6. T is a parallelopiped in which A, B, C and D are vertices of one face. And the face just above it has corresponding vertices $\mathrm { A } ^ { \prime } , \mathrm { B } ^ { \prime } , \mathrm { C } ^ { \prime } , \mathrm { D } ^ { \prime }$. T is now compressed to S with face ABCD remaining same and $\mathrm { A } ^ { \prime }$, $B ^ { \prime } , C ^ { \prime } , D ^ { \prime }$ shifted to $A ^ { \prime \prime } , B ^ { \prime \prime } , C ^ { \prime \prime } , D ^ { \prime \prime }$ in $S$. The volume of parallelopiped $S$ is reduced to $90 \%$ of T. Prove that locus of $\mathrm { A } ^ { \prime \prime }$ is a plane. Sol. Let the equation of the plane ABCD be $\mathrm { ax } + \mathrm { by } + \mathrm { cz } + \mathrm { d } = 0$, the point $\mathrm { A } ^ { \prime \prime }$ be $( \alpha , \beta , \gamma )$ and the height of the parallelopiped ABCD be h . $\Rightarrow \frac { | \mathrm { a } \alpha + \mathrm { b } \beta + \mathrm { c } \gamma + \mathrm { d } | } { \sqrt { \mathrm { a } ^ { 2 } + \mathrm { b } ^ { 2 } + \mathrm { c } ^ { 2 } } } = 0.9 \mathrm {~h} . \Rightarrow \mathrm { a } \alpha + \mathrm { b } \beta + \mathrm { c } \gamma + \mathrm { d } = \pm 0.9 \mathrm {~h} \sqrt { \mathrm { a } ^ { 2 } + \mathrm { b } ^ { 2 } + \mathrm { c } ^ { 2 } }$ ⇒ the locus of $\mathrm { A } ^ { \prime \prime }$ is a plane parallel to the plane ABCD .
Let $f ( x )$ be a differentiable function and $\int _ { 0 } ^ { t ^ { 2 } } x f ( x ) d x = \frac { 2 } { 5 } t ^ { 5 } , t > 0$. Then $f \left( \frac { 4 } { 25 } \right) =$
6. T is a parallelopiped in which A, B, C and D are vertices of one face. And the face just above it has corresponding vertices $\mathrm { A } ^ { \prime } , \mathrm { B } ^ { \prime } , \mathrm { C } ^ { \prime } , \mathrm { D } ^ { \prime }$. T is now compressed to S with face ABCD remaining same and $\mathrm { A } ^ { \prime }$, $B ^ { \prime } , C ^ { \prime } , D ^ { \prime }$ shifted to $A ^ { \prime \prime } , B ^ { \prime \prime } , C ^ { \prime \prime } , D ^ { \prime \prime }$ in $S$. The volume of parallelopiped $S$ is reduced to $90 \%$ of T. Prove that locus of $\mathrm { A } ^ { \prime \prime }$ is a plane.
Sol. Let the equation of the plane ABCD be $\mathrm { ax } + \mathrm { by } + \mathrm { cz } + \mathrm { d } = 0$, the point $\mathrm { A } ^ { \prime \prime }$ be $( \alpha , \beta , \gamma )$ and the height of the parallelopiped ABCD be h .\\
$\Rightarrow \frac { | \mathrm { a } \alpha + \mathrm { b } \beta + \mathrm { c } \gamma + \mathrm { d } | } { \sqrt { \mathrm { a } ^ { 2 } + \mathrm { b } ^ { 2 } + \mathrm { c } ^ { 2 } } } = 0.9 \mathrm {~h} . \Rightarrow \mathrm { a } \alpha + \mathrm { b } \beta + \mathrm { c } \gamma + \mathrm { d } = \pm 0.9 \mathrm {~h} \sqrt { \mathrm { a } ^ { 2 } + \mathrm { b } ^ { 2 } + \mathrm { c } ^ { 2 } }$\\
⇒ the locus of $\mathrm { A } ^ { \prime \prime }$ is a plane parallel to the plane ABCD .\\