The number of points of discontinuity of the function f(x) = left[fracx^22right] - left[sqrtxright], x in [0,4] , where [cdot] denotes the greatest integer function is

Numerical Answer Type:
Enter a numerical value Answer: 8 to 8 +4 marks

Solution & Explanation

### Related Formula The greatest integer function [u] changes value and experiences a step discontinuity at any point where its inner argument u takes on an integer value. ### Core Logic Analyze the potential points where either component function argument changes into an integer within the interval x in [0, 4]. 1. For left[fracx^22right]: fracx^22 can range from frac02 = 0 up to frac162 = 8. Integer values are reached at fracx^22 = 0, 1, 2, 3, 4, 5, 6, 7, 8, which means critical test locations are: x = 0, \, sqrt2, \, 2, \, sqrt6, \, sqrt8, \, sqrt10, \, sqrt12, \, sqrt14, \, 4 2. For [sqrtx]: sqrtx can range from sqrt0 = 0 to sqrt4 = 2. Integer values are reached at sqrtx = 0, 1, 2, which means critical test locations are: x = 0, \, 1, \, 4 ### Step 1: Audit Each Critical Point Combine the set of test points within domain boundaries (0, 4): x in \1, \, sqrt2, \, 2, \, sqrt6, \, sqrt8, \, sqrt10, \, sqrt12, \, sqrt14\ Let's evaluate the left and right hand limits at these specific values: - At x = 1: [sqrtx] steps up while left[fracx^22right] is constant implies Discontinuous. - At x = sqrt2: left[fracx^22right] steps up while [sqrtx] is constant implies Discontinuous. - At x = 2: Both functions experience an simultaneous integer step. Let's inspect: - f(2) = [2] - [sqrt2] = 2 - 1 = 1 - f(2^-) = [1.99] - [1.41] = 1 - 1 = 0 Since LHL neq value at point, it is Discontinuous. Continuing this verification down the full combined list confirms that none of the step jumps cancel each other out. ### Step 2: Sum the Discontinuity Points Counting all isolated inner points within (0, 4) yields exactly 8 locations: textTotal Points = 8 ### Pattern Recognition When two greatest integer functions drop steps simultaneously at the same point (like at x=2), always write out the explicit left and right limits manually, as simultaneous steps occasionally step in matching directions and maintain unexpected continuity. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Limits, Continuity and Differentiability

Reference Study Guides

More Limits, Continuity and Differentiability Previous-Year Questions — Page 8

Q29 jee_main_2024_30_jan_morning Differentiability
If the function f(x) = begincases frac1|x| & ,|x| geq 2 \\ ax^2 + 2b & ,|x| < 2 endcases is differentiable on mathbbR, then 48 (a + b) is equal to
Numerical Answer. Answer: 15 to 15

Solution

### Related Formula textContinuity at x=c: lim_x to c^- f(x) = lim_x to c^+ f(x) textDifferentiability at x=c: lim_x to c^- f'(x) = lim_x to c^+ f'(x) ### Core Logic Rewrite the piecewise function without absolute values: f(x) = begincases frac1x & , x geq 2 \\ ax^2 + 2b & , -2 < x < 2 \\ -frac1x & , x leq -2 endcases ### Step 1: Applying Continuity For f(x) to be continuous at x = 2: lim_x to 2^- (ax^2 + 2b) = lim_x to 2^+ frac1x a(2)^2 + 2b = frac12 Rightarrow 4a + 2b = frac12 quad dots (1) Because the function is even, continuity at x = -2 yields the exact same equation: 4a + 2b = 1/2. ### Step 2: Applying Differentiability Find the derivative f'(x) for piecewise sections: f'(x) = begincases -frac1x^2 & , x > 2 \\ 2ax & , -2 < x < 2 \\ frac1x^2 & , x < -2 endcases For f(x) to be differentiable at x = 2: lim_x to 2^- (2ax) = lim_x to 2^+ left(-frac1x^2right) 4a = -frac14 Rightarrow a = -frac116 ### Step 3: Finding variables and final target Substitute a back into equation (1): 4left(-frac116right) + 2b = frac12 -frac14 + 2b = frac12 Rightarrow 2b = frac34 Rightarrow b = frac38 We need to evaluate 48(a + b): 48left(-frac116 + frac38right) = 48left(frac-1 + 616right) = 48left(frac516right) = 3 times 5 = 15 ### Pattern Recognition Piecewise differentiability forces simultaneous linear equations matching function values and their first derivatives at boundary limits. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Continuity and Differentiability
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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