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 |
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