
3. (a) f (x) = 
Let 
Let  g (t) = t
3
 – 3t – 4t
2
 + 8
g (t) = t
3
 – 4t
2
 – 3t + 8
g' (t) = 3t
2
 – 8t – 3 = (t – 3) (3t + 1)
g' (t) = 0  ? t = 3  (t ? –1/3)
g'' (t) = 6t – 8
g'' (3) = 10 > 0  ?  g (3) is minimum
g (3) = 27 – 9 – 36 + 8 = – 10
4. (c) Given,  6y = x
3
 + 2
On differentiating w.r.t. t, we get
?
? 3x
2 
= 48 ? x
2
 = 16
?
When x = 4, then 6y = (4)
3
 + 2
? 6y = 64 + 2 ?  = 11
When x = – 4, then 6y = (– 4)
3
 + 2
? 6y = – 64 + 2 ? 
Hence, the required points on the curve are (4, 11) and 
5. (c) (a) x
2
y
2
 = 16a
4
 ? xy = 4a
2
y + xy' = 0
 ;  
? (a) is true.
(b)  i.e. x = 2
? (2, 1) is the point on the curve x
2
 = 4y at which
the normal is
y – 1 = – 1 (x – 2) i.e. x + y = 3   ?  (b) is true
(c) y = – 4x
2
, y = e
–x/2
The curves are non-intersecting
?  curves are not orthogonal i.e. (c) is false.
(d) y = x
3
 – 2ax
2
 + 2x + 5
 = 2x
2
 – 4ax + 2 = 2 (x
2
 – 2ax + 1)
= 2 (x – a)
2
 + 2 – 2a
2
 > 0
[ Q a ? (–1, 0) ? 0 < a
2
 < 1]
? (d) is true
6. (a) Let (x, y) be the one point of parabola,  y = x
2
 + 7x + 2
its distance from the line
y = 3x – 3 or 3x – y – 3 = 0 is
D =  =
= 
D = =
=  as   is + ive
 = 0 ?x = – 2
and hence y is – 8 i.e. point is (–2, – 8)
 =+ ive and hence min. at (–2, –8)
7. (b) f(x) = (a
2
 – 3a + 2) (cos
2
x/4 – sin
2
x/4)
+ (a – 1)  x + sin 1
? f(x) = (a –1) (a –2) cos x/2 + (a –1) x +sin1
? f ' (x) = – (a –1) (a – 2) sin  + (a –1)
Page 5


If  x = 0 then y = 0
If  then 
Therefore, the points are (0, 0) and .
2. (d) f '(x) = 4x – 1/x
f '(x) is monotonic increasing when f '(x) > 0
? 4x – 1/x > 0
?  > 0
? 
But x > 0, 4x
2
 – 1> 0 ? x
2
 > 1/4 ? | x | > 1/2
? x ? (1/2, 8 )
and x < 0, 4x
2
 – 1< 0 ? x
2
 < 1/4 ? | x | < 1/2
? x ? (–1/2, 0 )
? x ? (–1/2, 0) ? (1/2, 8)
3. (a) f (x) = 
Let 
Let  g (t) = t
3
 – 3t – 4t
2
 + 8
g (t) = t
3
 – 4t
2
 – 3t + 8
g' (t) = 3t
2
 – 8t – 3 = (t – 3) (3t + 1)
g' (t) = 0  ? t = 3  (t ? –1/3)
g'' (t) = 6t – 8
g'' (3) = 10 > 0  ?  g (3) is minimum
g (3) = 27 – 9 – 36 + 8 = – 10
4. (c) Given,  6y = x
3
 + 2
On differentiating w.r.t. t, we get
?
? 3x
2 
= 48 ? x
2
 = 16
?
When x = 4, then 6y = (4)
3
 + 2
? 6y = 64 + 2 ?  = 11
When x = – 4, then 6y = (– 4)
3
 + 2
? 6y = – 64 + 2 ? 
Hence, the required points on the curve are (4, 11) and 
5. (c) (a) x
2
y
2
 = 16a
4
 ? xy = 4a
2
y + xy' = 0
 ;  
? (a) is true.
(b)  i.e. x = 2
? (2, 1) is the point on the curve x
2
 = 4y at which
the normal is
y – 1 = – 1 (x – 2) i.e. x + y = 3   ?  (b) is true
(c) y = – 4x
2
, y = e
–x/2
The curves are non-intersecting
?  curves are not orthogonal i.e. (c) is false.
(d) y = x
3
 – 2ax
2
 + 2x + 5
 = 2x
2
 – 4ax + 2 = 2 (x
2
 – 2ax + 1)
= 2 (x – a)
2
 + 2 – 2a
2
 > 0
[ Q a ? (–1, 0) ? 0 < a
2
 < 1]
? (d) is true
6. (a) Let (x, y) be the one point of parabola,  y = x
2
 + 7x + 2
its distance from the line
y = 3x – 3 or 3x – y – 3 = 0 is
D =  =
= 
D = =
=  as   is + ive
 = 0 ?x = – 2
and hence y is – 8 i.e. point is (–2, – 8)
 =+ ive and hence min. at (–2, –8)
7. (b) f(x) = (a
2
 – 3a + 2) (cos
2
x/4 – sin
2
x/4)
+ (a – 1)  x + sin 1
? f(x) = (a –1) (a –2) cos x/2 + (a –1) x +sin1
? f ' (x) = – (a –1) (a – 2) sin  + (a –1)
If f(x) does not possess critical points, then
f '(x) ? 0 for any 
?a ? 1 and 
must not have any solution in R.
? a ?1 and  is not solvable in R
   
? a ?1 and | a – 2| < 2
? a ? 1  and –2 < a –2 < 2
? a ?1 and 0 < a < 4 ? a ? (0, 1) ? (1, 4).
8. (c) We have f(x) = x
3
 + ax
2
 + bx + 5 sin
2
 x
 f(x) is an increasing function
 
( sin 2x < 1)
 
[ for all real x if . a > 0 and discriminant < 0].
9. (d) We have equation of tangent to any curve f(x) at (x
1
, y
1
) is
Given curve is