question_answer

Let S be the set of all functions \[f:\left[ 0,1 \right]\to R,\]which are continuous on [0, 1] and differentiable on (0, 1). Then for every f in S, there exists a c \[\in \](0, 1), depending on f, such that   [JEE MAIN Held on 08-01-2020 Evening]

A) \[\left| f\left( c \right)-f\left( 1 \right)< \right|f'\left( c \right)|\]

B) \[\frac{f(1)-f(c)}{1-c}f'\left( c \right)\]

C) \[\left| f\left( c \right)-f\left( 1 \right)<\left( 1-c \right) \right|f'\left( c \right)|\]

D) \[\left| f\left( c \right)+f\left( 1 \right)<\left( 1+c \right) \right|f'\left( c \right)|\]

Correct Answer: A , B , C , D

Solution :

(Bonus)

Case - 1 If f(x) is non constant

By L.M.V.T in \[x\,\,\in (0,c)\,\,\,\,\,\,{f}'(\alpha )=\frac{f(c)-f(0)}{c}\]

\[\Rightarrow \,\,\,\,\,\,\,\,\,c\,f'\left( \alpha  \right)=f(c)-f(0)\]             .? (i)

By L.M.V.T in \[x\in \left( c,1 \right)\text{ }f'\left( \beta  \right)=\frac{f\left( 1 \right)-f\left( c \right)}{1-c}\]

\[\Rightarrow \,\,\,\,\,\,f'\left( \beta  \right)\left( 1-c \right)=f(1)-f(c)\]                .? (ii)

By (i) + (ii)

\[\Rightarrow \,\,\,\,\,cf'\alpha +\left( 1-c \right)f'\left( \beta  \right)=f\left( 1 \right)-f\left( 0 \right)\]        ?. (iii)

By L.M.V.T in \[\left( 0,\text{ }1 \right),f'\left( c \right)=f\left( 1 \right)-f\left( 0 \right)\]    ... (iv)

From equation (iii) and (iv)

\[f'\left( c \right)=cf'\left( \alpha  \right)+\left( 1-c \right)f'\left( \beta  \right)\]

\[\Rightarrow \,\,\,\,\,\,\left| f'(c) \right|=|cf'(\alpha )+(1-c)f'(\beta )|\]                       ... (v)

From equation (ii)

\[\Rightarrow \,\,\,\,\,\left| f'(c) \right|=|cf'(\alpha )+f(1)-f(c)|\]

\[\therefore \,\,\,\,\,\,\left| f'\left( c \right) \right|>\left| f\left( 1 \right)-f\left( c \right) \right|\]

Case 2 - If f(x) is taken as constant then options (2) will satisfy which is contradictory. So it can be given as BONUS

Solution :

(Bonus)

Case - 1 If f(x) is non constant

By L.M.V.T in \[x\,\,\in (0,c)\,\,\,\,\,\,{f}'(\alpha )=\frac{f(c)-f(0)}{c}\]

\[\Rightarrow \,\,\,\,\,\,\,\,\,c\,f'\left( \alpha  \right)=f(c)-f(0)\]             .? (i)

By L.M.V.T in \[x\in \left( c,1 \right)\text{ }f'\left( \beta  \right)=\frac{f\left( 1 \right)-f\left( c \right)}{1-c}\]

\[\Rightarrow \,\,\,\,\,\,f'\left( \beta  \right)\left( 1-c \right)=f(1)-f(c)\]                .? (ii)

By (i) + (ii)

\[\Rightarrow \,\,\,\,\,cf'\alpha +\left( 1-c \right)f'\left( \beta  \right)=f\left( 1 \right)-f\left( 0 \right)\]        ?. (iii)

By L.M.V.T in \[\left( 0,\text{ }1 \right),f'\left( c \right)=f\left( 1 \right)-f\left( 0 \right)\]    ... (iv)

From equation (iii) and (iv)

\[f'\left( c \right)=cf'\left( \alpha  \right)+\left( 1-c \right)f'\left( \beta  \right)\]

\[\Rightarrow \,\,\,\,\,\,\left| f'(c) \right|=|cf'(\alpha )+(1-c)f'(\beta )|\]                       ... (v)

From equation (ii)

\[\Rightarrow \,\,\,\,\,\left| f'(c) \right|=|cf'(\alpha )+f(1)-f(c)|\]

\[\therefore \,\,\,\,\,\,\left| f'\left( c \right) \right|>\left| f\left( 1 \right)-f\left( c \right) \right|\]

Case 2 - If f(x) is taken as constant then options (2) will satisfy which is contradictory. So it can be given as BONUS

Solution :

(Bonus)

Case - 1 If f(x) is non constant

By L.M.V.T in \[x\,\,\in (0,c)\,\,\,\,\,\,{f}'(\alpha )=\frac{f(c)-f(0)}{c}\]

\[\Rightarrow \,\,\,\,\,\,\,\,\,c\,f'\left( \alpha  \right)=f(c)-f(0)\]             .? (i)

By L.M.V.T in \[x\in \left( c,1 \right)\text{ }f'\left( \beta  \right)=\frac{f\left( 1 \right)-f\left( c \right)}{1-c}\]

\[\Rightarrow \,\,\,\,\,\,f'\left( \beta  \right)\left( 1-c \right)=f(1)-f(c)\]                .? (ii)

By (i) + (ii)

\[\Rightarrow \,\,\,\,\,cf'\alpha +\left( 1-c \right)f'\left( \beta  \right)=f\left( 1 \right)-f\left( 0 \right)\]        ?. (iii)

By L.M.V.T in \[\left( 0,\text{ }1 \right),f'\left( c \right)=f\left( 1 \right)-f\left( 0 \right)\]    ... (iv)

From equation (iii) and (iv)

\[f'\left( c \right)=cf'\left( \alpha  \right)+\left( 1-c \right)f'\left( \beta  \right)\]

\[\Rightarrow \,\,\,\,\,\,\left| f'(c) \right|=|cf'(\alpha )+(1-c)f'(\beta )|\]                       ... (v)

From equation (ii)

\[\Rightarrow \,\,\,\,\,\left| f'(c) \right|=|cf'(\alpha )+f(1)-f(c)|\]

\[\therefore \,\,\,\,\,\,\left| f'\left( c \right) \right|>\left| f\left( 1 \right)-f\left( c \right) \right|\]

Case 2 - If f(x) is taken as constant then options (2) will satisfy which is contradictory. So it can be given as BONUS

Solution :

(Bonus)

Case - 1 If f(x) is non constant

By L.M.V.T in \[x\,\,\in (0,c)\,\,\,\,\,\,{f}'(\alpha )=\frac{f(c)-f(0)}{c}\]

\[\Rightarrow \,\,\,\,\,\,\,\,\,c\,f'\left( \alpha  \right)=f(c)-f(0)\]             .? (i)

By L.M.V.T in \[x\in \left( c,1 \right)\text{ }f'\left( \beta  \right)=\frac{f\left( 1 \right)-f\left( c \right)}{1-c}\]

\[\Rightarrow \,\,\,\,\,\,f'\left( \beta  \right)\left( 1-c \right)=f(1)-f(c)\]                .? (ii)

By (i) + (ii)

\[\Rightarrow \,\,\,\,\,cf'\alpha +\left( 1-c \right)f'\left( \beta  \right)=f\left( 1 \right)-f\left( 0 \right)\]        ?. (iii)

By L.M.V.T in \[\left( 0,\text{ }1 \right),f'\left( c \right)=f\left( 1 \right)-f\left( 0 \right)\]    ... (iv)

From equation (iii) and (iv)

\[f'\left( c \right)=cf'\left( \alpha  \right)+\left( 1-c \right)f'\left( \beta  \right)\]

\[\Rightarrow \,\,\,\,\,\,\left| f'(c) \right|=|cf'(\alpha )+(1-c)f'(\beta )|\]                       ... (v)

From equation (ii)

\[\Rightarrow \,\,\,\,\,\left| f'(c) \right|=|cf'(\alpha )+f(1)-f(c)|\]

\[\therefore \,\,\,\,\,\,\left| f'\left( c \right) \right|>\left| f\left( 1 \right)-f\left( c \right) \right|\]

Case 2 - If f(x) is taken as constant then options (2) will satisfy which is contradictory. So it can be given as BONUS