JEE Main · 2020 · 08 Jan 2020 · Shift 2 · Continuity and Differentiability

Test the options against specific $f$. For $f(x)=x^2$, option D needs $\frac{1-c^2}{1-c}=1+c=2c$, i.e. $c=1\notin (0,1)$ — D fails. Option A fails for $f(x)=x$ (both sides equal $1-c$, never strict); option B fails for a constant $f$ (right side $=0$). Option C, $|f(c)-f(1)|<|f^{\prime }(c)|$, holds for every $f\in S$: by the mean value theorem $|f(1)-f(c)|=(1-c)|f^{\prime }(\xi )|$ with $0<1-c<1$. Hence C.