jee-main 2021 Q77

jee-main · India · session4_31aug_shift1 Chain Rule Chain Rule Combined with Fundamental Theorem of Calculus

Let $f$ be a non-negative function in $[ 0,1 ]$ and twice differentiable in $( 0,1 )$. If $\int _ { 0 } ^ { x } \sqrt { 1 - \left( f ^ { \prime } ( t ) \right) ^ { 2 } } \mathrm { dt } = \int _ { 0 } ^ { x } f ( \mathrm { t } ) \mathrm { dt } , 0 \leq x \leq 1$ and $f ( 0 ) = 0$, then $\lim _ { x \rightarrow 0 } \frac { 1 } { x ^ { 2 } } \int _ { 0 } ^ { x } f ( \mathrm { t } ) \mathrm { dt } :$
(1) does not exist
(2) equals 0
(3) equals 1
(4) equals $\frac { 1 } { 2 }$

Let $f$ be a non-negative function in $[ 0,1 ]$ and twice differentiable in $( 0,1 )$. If $\int _ { 0 } ^ { x } \sqrt { 1 - \left( f ^ { \prime } ( t ) \right) ^ { 2 } } \mathrm { dt } = \int _ { 0 } ^ { x } f ( \mathrm { t } ) \mathrm { dt } , 0 \leq x \leq 1$ and $f ( 0 ) = 0$, then $\lim _ { x \rightarrow 0 } \frac { 1 } { x ^ { 2 } } \int _ { 0 } ^ { x } f ( \mathrm { t } ) \mathrm { dt } :$\\
(1) does not exist\\
(2) equals 0\\
(3) equals 1\\
(4) equals $\frac { 1 } { 2 }$

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