Q81.Let α > 0 . If ∫α0 √x+α−√xx (1) 2 (2) 2√2 (3) 4 (4) √2 = sin t ∫xπ x > 0 then ϕ′( 4 ) is equal to √x
What This Question Tests
This problem requires finding the absolute minimum value of a function involving the greatest integer function (GIF) over a given interval, by analyzing the behavior of the quadratic term and the GIF separately in different sub-intervals.
Concepts Tested
Formulas Used
Minimum of ax^2+bx+c at x=-b/2a
📚 NCERT Sections This Tests
12.5 — A Hydrogen Atom Initially In The Ground Level Absorbs A Photon,
Physics Class 12 · Chapter 12
12.5 A hydrogen atom initially in the ground level absorbs a photon, which excites it to the n = 4 level. Determine the wavelength and frequency of photon.
9.15 — Apply Mirror Equation And The Condition:
Physics Class 12 · Chapter 9
9.15 Apply mirror equation and the condition: (a) f < 0 (concave mirror); u < 0 (object on left) (b) f > 0; u < 0 (c) f > 0 (convex mirror) and u < 0 (d) f < 0 (concave mirror); f < u < 0 to deduce the desired result.
14.2 — Which Of The Statements Given In Exercise 14.1 Is True For P-Type
Physics Class 12 · Chapter 14
14.2 Which of the statements given in Exercise 14.1 is true for p-type semiconductos.
📋 Question Details
- Chapter
- Applications of Derivatives
- Topic
- Absolute minimum of a function with GIF
- Year
- 2023
- Shift
- 31 Jan Shift 2
- Q Number
- Q81
- Type
- MCQ
- NCERT Ref
- Class 12 Mathematics Ch 6: Applications of Derivatives
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Q81.Given P(x) = x4 + ax3 + bx2 + cx + d such that x = 0 is the only real root of P ′(x) = 0 . If P(−1) < P(1), then in the interval [−1, 1] (1) P(−1) is the minimum and P(1) is the (2) P(−1) is not minimum but P(1) is the maximum maximum of P of P (3) P(−1) is the minimum and P(1) is not the (4) neither P(−1) is the minimum nor P(1) is the maximum of P maximum of P