Question 26 of 70advanced🔧 ApplyShort Answer2 marks

Suppose f(x)={a+bx,x<14,x=1b−ax,x>1f(x) = \begin{cases} a + bx, & x < 1 \\ 4, & x = 1 \\ b - ax, & x > 1 \end{cases} and if lim⁡x→1f(x)=f(1)\lim_{x \to 1} f(x) = f(1) what are possible values of aa and bb?

Correct Answer

a = 0 and b = 4

Exercise: EXERCISE 12.1 | Q: 28 | (Page Page 22)
For More Understanding

Explanation

This question tests the concept of existence of limit at a point. From the context, for a limit to exist at x = a, the left hand limit must equal the right hand limit. Additionally, for the function to be continuous (which is what lim(x→1) f(x) = f(1) implies), this common limit must also equal the value of the function at that point.

Solution Steps

  1. Step 1: For the limit to exist at x = 1, left hand limit must equal right hand limit.

  2. Step 2: Left hand limit: lim(x→1⁻) f(x) = a + b(1) = a + b

  3. Step 3: Right hand limit: lim(x→1⁺) f(x) = b - a(1) = b - a

  4. Step 4: Equating both limits: a + b = b - a, which gives 2a = 0, so a = 0.

  5. Step 5: Since lim(x→1) f(x) = f(1) = 4, we have a + b = 4. With a = 0, we get b = 4.

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