Question

If the function f(x)= 3ax+b, x>1 11. 5ax-2b, x = 1 is continuous at x = 1, then find the values of a and b x​

Answer 1

It looks like the function might be defined by

`f(x) = \begin{cases} 3a{}x + b & \text{if } x > 1 \\ 11 & \text{if } x = 1 \\ 5a{}x - 2b & \text{if } x < 1 \end{cases}`

To ensure continuity at
`x=1`, we need both one-sided limits to exist and have the same value.

`\displaystyle \lim_(x\to1^-) f(x) = \lim_(x\to1) (5a{}x - 2b) = 5a - 2b`

`\displaystyle \lim_(x\to1^+) f(x) = \lim_(x\to1) (3a{}x + b) = 3a + b`

Both limit values must be equal to
`f(1) = 11`, so that

`\begin{cases} 5a - 2b = 11 \\ 3a + b = 11 \end{cases}`

Eliminating
`b`, we have

`(5a - 2b) + 2 (3a + b) = 11 + 2\cdot11 \implies 11a = 33 \implies \boxed{a=3}`

Solving for
`b`, we get

`5\cdot3 - 2b = 11 \implies -2b = -4 \implies \boxed{b=2}`