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MathsMediumMCQ2022 · 26 Jun Shift 1

Q73.The sum of the absolute minimum and the absolute maximum values of the function f(x) = 3x −x2 + 2 −x in the interval [−1, 2] is (1) √17+3 (2) √17+5 2 2 (3) 5 (4) 9−√17 2

What This Question Tests

This question requires finding the absolute maximum and minimum values of a function on a closed interval by evaluating the function at critical points and endpoints.

Concepts Tested

Absolute maximum and minimumDerivative of a functionCritical points in an interval

Formulas Used

f'(x) = 0 for critical points

Comparison of function values at critical points and endpoints

📚 NCERT Sections This Tests

9.18For Fixed Distance S Between Object And Screen, The Lens Equation

Physics Class 12 · Chapter 9

69% match

9.18 For fixed distance s between object and screen, the lens equation does not give a real solution for u or v if f is greater than s/4. Therefore, fmax = 0.75 m.

9.17(A) Sin I¢C = 1.44/1.68 Which Gives I¢C = 59°. Total Internal Reflection

Physics Class 12 · Chapter 9

69% match

9.17 (a) sin i¢c = 1.44/1.68 which gives i¢c = 59°. Total internal reflection takes place when i > 59° or when r < rmax = 31°. Now, (sin i /sin r max max ) = 1.68 , which gives imax ~ 60°. Thus, all incident rays of angles in the range 0 < i < 60° will suffer total internal reflections in the pipe. (If the length of the pipe is finite, which it is in practice, there will be a lower limit on i determined by the ratio of the diameter to the length of the pipe.) (b) If there is no outer coating, i¢c = sin–1(1/1.68) = 36.5°. Now, i = 90° will have r = 36.5° and i¢ = 53.5° which is greater than i¢c. Thus, all incident rays (in the range 53.5° < i < 90°) will suffer total internal reflections.

9.15Apply Mirror Equation And The Condition:

Physics Class 12 · Chapter 9

69% match

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.