Question

6. A function f is defined as f (x) = x2 + f(x - 1) with f(0) = 0, find the value of
the function when I is 6​

Answer 1

Answer:

F(x + 2) = 2F(x) - F(x + 1) ...(i)

Putting x = 0, we get

F(2) = 2F(0) - F(1)

F(2) = 2(2) - 3

F(2) = 4 - 3 F(2) = 1 Putting x = 1, in eq. (i), we get

F(3) = 2F(1) - F(2)

= 2(3) - 1

F(3) = 5

Putting x = 2, in eq. (i), we get

F(4) = 2F(2) - F(3)

= 2(1) - 5

F(4) = - 3

Putting x = 3, in eq. (i), we get

F(5) = 2F(3) - F(4)

= 2(5) + 3

F(5) = 13

Answer 2

​  

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​  

 

x  

2

+ax+b,

3x+2,

2ax+5b,

​  

 

if 0≤x≤2

if 2≤x≤4

if 4≤x≤8

​  

 

Given, f(x) is continuous in [0,8].

∴ f(x) is continuous at x=2 and x=4.

At x=2, we have

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find the value of I 6 ?

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∴ f(x) is continuous at x=2 and x=4.

At x=2, we have

lim

​  

f(x)=  

h→0

lim

​  

f(2−h)=  

h→0

lim

​  

[(2−h)  

2

+a(2−h)+b]=4+2a+b

x→2  

+

 

lim

​  

f(x)=  

h→0

lim

​  

f(2+h)=  

h→0

lim

​  

[3(2+h)+2]=8

∴  

x→2  

−

 

lim

​  

f(x)=  

x→2  

+

 

lim

​  

f(x)

⇒4+2a+b=8

⇒2a+b=4.....(1)

Also, at x=4

x→4  

−

 

lim

​  

f(x)=  

h→0

lim

​  

f(4−h)=  

h→0

lim

​  

[3(4−h)+2]=14

x→4  

+

 

lim

​  

f(x)=  

h→0

lim

​  

f(4+h)=  

h→0

lim

​  

[2a(4+h)+5b]=8a+5b

∴  

x→4  

+

 

lim

​  

f(x)=  

x→4  

−

 

lim

​  

f(x)

∴ 8a+5b=14.....(2)

Solving equation (1) and (2), we get

a=3 and b=−2