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    The function f(x)=2x+∣x−1∣at x=1 isf(x) = 2x + |x - 1| \quad \text{at } x = 1 \text{ is}f(x)=2x+∣x−1∣at x=1 is​​
    Question

    The function f(x)=2x+x1at x=1 isf(x) = 2x + |x - 1| \quad \text{at } x = 1 \text{ is}​​

    A.

    differentiable

    B.

    continuous

    C.

    not differentiable and has infinite value

    D.

    minimum in open interval (0, 2)

    Correct option is B

    Solution: f(x)=2x+x1Case 1: x>1=>f(x)=2x+(x1)=3x1Case 2: x<1=>f(x)=2x(x1)=x+1Left-hand derivative at x=1:f(x)=1Right-hand derivative at x=1:f(x)=3Since fL(1)fR(1), function is not differentiable at x=1However, f(x) is continuous at x=1Correct answer: (B) Continuous\text{Solution: }\\ f(x) = 2x + |x - 1| \\[6pt]\text{Case 1: } x > 1 \Rightarrow f(x) = 2x + (x - 1) = 3x - 1 \\[4pt]\text{Case 2: } x < 1 \Rightarrow f(x) = 2x - (x - 1) = x + 1 \\[8pt]\text{Left-hand derivative at } x = 1: f'(x) = 1 \\\text{Right-hand derivative at } x = 1: f'(x) = 3 \\[8pt]\text{Since } f'_L(1) \ne f'_R(1), \text{ function is not differentiable at } x = 1 \\[6pt]\text{However, } f(x) \text{ is continuous at } x = 1 \\[12pt]\boxed{\text{Correct answer: (B) Continuous}}​​

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