




Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class
Ex 6.3
Ex 6.3, 1 (ii)
Ex 6.3, 1 (iii) Important You are here
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
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 May 29, 2023 by Teachoo
Ex 6.3,1 (Method 1) Find the maximum and minimum values, if any, of the following functions given by (iii) π (π₯) = β(π₯ β 1)^2+10 f(π₯)=β(π₯β1)^2+10 Finding fβ (x) Diff w.r.t π₯ fβ(π₯)=π(β(π₯β1)^2+10)/ππ₯ fβ(π₯) = β2(π₯β1)(π(π₯β1)/ππ₯)+0 fβ(π₯) = β2(π₯β1)(1β0) + 0 fβ(π₯)=β2(π₯β1) Putting fβ(π)=π β2(π₯β1)=0 (π₯β1)=0 π₯=1 Hence, π₯=1 is point of Maxima & No point of Minima Thus, f(π₯) has maximum value at π₯=1 Putting x=1 in f(x) f(π₯)=β(π₯β1)^2+10 f(1)=β(1β1)^2+10 = 0 + 10 = 10 Maximum value of f(π₯) is 10 There is no minimum value of f(π) Ex 6.3,1 (Method 2) Find the maximum and minimum values, if any, of the following functions given by (iii) π (π₯) = β (π₯ β 1)^2+10 f(π₯)=β(π₯β1)^2+10 Finding fβ(x) Diff w.r.t π₯ fβ(π₯)=π(β(π₯β1)^2+10)/ππ₯ fβ(π₯) = β2(π₯β1)(π(π₯β1)/ππ₯)+0 fβ(π₯) = β2(π₯β1)(1β0) + 0 fβ(π₯)=β2(π₯β1) Putting fβ(π)=π β2(π₯β1)=0 (π₯β1)=0 π₯=1 Finding fββ(π) fβ(π₯)=β2(π₯β1) Again diff w.r.t π₯ fββ(π₯)=π(β2(π₯ β 1))/ππ₯ =β2 π(π₯ β 1)/ππ₯ =β2(1β0) =β2 Since fββ(π₯) < 0 for π₯=1 Hence, f(π₯) has Maximum value at π₯=1 Finding maximum value of f(π) f(π₯)=β(π₯β1)^2+10 Putting π₯=1 f(π₯) =β(1β1)^2+10 = 0 + 10 = 10 Maximum value of f(π) is 10 There is no minimum value of f(π)