Ex 6.5,2 - Chapter 6 Class 12 Application of Derivatives

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Ex 6.5,2 (Method 1) Find the maximum and minimum values, if any, of the following functions given by (𝑖) 𝑓 (𝑥)=|𝑥 + 2| –1 𝑓 (𝑥)=|𝑥 + 2| –1 Minimum value of |𝑥 + 2|=0 Minimum value of f(𝑥)= minimum value of |𝑥 + 2| –1 =0−1 =−1 Hence minimum value of f(𝒙)=−𝟏 And there is no maximum value of f(x) Ex 6.5,2 (Method 2) Find the maximum and minimum values, if any, of the following functions given by (𝑖) 𝑓 (𝑥)=|𝑥 + 2| –1 𝑓 (𝑥)=|𝑥 + 2| –1 f(𝑥)= {█((𝑥+2)−1" w" ℎ𝑒𝑟𝑒 𝑥≥−2@"–" (𝑥+2)−1 𝑤ℎ𝑒𝑟𝑒 " " 𝑥<−2" " )┤ f(𝑥)= {█(𝑥+1" , " 𝑤ℎ𝑒𝑟𝑒 𝑥≥−2@−𝑥−3" , where " 𝑥<−2)┤ We check sign of f’(𝑥) at 𝑥=−2 For 𝒙<−𝟐 f(𝑥)=−𝑥−3 f’(𝑥)=−1 ∴ f’(𝑥)<0 for any value of 𝑥<−2 For x > –2 f(𝑥) =𝑥+1 f’(𝑥) =1 ∴ f’(𝑥)=1>0 for any value of 𝑥>−2 Thus, at x = –2, f’(x) changes sign from negative to positive ⇒ 𝑥=−2 is point of minima of f(𝑥) Finding minimum value of f(𝑥) Putting 𝑥=−2 f(𝑥) =|𝑥 + 2| –1 f(2)=|−2+ 2| –1 = 0 – 1 = −1 Minimum value of f(𝑥) is –1 at 𝑥=−2 f(𝑥) has no maximum value Ex 6.5,2 (Method 1) Find the maximum and minimum values, if any, of the following functions given by (ii) 𝑓(𝑥)= – | 𝑥 +1|+3 f(𝑥)= – | 𝑥 +1|+3 We know that | 𝑥 +1|≥0 So, −| 𝑥 +1|≤0 Maximum value of g(𝑥) = maximum value of – | 𝑥 +1|+3 = 0 + 3 = 3 Hence maximum value of f(x) is 3 And there is no minimum value of f(x) Ex 6.5,2 (Method 2) Find the maximum and minimum values, if any, of the following functions given by (ii) 𝑓(𝑥)= – |𝑥+1|+3 g(𝑥)= –|𝑥+1|+3 g(𝑥)= {█("–" (𝑥+1)+3" " 𝑥≥−1@"–" (−(𝑥+1))+3" " 𝑥<−1)┤ g(𝑥)= {█(−𝑥+2" " 𝑥≥−1@𝑥+4" " 𝑥<−1)┤ We check sign of f’(𝑥) at 𝑥=−1 For 𝒙<−𝟏 f(𝑥)=𝑥+4 f’(𝑥)=1 ∴ f’(𝑥)>0 for any value of 𝑥<−1 For x > –1 f(𝑥) =−𝑥+2 f’(𝑥) =−1 ∴ f’(𝑥)=<0 for any value of 𝑥>−1 Thus, at x = –1, f’(x) changes sign from positive to negative ⇒ 𝑥=−1 is point of maxima of f(𝑥) Finding maximum value of f(𝑥) Putting 𝑥=−1 f(𝑥)=− | 𝑥 +1|+3 f(𝑥)=− | 𝑥 +1|+3 f(−1) =− |−1 +1|+3 =− |0|+3 = 3 Hence Maximum value is 3 And there is no minimum value of f(x) Ex 6.5,2 Find the maximum and minimum values, if any, of the following functions given by (iii) ℎ(𝑥)= sin ⁡(2𝑥)+ 5 ℎ(𝑥)= sin ⁡(2𝑥)+ 5 We know that –1 ≤ sin θ ≤ 1 –1 ≤ sin 2𝑥 ≤ 1 Adding 5 both sides –1 + 5 ≤ sin2𝑥 + 5 ≤ 1 + 5 4 ≤ sin 2𝑥 + 5 ≤ 6 4 ≤ f(𝑥)≤6 Hence Maximum value of f(𝒙)=𝟔 & Minimum value of f(𝒙)=𝟒 Ex 6.5,2 Find the maximum and minimum values, if any, of the following functions given by (iv) 𝑓 (𝑥)=|sin⁡4𝑥+3| 𝑓 (𝑥)=| sin⁡4𝑥+3| We know that –1 ≤ sin θ ≤ 1 So, –1 ≤ sin 4𝑥 ≤ 1 Adding 3 both sides –1 + 3 ≤ sin 4𝑥 + 3 ≤ 1 + 3 2 ≤ sin 4𝑥 +3 ≤ 4 Taking modulus |2| ≤ | sin⁡4𝑥+3| ≤ |4| 2 ≤ | sin⁡4𝑥+3| ≤ |4| 2 ≤ f(𝑥)≤4 Hence Maximum value of f(𝒙) is 4 & Minimum value of f(𝒙) is 2 Ex 6.5,2 Find the maximum and minimum values, if any, of the following functions given by (v) ℎ(𝑥)=𝑥 + 1 , 𝑥 ∈ (−1 , 1) Drawing graph of f(𝑥)=𝑥+1 ℎ(𝑥) have Maximum value of point closest to x = 1 & Minimum value of point closest to x = –1 but its not possible to locate such points Thus the given function has neither the maximum value nor minimum value