Maths Continuity and Differentiability

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New answer posted

3 weeks ago

0 Follower 2 Views

A
alok kumar singh

Contributor-Level 10

  f ( x ) = 4 x 3 3 x 2 6 2 s i n x + ( 2 x 1 ) c o s x

f ' ( x ) = 2 x 2 x 2 c o s x + 2 c o s x ( 2 x 1 ) s i n x

= (2x1) (xsinx)0forx0  

f ' ( x ) 0 x 1 / 2           

f ( x ) is increasing in [ 1 2 , )  

New answer posted

3 weeks ago

0 Follower 2 Views

V
Vishal Baghel

Contributor-Level 10

h' (x) = f' (x) – 2f (x).f' (x) + 3 (f (x)².f' (x)
= f' (x)/3 (f (x) - 1/3)² + 2/9)
∴ h' (x) ≥ 0 ⇔ f' (x) ≥ 0

New answer posted

3 weeks ago

0 Follower 2 Views

V
Vishal Baghel

Contributor-Level 10

g (x) = (f (nf (x) – n)?
g' (x) = n (f (nf (x) – n)? ¹ . f' (nf (x) – n) . n . f' (x)
∴ g' (0) = 0
⇒ 4 = n (f (nf (0) – n)? ¹ . f' (nf (0) – n) . nf' (0)
⇒ 4 = n (f (0)? ¹ . f' (0) . nf' (0)
⇒ 4 = n . 1 . (-1) . n (-1)
n² = 4
⇒ n = 2

New answer posted

3 weeks ago

0 Follower 2 Views

A
alok kumar singh

Contributor-Level 10

f (x)= {sinx, 0? x? }
f' (x)= {cosx, 0

New answer posted

a month ago

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A
alok kumar singh

Contributor-Level 10

The function f (x) is non-differentiable at x=1, 3, 5.
Σ f (f (x) = f (f (1) + f (f (3) + f (f (5).
Assuming f (x) is defined such that f (1)=1, f (3)=1, f (5)=1 (based on context of absolute value functions).
Then Σ f (f (x) = f (1) + f (1) + f (1) = 1 + 1 + 1 = 3.

New answer posted

a month ago

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R
Raj Pandey

Contributor-Level 9

Given f (g (x) is defined piecewise:
f (g (x) =
x³ + 2 ; x < 0
x? ; 0? x < 1
(3x - 2)² ; x? 1
fog (x) is discontinuous at x = 0.? non differentiable.
fog (x) is not differentiable at x = 0.

 

New answer posted

a month ago

0 Follower 4 Views

A
alok kumar singh

Contributor-Level 10

f (x) = {x+a, if x<0; |x-1|, if x0}
g (x) = {x+1, if x<0; (x-1)+b, if x0}
g (f (x) must be continuous. The potential points of discontinuity are where the definitions of f (x) and g (f (x) change. This is at x=0 and where f (x)=0.
f (x)=0 when x=-a (if a>0) or when x=1.
Continuity at x = 0:
lim (x→0? ) g (f (x) = lim (x→0? ) g (x+a). Since a could be anything, let's analyze f (0? )=a. So, lim is g (a).
lim (x→0? ) g (f (x) = g (f (0? ) = g (|0-1|) = g (1). Since 1≥0, g (1) = (1-1)²+b = b.
g (f (0) = g (|0-1|) = g (1) = b.
So we need g (a) = b.
Case 1: a < 0. g (a) = a+1. So a+1=b.
Case 2: a ≥ 0. g (a) = (a-1)²+b. So (a-1)²+b=b => (a-1)²=0 => a=1.
Now consider continuity at x=-a (assuming a>0).
l

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New answer posted

a month ago

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A
alok kumar singh

Contributor-Level 10

Kindly go through the solution

 

New answer posted

a month ago

0 Follower 7 Views

A
alok kumar singh

Contributor-Level 10

Kindly go through the solution

 

New answer posted

a month ago

0 Follower 5 Views

A
alok kumar singh

Contributor-Level 10

Kindly go through the solution 

 

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