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 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 wℎ𝑒𝑟𝑒 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 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 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 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 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 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