For what value of is the function f(x) = {[λ(x^2-2x),if x≤0][4x+1,if x>0 continuous at x = 0 ? What about continuity at x = ± 1? ​

Question

For what value of is the function\(f(x) = \begin{cases}\lambda(x^2-2x)&,\quad if\,x ≤0\\4x+1&,\quad if\,x>0\end{cases} \) continuous at x = 0 ? 

What about continuity at x = ± 1?

Answer 1

We have to find the value of 'λ' 

Such that f(x) is continuous at x = 0 

If f(x) is be continuous at x = 0,then, 

f(0)^– = f(0)⁺ = f(0)

⇒ 4(0) + 1 

⇒ 1 ...(2) 

From (1) & (2),we get 

f(0)^– = f(0)⁺, 

Hence, 

f(x) is not continuous at x = 0 

We also have to find out the continuity at point ±1

For f(x) is be continuous at x = 1,

Then,

f(0)^– = f(0)⁺ = f(0)

⇒ (5 + 4×0) 

⇒ 5 ...(2) 

From (1) & (2),we get 

f(0)^– = f(0)⁺, 

i.e, - λ = 5

⇒ λ = – 5

Hence, 

f(x) is continuous at x = 1,when λ = –5 

Similarly, 

For f(x) is be continuous at x = –1, 

Then, 

f(–1)^– = f(–1)⁺ = f(–1)

⇒ (–3 + 4×0)

⇒ –3 ...(2)

From (1) & (2),we get, 

f(–1)^– = f(–1)⁺

i.e, –3λ = -3

⇒ λ = 1

Hence, 

f(x) is continuous at x = 1,

when λ = 1.