Curve y = √3x−2

dy

dx

=

1

2

(3x−2)−

1

2

×3
⇒

dy

dx

=

3

2√(3x−2)

...(1)
Since, the tangent is parallel to the line 4x−2y = - 5
Therefore, slope of tangent can be obtained from equation
y =

4x

2

+

5

2

Slope = 2
⇒

dy

dx

= 2
Comparing equations (1) and (2), we have,

3

2

×

1

√(3x−2)

= 2

1

√(3x−2)

=

4

3

⇒

1

3x−2

=

16

9

⇒ 9 = 48x - 32
⇒ x =

41

48

We have y = √(3x−2)
Thus, substituting the value of x in the above equation,
y = √3×

41

48

−2
⇒ y = √

41

16

−2
⇒ y = √

41−32

16

⇒ y = √

9

16

⇒ y =

3

4

Equation of tangent is
(y−

3

4

) = 2(x−

41

48

)
⇒ (y−

3

4

) = 2x−

41

24

⇒ y = 2x−

41

24

+

3

4

⇒ y = 2x−

41

24

+

18

24

⇒ y = 2x−

23

24

⇒ 24y=48x−23
⇒ 48x−24y−23+0
OR
f(x) = x3+

1

x3

,x≠0
⇒ f′(x)=3x2−3x−4=3(x2−

1

x4

)
⇒ f′(x)=3x2−3x4=

3

x4

(x6−1)
⇒ f′(x)=

3

x4

(x2−1)(x4+x2+1)
⇒ f'(x) = 3 (

x4+x2+1

x4

)(x2−1)
(i) For an increasing function, we should have,
f ' (x) > 0
⇒ 3 (

x4+x2+1

x4

)(x2−1) > 0
⇒ (x2−1) > 0 [Since 3 (

x4+x2+1

x4

) > 0]
⇒ (x - 1) (x + 1) > 0
⇒ x ∊ (- ∞ , - 1) ∪ x ∊ (1 , ∞)
So, f(x) is increasing on (- ∞ , - 1) ∪ (1 , ∞)
(ii) For a decreasing function, we should have f’(x) < 0
f ' (x) < 0
⇒ 3 (

x4+x2+1

x4

)(x2−1) < 0
⇒ (x2−1) < 0 [Since 3 (

x4+x2+1

x4

)(x2−1) > 0]
⇒ (x - 1) (x + 1) < 0
⇒ ∊ (- 1 , 0) ∪ x ∊ (0 , 1)
So f(x) is decreasing on (- 1 , 0) ∪ (0 , 1)