Let f: R to (0, infty) be a twice differentiable function such that f(3) = 18 , f'(3) = 0 and f''(3) = 4 . Then lim_x to 1 left( log_e left( fracf(2 + x)f(3) right)^frac18(x-1)^2 right) is equal to:

Solution & Explanation

### Related Formula For a limit of 1^infty form, lim_x to a [g(x)]^h(x) equals: e^lim_x to a h(x)[g(x) - 1] ### Core Logic Let T = lim_x to 1 left( fracf(x + 2)f(3) right)^frac18(x - 1)^2. As x to 1, fracf(x+2)f(3) to fracf(3)f(3) = 1. The exponent goes to infty. This is a standard 1^infty form. T = e^lim_x to 1 frac18(x - 1)^2 left( fracf(x + 2) - f(3)f(3) right) ### Step 1: Simplify Exponent Limit Given f(3) = 18: T = e^lim_x to 1 frac18(x - 1)^2 left( fracf(x + 2) - f(3)18 right) T = e^lim_x to 1 fracf(x + 2) - f(3)(x - 1)^2 This is a frac00 form limit. ### Step 2: Apply L'Hopital's Rule Differentiate numerator and denominator w.r.t x: T = e^lim_x to 1 fracf'(x + 2)2(x - 1) This is still a frac00 form since f'(3) = 0. Apply L'Hopital's Rule again: T = e^lim_x to 1 fracf''(x + 2)2 Substitute x = 1: T = e^fracf''(3)2 ### Step 3: Final Calculation Given f''(3) = 4: T = e^frac42 = e^2 The question asks for log_e(T): log_e(T) = log_e(e^2) = 2 ### Pattern Recognition When expanding f(x) around an extrema (f'(a)=0) inside a 1^infty limit form, double L'Hopital isolates the second derivative divided by the Taylor factorial (2! = 2). The limit elegantly shrinks directly to f''(a)/2. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Limits and Derivatives Class 12 Maths: Limits, Continuity and Differentiability

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More Limits, Continuity and Differentiability Previous-Year Questions — Page 2

Q73 jee_main_2025_03_april_evening Evaluation of Limits
If lim_x to 0 left( fractan xx right)^frac1x^2 = p, then 96log_e p is equal to
Numerical Answer. Answer: 32 to 32

Solution

### Related Formula For a limit of 1^infty form, where lim f(x) = 1 and lim g(x) = infty: lim [f(x)]^g(x) = e^lim g(x)(f(x) - 1) Taylor expansion of tan x: tan x = x + fracx^33 + frac2x^515 + dots ### Core Logic Here, lim_x to 0 fractan xx = 1 and lim_x to 0 frac1x^2 = infty. p = e^lim_x to 0 frac1x^2left(fractan xx - 1right) = e^lim_x to 0 fractan x - xx^3 ### Step 1: Finding limit exponent Using expansion of tan x: lim_x to 0 fracleft(x + fracx^33 + dotsright) - xx^3 = lim_x to 0 fracfracx^33 + O(x^5)x^3 = frac13 Thus: p = e^1/3 96log_e p = 96left(frac13right) = 32 ### Pattern Recognition Limits of forms like 1^infty always boil down to evaluating standard polynomials in the exponent. Using Taylor series expansion rather than L'Hopital's rule directly avoids taking multiple heavy derivatives. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Limits, Continuity and Differentiability
Q51 jee_main_2025_07_april_morning Evaluation of Limits
lim_xto 0^+fractanleft(5(x)^frac13right)log_e(1 + 3x^2)left(tan^-13sqrtxright)^2left(e^5(x)^frac43 - 1right) is equal to
  • A. frac115
  • B. 1
  • C. frac13
  • D. frac53

Solution

### Related Formula Standard limits commands: lim_f(x)to 0 fractan(f(x))f(x) = 1 lim_f(x)to 0 fracln(1+f(x))f(x) = 1 lim_f(x)to 0 fractan^-1(f(x))f(x) = 1 lim_f(x)to 0 frace^f(x)-1f(x) = 1 ### Core Logic We can rewrite the limit by grouping each term with its standard balancing factor: lim_xto 0^+ left(fractanleft(5x^1/3right)5x^1/3right) cdot left(frac3sqrtxtan^-13sqrtxright)^2 cdot left(fraclog_e(1 + 3x^2)3x^2right) cdot left(frac5x^4/3e^5x^4/3 - 1right) times frac5x^1/3 cdot 3x^2(3sqrtx)^2 cdot 5x^4/3 ### Step 1: Simplify the Compensating Factor Evaluate the remaining algebraic factor: frac5x^1/3 cdot 3x^29x cdot 5x^4/3 = frac15x^7/345x^7/3 = frac1545 = frac13 Since all individual standard limit terms approach 1, the value of the limit is exactly: 1 cdot 1^2 cdot 1 cdot 1 cdot frac13 = frac13 ### Pattern Recognition Shortcut: For standard limits involving tan(u), ln(1+u), tan^-1(u), and e^u-1 as u to 0, replace each function directly with its argument: frac(5x^1/3)(3x^2)(3sqrtx)^2(5x^4/3) = frac15x^7/345x^7/3 = frac13 ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Limits and Derivatives Class 12 Mathematics: Limits, Continuity and Differentiability
Q71 jee_main_2025_07_april_morning Points of Discontinuity
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. Answer: 8 to 8

Solution

### 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
Q70 jee_main_2025_08_april_evening Standard Limits and Expansion
Given below are two statements : Statement I : lim _ mathrm x rightarrow 0 left(frac tan^ - 1 mathrm x + log_ mathrm e sqrt frac 1 + mathrm x1 - mathrm x - 2 mathrm xmathrm x ^ 5right) = frac 25 Statement II: lim_xto 1left(x^frac21 - xright) = frac1e^2 In the light of the above statements, choose the correct answer from the options given below:
  • A. Statement I is false but Statement II is true
  • B. Statement I is true but Statement II is false
  • C. Both Statement I and Statement II are false.
  • D. Both Statement I and Statement II are true.

Solution

### Related Formula log(1+x) = x - fracx^22 + fracx^33 - dots lim_x to a u^v = e^lim (u-1)v ### Core Logic Verify Statement I via high-order Taylor polynomial series tracking, and determine Statement II values by resolving standard exponential limit properties. ### Step 1: Expand Statement I Sequence Polynomials tan^-1x = x - fracx^33 + fracx^55 - dots frac12[ln(1+x) - ln(1-x)] = x + fracx^33 + fracx^55 + dots Summing terms together and subtracting 2x leaves: lim_x to 0 frac2x^5/5 + dotsx^5 = frac25 quad text(Statement I is true) ### Step 2: Verify Statement II Limit Power Structure Evaluating the 1^infty form configuration style: e^lim_x to 1 left(frac21-xright)(x-1) = e^-2 = frac1e^2 quad text(Statement II is true) ### Step 3: State Conclusion Both statements are correct. ### Pattern Recognition When a limit features a power factor of 5 in the denominator, you must track expansion variables through the 5th degree term to guarantee accuracy. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Limits
Q55 jee_main_2025_29_jan_evening Differentiability of Modulus Functions
Let the function f(x) = (x^2 - 1)|x^2 - ax + 2| + cos |x| be not differentiable at the two points x = alpha = 2 and x = beta. Then the distance of the point (alpha, beta) from the line 12x + 5y + 10 = 0 is equal to:
  • A. 3
  • B. 4
  • C. 2
  • D. 5

Solution

### Core Logic The expression cos|x| is everywhere differentiable. Thus, non-differentiability relies completely on the modulus function containing the quadratic factor, i.e., |x^2 - ax + 2|. Non-differentiability points generally happen where: x^2 - ax + 2 = 0 ### Step 1: Evaluation of Roots Given one of the roots is alpha = 2: 2^2 - a(2) + 2 = 0 implies 6 - 2a = 0 implies a = 3 Substituting a=3 gives the other root beta = 1. However, evaluating differentiability at x=1 reveals properties that invalidate standard options. ### Step 2: Conclusion Due to a technical contradiction in the configuration of the differentiable constraints at x=1, this question was officially dropped by NTA. ### Pattern Recognition If a quadratic inside a modulus has distinct real roots, it normally creates non-differentiable sharp turns unless a repeated multiplying factor outside cancels it out. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Limits, Continuity and Differentiability

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