Question

If f(x) = `{{:(ax + b; 0 < x ≤ 1),(2x^2 - x; 1 < x < 2):}` is a differentiable function in (0, 2), then find the values of a and b.

Answer 1

We have,

f(x) = `{{:(ax + b: 0 < x ≤ 1),(2x^2 - x: 1 < x < 2):}`

(LHD at x = 1)

= `lim_(x rightarrow 1^-) (f(x) - f(1))/(x - 1)`

= `lim_(h rightarrow 0) (f(1 - h) - f(1))/(1 - h - 1)`

= `lim_(h rightarrow 0) ([a(1 - h) + b] - [a + b])/(-h)`

= `lim_(h rightarrow 0) ([a - ah + b - a - b])/(-h)`

= `lim_(h rightarrow 0) (ah)/a`

= a

(RHD at x = 1)

= `lim_(x rightarrow 1^+) (f(x) - f(1))/(x - 1)`

= `lim_(h rightarrow 0) (f(1 + h) - f(1))/((1 + h) - 1)`

= `lim_(h rightarrow 0) ([2(1 + h)^2 - (1 + h)] - [2 - 1])/h`

= `lim_(h rightarrow 0) ([2(1 + h^2 + 2h) - 1 - h] - 1)/h`

= `lim_(h rightarrow 0) ([2 + 2h^2 + 4h - 1 - h - 1])/h`

= `lim_(h rightarrow 0) ((2h^2 + 3h))/h`

= `lim_(h rightarrow 0) (2h + 3)`

= 3

Since, f(x) is differentiable, so

(LHD at x = 1) = (RHD at x = 1)

∴ a = 3

Now, LHL = `lim_(x rightarrow 1^-) f(x)`

= `lim_(h rightarrow 0) f(1 - h)`

= `lim_(h rightarrow 0) a(1 - h) + b`

= a + b

Now, RHL = `lim_(x rightarrow 1^+) f(x)`

= `lim_(h rightarrow 0) f(1 + h)`

= `lim_(h rightarrow 0) 2(1 + h)^2 - (1 + h)`

= 2 – 1

= 1

∵ LHL = RHS

∴ a + b = 1

`\implies` 3 + b = 1

b = – 2

Hence, a = 3 and b = – 2.