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Ex 6.3
Ex 6.3, 1 (ii)
Ex 6.3, 1 (iii) Important
Ex 6.3, 1 (iv)
Ex 6.3, 2 (i)
Ex 6.3, 2 (ii) Important
Ex 6.3, 2 (iii)
Ex 6.3, 2 (iv) Important
Ex 6.3, 2 (v) Important
Ex 6.3, 3 (i)
Ex 6.3, 3 (ii)
Ex 6.3, 3 (iii)
Ex 6.3, 3 (iv) Important You are here
Ex 6.3, 3 (v)
Ex 6.3, 3 (vi)
Ex 6.3, 3 (vii) Important
Ex 6.3, 3 (viii)
Ex 6.3, 4 (i)
Ex 6.3, 4 (ii) Important
Ex 6.3, 4 (iii)
Ex 6.3, 5 (i)
Ex 6.3, 5 (ii)
Ex 6.3, 5 (iii) Important
Ex 6.3, 5 (iv)
Ex 6.3,6
Ex 6.3,7 Important
Ex 6.3,8
Ex 6.3,9 Important
Ex 6.3,10
Ex 6.3,11 Important
Ex 6.3,12 Important
Ex 6.3,13
Ex 6.3,14 Important
Ex 6.3,15 Important
Ex 6.3,16
Ex 6.3,17
Ex 6.3,18 Important
Ex 6.3,19 Important
Ex 6.3, 20 Important
Ex 6.3,21
Ex 6.3,22 Important
Ex 6.3,23 Important
Ex 6.3,24 Important
Ex 6.3,25 Important
Ex 6.3, 26 Important
Ex 6.3, 27 (MCQ)
Ex 6.3,28 (MCQ) Important
Ex 6.3,29 (MCQ)
Last updated at June 12, 2023 by Teachoo
Ex 6.3, 3 Find the local maxima and local minima, if any, of the following functions. Find also the local maximum and the local minimum values, as the case may be: (iv) f (𝑥)=sin𝑥 –cos𝑥, 0<𝑥<2 𝜋f (𝑥)=sin𝑥 –cos𝑥, 0<𝑥<2 𝜋 Finding f’(𝒙) f’(𝑥)=cos𝑥−(−sin𝑥 ) f’(𝑥)=cos𝑥+sin𝑥 Putting f’(𝒙)=𝟎 cos𝑥+sin𝑥 = 0 cos𝑥=−sin𝑥 1=(−sin𝑥)/cos𝑥 (−sin𝑥)/cos𝑥 =1 – tan 𝑥=1 tan 𝑥=−1 Since 0 < 𝑥 < 2π & tan 𝑥 is negative tan θ lies in either 2nd or 4th quadrant So, value of 𝑥 is 𝑥=3𝜋/4 𝑜𝑟 7𝜋/4 Now finding f’’(𝑥) f’’(𝑥)=𝑑(cos𝑥 + sin𝑥 )/𝑑𝑥 f’’(𝑥)=−sin𝑥+cos𝑥 Putting 𝒙 = 𝟑𝝅/𝟒 f’’(3𝜋/4)=−𝑠𝑖𝑛(3𝜋/4)+𝑐𝑜𝑠(3𝜋/4) =−𝑠𝑖𝑛(𝜋−𝜋/4)+𝑐𝑜𝑠(𝜋− 𝜋/4) =−𝑠𝑖𝑛(𝜋/4)+(−𝑐𝑜𝑠 𝜋/4) =(−1)/√2−1/√2 =(−2)/√2 =−√2 < 0 Hence f’’(𝑥)<0 when 𝑥 = 3𝜋/4 Thus 𝑥 = 3𝜋/4 is point of local maxima ∴ f(𝑥) is maximum value at 𝑥 = 3𝜋/4 The local maximum value is f(𝑥)=sin𝑥−cos𝑥 f(3𝜋/4)=𝑠𝑖𝑛(3𝜋/4)−𝑐𝑜𝑠(3𝜋/4) =𝑠𝑖𝑛(𝜋−𝜋/4)−𝑐𝑜𝑠(𝜋−𝜋/4) =𝑠𝑖𝑛(𝜋/4)−(−𝑐𝑜𝑠 𝜋/4) =𝑠𝑖𝑛 𝜋/4+𝑐𝑜𝑠 𝜋/4 =1/√2+1/√2 =2/√2 =√2 Now, for 𝒙 = 𝟕𝝅/𝟒 f’’(𝑥)=−sin𝑥+cos𝑥 Putting 𝑥 = 7𝜋/4 f’’(7𝜋/4)=−sin(7𝜋/4)+cos(7𝜋/4) f’’(7𝜋/4)=−𝑠𝑖𝑛(2𝜋−𝜋/4)+𝑐𝑜𝑠(2𝜋−𝜋/4) =−(−𝑠𝑖𝑛(𝜋/4))+𝑐𝑜𝑠(𝜋/4) =𝑠𝑖𝑛 𝜋/4+𝑐𝑜𝑠 𝜋/4 =1/√2 + 1/√2 =2/√2 =√2 > 0 f’’(𝑥)>0 when 𝑥 = 7𝜋/4 Thus 𝑥 = 7𝜋/4 is point of local minima f(𝑥) has minimum value at 𝑥 = 7𝜋/4 Local minimum value is f(𝑥)=𝑠𝑖𝑛𝑥−𝑐𝑜𝑠𝑥 f(7𝜋/4)=𝑠𝑖𝑛(7𝜋/4)−𝑐𝑜𝑠(7𝜋/4) =𝑠𝑖𝑛(2𝜋−𝜋/4)−𝑐𝑜𝑠(2𝜋−𝜋/4) =−𝑠𝑖𝑛(𝜋/4)−𝑐𝑜𝑠(𝜋/4) =(−1)/√2 − 1/√2 =(−2)/√2 =−√2 Thus, f(𝑥) is maximum at x = 𝟑𝝅/𝟒 and maximum value is √𝟐 & f(𝑥) is minimum at x = 𝟕𝝅/𝟒 and maximum value is –√𝟐