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MathsHardMCQ2023 · 31 Jan Shift 2

Q79.Let f : R −{2, 6} →R be real valued function defined as f(x) = x+2x+1 . Then range of f is x2−8x+12 (1) (−∞, −214 ] ∪[ 214 , ∞) (2) (−∞, −214 ] ∪[0, ∞) (3) (−∞, −214 ) ∪(0, ∞) (4) (−∞, −214 ] ∪[1, ∞)

What This Question Tests

This question involves manipulating inverse trigonometric functions, specifically sin⁻¹(sinx) and cos⁻¹(sinx), over a given interval to solve an inequality, then using the properties of sin⁻¹x + cos⁻¹x to simplify another equation and solve for alpha.

Concepts Tested

Properties of sin⁻¹(sinx) and cos⁻¹(sinx)Domain and range of inverse trigonometric functionsSolving inequalities

Formulas Used

sin⁻¹(sinx) = x (for x ∈ [-π/2, π/2])

cos⁻¹(cosx) = x (for x ∈ [0, π])

sin⁻¹(x) + cos⁻¹(x) = π/2

📚 NCERT Sections This Tests

9.15Apply Mirror Equation And The Condition:

Physics Class 12 · Chapter 9

70% 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.

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.

9.18For 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.