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Let m and n be the number of points at which the function f(x) = max \x, x^3, x^5, dots, x^21\ for x in mathbbR is not differentiable and not continuous, respectively. Then m + n is equal to

Numerical Answer Type:
Enter a numerical value Answer: 3 to 3 +4 marks

Solution & Explanation

### Related Formula A function is non-differentiable at sharp corner transition points where left-hand and right-hand derivatives do not match. ### Core Logic Analyze the behavior of powers of x across significant transition domains: For x < -1: x is the largest because higher odd powers of negative fractions decrease rapidly (x > x^3 > x^5...). For -1 le x < 0: x^21 is largest (closest to zero from below). For 0 le x < 1: x is largest. For x ge 1: x^21 is largest. f(x) = begincases x, & x < -1 \\ x^21, & -1 le x < 0 \\ x, & 0 le x < 1 \\ x^21, & x ge 1 endcases ### Step 1: Continuity and Differentiability Checks At critical intersection boundaries x = -1, 0, 1, f(x) matches continuous values perfectly, so n = 0. Now check derivative transitions f'(x): f'(x) = begincases 1, & x < -1 \\ 21x^20, & -1 < x < 0 \\ 1, & 0 < x < 1 \\ 21x^20, & x > 1 endcases At x = -1: textLHD = 1, textRHD = 21(-1)^20 = 21 implies textNon-differentiable. At x = 0: textLHD = 0, textRHD = 1 implies textNon-differentiable. At x = 1: textLHD = 1, textRHD = 21(1)^20 = 21 implies textNon-differentiable. ### Step 2: Conclusion Thus, the function is non-differentiable at exactly 3 points (x = -1, 0, 1), so m = 3. Since n = 0: m + n = 3 + 0 = 3 ### Pattern Recognition Maximum boundary tracking curves for standard power elements always form continuous shapes but introduce non-differentiable sharp corners at every intersection crossover point. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Limits, Continuity and Differentiability

Reference Study Guides

Limits Continuity And Differentiability

More Limits, Continuity and Differentiability Previous-Year Questions — Page 5

Q65 2025 Limits of Special Series
The value of lim_nto inftyleft(sum_K = 1^nfrack^3 + 6k^2 + 11k + 5(k + 3)!right) is:
  • A. \frac{4}{3}
  • B. 2
  • C. \frac{7}{3}
  • D. \frac{5}{3}

Solution

### Related Formula sum_k=1^infty left( frac1k! - frac1(k+3)! right) implies textTelescoping Series simplification ### Core Logic Rewrite the numerator polynomial to establish factor terms matching the factorial expansion base (k+3): k^3 + 6k^2 + 11k + 5 = (k^3 + 6k^2 + 11k + 6) - 1 = (k+1)(k+2)(k+3) - 1 ### Step 1: Simplify General Term T_k = frac(k+1)(k+2)(k+3)(k+3)! - frac1(k+3)! T_k = frac1k! - frac1(k+3)! This creates a clean telescoping layout format structure. ### Step 2: Sum the Series Writing out expanded \partial sums up to infinity: S = left( frac11! + frac12! + frac13! + frac14! + dots right) - left( frac14! + frac15! + frac16! + dots right) All higher terms cancel out systematically, leaving exactly the leading remaining fragments: S = frac11! + frac12! + frac13! = 1 + frac12 + frac16 = frac106 = frac53 ### Pattern Recognition Whenever factorials dominate fraction denominators, manipulate structural terms to align components via Telescoping sums (V_n - V_n-k). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Limits, Continuity and Differentiability Class 11 Mathematics: Sequences and Series

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