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Let g(x) be a linear function and f(x) = begincases g(x) & , x le 0 \\ left(frac1+x2+xright)^frac1x & , x > 0 endcases is continuous at x = 0. If f'(1) = f(-1), then the value of g(3) is

Solution & Explanation

### Core Logic Let g(x) = ax + b. Since f(x) is continuous at x = 0: lim_x to 0^+ f(x) = f(0) lim_x to 0 left(frac1+x2+xright)^frac1x = b As x to 0, the base approaches frac12, and exponent approaches infty. Thus, left(frac12right)^infty = 0. So, b = 0. Thus, g(x) = ax. ### Step 1: Calculate Derivative For x > 0, f(x) = left(frac1+x2+xright)^frac1x. Let y = f(x). ln y = frac1x lnleft(frac1+x2+xright) Differentiating both sides w.r.t x: frac1y y' = -frac1x^2 lnleft(frac1+x2+xright) + frac1x cdot frac2+x1+x cdot frac1(2+x) - (1+x)1(2+x)^2 y' = y left[ -frac1x^2 lnleft(frac1+x2+xright) + frac1x(1+x)(2+x) right] ### Step 2: Apply Condition At x=1, y = f(1) = frac23. f'(1) = frac23 left[ -1 lnleft(frac23right) + frac16 right] = -frac23 lnleft(frac23right) + frac19 Also f(-1) = g(-1) = -a. Given f'(1) = f(-1) implies -a = -frac23 lnleft(frac23right) + frac19. a = frac23 lnleft(frac23right) - frac19 ### Step 3: Evaluate g(3) g(3) = 3a = 2 lnleft(frac23right) - frac13 g(3) = lnleft(frac49right) - frac13 = lnleft(frac49right) - ln(e^1/3) = lnleft(frac49e^1/3right) ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Continuity and Differentiability Class 12 Maths: Application of Derivatives

Reference Study Guides

Continuity And Differentiability

More Continuity and Differentiability Previous-Year Questions — Page 8

Q10 jee_main_2024_31_jan_evening Limits of Functions
Let f: mathbbR to (0, infty) be strictly increasing function such that lim_x to infty fracf(7x)f(x) = 1. Then, the value of lim_x rightarrow infty left[ fracf(5x)f(x) - 1 right] is equal to
  • A. 4
  • B. 0
  • C. 7/5
  • D. 1

Solution

### Related Formula textSandwich / Squeeze Theorem: If g(x) le h(x) le k(x) text and lim g(x) = lim k(x) = L, text then lim h(x) = L ### Core Logic Since f is a strictly increasing function mapping to (0,infty): For x > 0, we have x < 5x < 7x. Thus, f(x) < f(5x) < f(7x). Divide everything by f(x) (which is strictly positive): 1 < fracf(5x)f(x) < fracf(7x)f(x) Take the limit as x to infty: lim_xtoinfty 1 le lim_xtoinfty fracf(5x)f(x) le lim_xtoinfty fracf(7x)f(x) 1 le lim_xtoinfty fracf(5x)f(x) le 1 Therefore, lim_xtoinfty fracf(5x)f(x) = 1. The required value is: lim_x rightarrow infty left[ fracf(5x)f(x) - 1 right] = 1 - 1 = 0 ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Limits and Derivatives Class 12 Maths: Continuity and Differentiability
Q14 jee_main_2024_31_jan_evening Differentiability
Consider the function f:(0,infty)to mathbbR defined by f(x) = e^-|log_ex|. If m and n be respectively the number of points at which f is not continuous and f is not differentiable, then m + n is
  • A. 0
  • B. 3
  • C. 1
  • D. 2

Solution

### Core Logic
Differentiability diagram for Q14 - JEE Main 2024 Evening
Differentiability diagram for Q14 - JEE Main 2024 Evening
The function is f(x) = e^-|ln x|. Rewrite piecewise for (0, infty): f(x) = begincases e^-(-ln x) & textif 0 < x < 1 \\ e^-ln x & textif x ge 1 endcases f(x) = begincases e^ln x = x & textif 0 < x < 1 \\ frac1e^ln x = frac1x & textif x ge 1 endcases Check continuity at x = 1: lim_x to 1^- f(x) = lim_x to 1^- x = 1 lim_x to 1^+ f(x) = lim_x to 1^+ frac1x = 1 f(1) = 1. The function is continuous everywhere on (0, infty). Thus, m = 0. Check differentiability at x = 1: LHD = lim_x to 1^- f'(x) = 1 RHD = lim_x to 1^+ f'(x) = -frac1x^2Big|_x=1 = -1 Since LHD neq RHD, the function is not differentiable at x = 1. Thus, n = 1. Finally, m + n = 0 + 1 = 1. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Continuity and Differentiability
Q27 jee_main_2024_31_jan_evening Maclaurin Series / L'Hopital
If lim_x to 0 fracax^2e^x - b log_e(1 + x) + cxe^-xx^2sin x = 1, then 16(a^2 + b^2 + c^2) is equal to
Numerical Answer. Answer: 81 to 81

Solution

### Related Formula e^x = 1 + x + fracx^22! + dots ln(1+x) = x - fracx^22 + fracx^33 - dots sin x approx x implies x^2sin x approx x^3 ### Core Logic Expand the numerator terms using Maclaurin series around x=0: ax^2 left(1 + x + fracx^22 + dotsright) - b left(x - fracx^22 + fracx^33 - dotsright) + cx left(1 - x + fracx^22 - fracx^36 + dotsright) Denominator behavior is x^3. Group by powers of x: Coefficient of x: -b + c = 0 implies c = b Coefficient of x^2: a + fracb2 - c = 0 implies a = c - fracb2 = fracb2 Coefficient of x^3: a - fracb3 + fracc2 = 1 Substitute a = b/2 and c = b into the x^3 equation: fracb2 - fracb3 + fracb2 = 1 b - fracb3 = 1 implies frac2b3 = 1 implies b = frac32 This gives c = frac32 and a = frac34. Calculate the required value: 16(a^2 + b^2 + c^2) = 16left(frac916 + frac94 + frac94right) = 9 + 36 + 36 = 81 ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Limits and Derivatives
Q7 jee_main_2024_31_jan_morning Exponential Limits
lim_xto 0frace^2|sin x| - 2|sin x| - 1x^2
  • A. textis equal to -1
  • B. textdoes not exist
  • C. textis equal to 1
  • D. textis equal to 2

Solution

### Core Logic Evaluate lim_x to 0 frace^2|sin x| - 2|sin x| - 1x^2. Multiply and divide by |sin x|^2: = lim_x to 0 frace^2|sin x| - 2|sin x| - 1|sin x|^2 times fracsin^2 xx^2 ### Step 1: Substitution and L'Hôpital Let |sin x| = t. As x to 0, t to 0. lim_t to 0 frace^2t - 2t - 1t^2 times lim_x to 0 fracsin^2 xx^2 The second limit evaluates to 1. Using L'Hôpital's Rule on the first limit: = lim_t to 0 frac2e^2t - 22t = lim_t to 0 frac4e^2t2 = 2 ### Step 2: Final Result 2 times 1 = 2 ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Limits and Derivatives

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