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For what value of α\alpha, the function ff defined by f(x)={α(x2−2x+1),if x≤02x+1,if x>0f(x) = \begin{cases} \alpha(x^2 - 2x + 1), & \text{if } x \leq 0 \\ 2x + 1, & \text{if } x > 0 \end{cases} is continuous at x=0x = 0?

Solution

✅ Correct Option: 1

For the function to be continuous at x=0x = 0, the left-hand limit, right-hand limit, and the value at x=0x = 0 must all be equal.


Since x=0x = 0 satisfies x≤0x \leq 0, the first piece of the function applies:

f(0)=α(02−2(0)+1)f(0) = \alpha(0^2 - 2(0) + 1)

f(0)=α(1)f(0) = \alpha(1)

f(0)=αf(0) = \alpha


For x>0x > 0, the function is f(x)=2x+1f(x) = 2x + 1.

The right-hand limit as xx approaches 00:

lim⁡x→0+f(x)=2(0)+1\lim_{x \to 0^+} f(x) = 2(0) + 1

lim⁡x→0+f(x)=1\lim_{x \to 0^+} f(x) = 1


For x<0x < 0, the function is f(x)=α(x2−2x+1)f(x) = \alpha(x^2 - 2x + 1).

The left-hand limit as xx approaches 00:

lim⁡x→0−f(x)=α(02−2(0)+1)\lim_{x \to 0^-} f(x) = \alpha(0^2 - 2(0) + 1)

lim⁡x→0−f(x)=α\lim_{x \to 0^-} f(x) = \alpha


For continuity at x=0x = 0:

f(0)=lim⁡x→0+f(x)=lim⁡x→0−f(x)f(0) = \lim_{x \to 0^+} f(x) = \lim_{x \to 0^-} f(x)

α=1=α\alpha = 1 = \alpha

Therefore, α=1\alpha = 1

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