Search for: The set of all point., where the function f(x)=x1+|x| is differentiable, IsThe set of all point., where the function f(x)=x1+|x| is differentiable, IsA−∞,∞B(0,∞)C(−∞,0)−(0,∞)D(−∞,−1)∪(−1−∞) Register to Get Free Mock Test and Study Material +91 Verify OTP Code (required) I agree to the terms and conditions and privacy policy. Solution:We have,f(x)=x1+|x|=g(x)h(x), where g(x)=x,h(x)=1+|x|. Tearly, g(x) is differentiable on (−∞,∞). The domain of h(x) is −x,∞). Since |x| is not differentiable at x=0. Therefore, h(x) not differentiable at x=0. Thus, f(x) is differentiable on −∞,0)∪(0,∞) At x=0, we have limx→0 f(x)−f(0)x−0=limx→0 x1+|x|x=limx→0 11+|x|=1 Therefore, f(x) is differentiable at x=0. Hence, f is differentiable on (−∞,∞). Post navigationPrevious: The value of limn→∞ 11.3+13.5+15.7+….+1(2n+1)(2n+3), isNext: If function f (x) given by f(x)=(sinx)1/(π−2x),x≠π/2λ,x=π/2is continuous at x=π2, then, λ=Related content JEE Main 2023 Question Papers with Solutions JEE Main 2024 Syllabus Best Books for JEE Main 2024 JEE Advanced 2024: Exam date, Syllabus, Eligibility Criteria JEE Main 2024: Exam dates, Syllabus, Eligibility Criteria JEE 2024: Exam Date, Syllabus, Eligibility Criteria NCERT Solutions For Class 6 Maths Data Handling Exercise 9.3 JEE Crash Course – JEE Crash Course 2023 NEET Crash Course – NEET Crash Course 2023 JEE Advanced Crash Course – JEE Advanced Crash Course 2023