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Let (2, 3) be the largest open interval in which the function f(x)=2log_e(x-2)-x^2+ax+1 is strictly increasing and (b, c) be the largest open interval, in which the function g(x)=(x-1)^3(x+2-a)^2 is strictly decreasing. Then 100(a+b-c) is equal to: [cite: 3335, 3336, 3337]

Solution & Explanation

### Related Formula A differentiable function is strictly increasing where its first derivative is positive (f'(x) ge 0) and strictly decreasing where its first derivative is negative (g'(x) le 0). ### Step 1: Differentiate f(x) to solve for a Find f'(x) : f'(x) = frac2x-2 - 2x + a ge 0 Since (2,3) is the largest open interval of increasing behavior [cite: 3335, 3336], the transition root occurs at the upper boundary x=3 : f'(3) = 0 Rightarrow frac23-2 - 2(3) + a = 0 Rightarrow 2 - 6 + a = 0 Rightarrow a = 4 [cite: 3989, 3990, 3991] ### Step 2: Differentiate g(x) to solve for interval (b, c) Substitute a = 4 into g(x) [cite: 4002, 4003]: g(x) = (x-1)^3(x + 2 - 4)^2 = (x-1)^3(x-2)^2 Compute g'(x) using the product rule : g'(x) = 3(x-1)^2(x-2)^2 + (x-1)^3 cdot 2(x-2) g'(x) = (x-1)^2(x-2)left[3(x-2) + 2(x-1)right] = (x-1)^2(x-2)(5x - 8) [cite: 4005, 4006] For g(x) to be strictly decreasing, set g'(x) < 0 : Since (x-1)^2 ge 0, we require (x-2)(5x-8) < 0 Rightarrow x in left(frac85, 2right) [cite: 4006, 4007]. Thus, the interval is (b, c) = left(frac85, 2right), yielding b = frac85 = 1.6 and c = 2[cite: 3336, 4007]. ### Step 3: Compute Target Value Evaluate the objective expression [cite: 3337, 4008]: 100(a + b - c) = 100left(4 + frac85 - 2right) = 100(3.6) = 360 ### Pattern Recognition Boundary parameters of maximal monotonic intervals are always the exact zero-crossings of the derivative function. Setting f'(3) = 0 immediately establishes a=4 without secondary algebraic transformations. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Application of Derivatives

Reference Study Guides

Application Of Derivatives

More Application of Derivatives Previous-Year Questions — Page 2

Q51 2025 Maxima and Minima
Let a > 0. If the function f(x) = 6x^3 - 45ax^2 + 108a^2x + 1 attains its local maximum and minimum values at the points x_1 and x_2 respectively such that x_1x_2 = 54, then a + x_1 + x_2 is equal to :-
  • A. 15
  • B. 18
  • C. 24
  • D. 13

Solution

### Related Formula For a function f(x) to have a local maximum or minimum at a point, its first derivative must vanish at that point: f'(x) = 0 ### Core Logic Differentiating the given function f(x) = 6x^3 - 45ax^2 + 108a^2x + 1 with respect to x: f'(x) = 18x^2 - 90ax + 108a^2 = 0 Dividing the entire equation by 18: x^2 - 5ax + 6a^2 = 0 Factoring the quadratic equation: (x - 2a)(x - 3a) = 0 Thus, the critical points are x = 2a and x = 3a. Since a > 0, we assign x_1 = 2a and x_2 = 3a. ### Step 1: Finding the value of a Given that the product of the roots x_1x_2 = 54: (2a)(3a) = 54 6a^2 = 54 implies a^2 = 9 Since a > 0, we have a = 3. ### Step 2: Calculating the final expression Substituting a = 3 back to find x_1 and x_2: x_1 = 2(3) = 6 x_2 = 3(3) = 9 Now, evaluating a + x_1 + x_2: a + x_1 + x_2 = 3 + 6 + 9 = 18 ### Pattern Recognition When critical points are expressed in terms of a parameter, relate the given root condition (x_1x_2 = 54) directly to the product of roots formula (fracca) of the simplified quadratic equation to save factoring time. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Application of Derivatives
Q62 2025 Maxima and Minima
Consider the region R = left\(x,y)colon x leq y leq 9 - frac113x^2, \, x geq 0right\. The area of the largest rectangle with sides parallel to the coordinate axes inscribed in R is :
  • A. frac625111
  • B. frac730119
  • C. frac567121
  • D. frac821123

Solution

### Related Formula The area of a rectangle bounded between an upper function curve y_2(x) and lower function curve y_1(x) spanning width x = t is formulated as: A(t) = t cdot [y_2(t) - y_1(t)] ### Core Logic The region is bounded below by the line y = x and above by the downward opening parabola y = 9 - frac113x^2 in the first quadrant:
Maxima and Minima
Maxima and Minima
Let a vertex of the rectangle lie on the upper parabolic boundary at x = t. The corresponding height span of the rectangle is bounded by the line y = t at the base:
Maxima and Minima
Maxima and Minima
Thus, the area equation as a function of variable parameter t is: A(t) = t cdot left( 9 - frac113t^2 - t right) = 9t - t^2 - frac113t^3 ### Step 1: Differentiate to Identify Critical Values Differentiate the area function with respect to t and equate to zero: fracdAdt = 9 - 2t - 11t^2 = 0 11t^2 + 2t - 9 = 0 11t^2 + 11t - 9t - 9 = 0 +(11t - 9)(t + 1) = 0 Since x \geq 0, we reject the negative root t = -1. This isolates the physical critical point at: t = frac911 ### Step 2: Evaluate Maximum Inscribed Area Substitute t = \frac{9}{11} back into the factored area equation formulation: A_textmax = frac911 cdot left( 9 - frac911 - frac113left(frac911right)^2 right) A_textmax = frac911 cdot left( 9 - frac911 - frac2711 right) A_textmax = frac911 cdot left( 9 - frac3611 right) = frac911 cdot frac6311 = frac567121$ ### Pattern Recognition For optimized area allocation inside custom functional borders, setting the optimization parameter strictly to the horizontal coordinates simplifies high-degree polynomials down to standard derivative templates. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Application of Derivatives

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