Maximum and Minimum value of Trigonometric functions and Quadratic functions are the important one for Advanced Maths topic. SSC and other competitive exams usually weighs more importance and it will be easy to solve these types of problem, if proper rules and concepts are followed.
Limit values of Trigonometric Functions:
(i) -1 ≤ sinθ ≤ 1
(ii) -1 ≤ cosθ ≤ 1
(iii) -∞ ≤ tanθ ≤ ∞
(iv) -∞ ≤ cotθ ≤ ∞
(v) cosecθ ≤ -1 or cosecθ ≥ 1
(vi) secθ ≥ 1 or secθ ≤ -1
(vii) 0 ≤ sin2θ ≤ 1
(viii) 0 ≤ cos2θ ≤ 1
(ix) sin2θ ≤ |sinθ| ≤ 1
(x) cos2θ ≤ |cosθ| ≤ 1
(xi) sin2nθ + cos2nθ ≤ 1
Maximum and Minimum value of asinθ + bcosθ
(i) Maximum value is +√(a2 + b2)
(ii) Minimum value is -√(a2 + b2)
Minimum value of ax2 + b/x2 = 2√ab
Minimum value of asecθ + bcosecθ = (√a + √b)2
Maximum value of sinnθ.consnθ = (1/2)n
Maxima and Minima of Quadratic Equations:
- You might have seen the questions to find Maximum and Minimum value of any quadratic equation. These questions in the exam can be solved through direct formula based on it.
For example: Find the Maximum and Minimum value of x2 - 5x +6.
Now, f(x) = x2 - 5x + 6
Rules:
- The maximum or minimum of any quadratic equations occur at the x = -b/2a.
f(x) = ax2 + bx + c
Minima / Maxima occur at x = -b/2a
Minima / Maxima value of function f(x) = (4ac - b2)/4a
Conditions:
(i) If a is negative, the maximum value of the function
(ii) If a is positive, the minimum value of the function
Hence, for given equation i.e. f(x) = x2 - 5x + 6, the minimum value will be f(5/2)2 = (5/2) - 5(5/2) - 6 = -13/4
Limit values of Trigonometric Functions:
(i) -1 ≤ sinθ ≤ 1
(ii) -1 ≤ cosθ ≤ 1
(iii) -∞ ≤ tanθ ≤ ∞
(iv) -∞ ≤ cotθ ≤ ∞
(v) cosecθ ≤ -1 or cosecθ ≥ 1
(vi) secθ ≥ 1 or secθ ≤ -1
(vii) 0 ≤ sin2θ ≤ 1
(viii) 0 ≤ cos2θ ≤ 1
(ix) sin2θ ≤ |sinθ| ≤ 1
(x) cos2θ ≤ |cosθ| ≤ 1
(xi) sin2nθ + cos2nθ ≤ 1
Maximum and Minimum value of asinθ + bcosθ
(i) Maximum value is +√(a2 + b2)
(ii) Minimum value is -√(a2 + b2)
Minimum value of ax2 + b/x2 = 2√ab
Minimum value of asecθ + bcosecθ = (√a + √b)2
Maximum value of sinnθ.consnθ = (1/2)n
Maxima and Minima of Quadratic Equations:
- You might have seen the questions to find Maximum and Minimum value of any quadratic equation. These questions in the exam can be solved through direct formula based on it.
For example: Find the Maximum and Minimum value of x2 - 5x +6.
Now, f(x) = x2 - 5x + 6
Rules:
- The maximum or minimum of any quadratic equations occur at the x = -b/2a.
f(x) = ax2 + bx + c
Minima / Maxima occur at x = -b/2a
Minima / Maxima value of function f(x) = (4ac - b2)/4a
Conditions:
(i) If a is negative, the maximum value of the function
(ii) If a is positive, the minimum value of the function
Hence, for given equation i.e. f(x) = x2 - 5x + 6, the minimum value will be f(5/2)2 = (5/2) - 5(5/2) - 6 = -13/4
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