Most Important Selected Qs for JEE AdvancedMathematicsApplication of Derivatives
Let f(x)=a b x+b 1-a^2 x+c , where |a| 1, b 0 then
Options
- Amaximum value of f(x) is b if c=0
- Bdifference of maximum and minimum value of f(x) is 2 b
- Cf(x)=c if x=- ⁻¹ a
- Df(x)=c if x= ⁻¹ a
Correct answer
C. f(x)=c if x=- ⁻¹ a
Step-by-step solution
aligned & f(x)= a^2 b^2+b^2-b^2 a^2 (x+ )+c=b (x+ )+c where = b 1-a^2 a b = 1-a^2 a . & f(x)_ -f(x)_ mn =c+b-(c-b)=2 b & Also, = ⁻¹ a & at x=- ⁻¹ a & f(x)=c . aligned