jee-advanced 2004 Q17

jee-advanced · India · mains Stationary points and optimisation Prove or Verify an Algebraic Inequality (AM-GM, Cauchy-Schwarz, etc.)

17. Prove that $\sin \mathrm { x } + 2 \mathrm { x } \geq \frac { 3 \mathrm { x } \cdot ( \mathrm { x } + 1 ) } { \pi } \forall \mathrm { x } \in \left[ 0 , \frac { \pi } { 2 } \right]$. (Justify the inequality, if any used).
Sol. Let $\mathrm { f } ( \mathrm { x } ) = 3 \mathrm { x } ^ { 2 } + ( 3 - 2 \pi ) \mathrm { x } - \pi \sin \mathrm { x }$ $\mathrm { f } ( 0 ) = 0 , \mathrm { f } \left( \frac { \pi } { 2 } \right) = - \mathrm { ve }$ $\mathrm { f } ^ { \prime } ( \mathrm { x } ) = 6 \mathrm { x } + 3 - 2 \pi - \pi \cos \mathrm { x }$ $\mathrm { f } ^ { \prime \prime } ( \mathrm { x } ) = 6 + \pi \sin \mathrm { x } > 0$ $\Rightarrow \mathrm { f } ^ { \prime } ( \mathrm { x } )$ is increasing function in $\left[ 0 , \frac { \pi } { 2 } \right]$ ⇒ there is no local maxima of $\mathrm { f } ( \mathrm { x } )$ in $\left[ 0 , \frac { \pi } { 2 } \right]$ ⇒ graph of $\mathrm { f } ( \mathrm { x } )$ always lies below the x -axis [Figure] in $\left[ 0 , \frac { \pi } { 2 } \right]$. $\Rightarrow \mathrm { f } ( \mathrm { x } ) \leq 0$ in $\mathrm { x } \in \left[ 0 , \frac { \pi } { 2 } \right]$. $3 \mathrm { x } ^ { 2 } + 3 \mathrm { x } \leq 2 \pi \mathrm { x } + \pi \sin \mathrm { x } \Rightarrow \sin \mathrm { x } + 2 \mathrm { x } \geq \frac { 3 \mathrm { x } ( \mathrm { x } + 1 ) } { \pi }$.

If the line $2 x + \sqrt { 6 } y = 2$ touches the hyperbola $x ^ { 2 } - 2 y ^ { 2 } = 4$, then the point of contact is

17. Prove that $\sin \mathrm { x } + 2 \mathrm { x } \geq \frac { 3 \mathrm { x } \cdot ( \mathrm { x } + 1 ) } { \pi } \forall \mathrm { x } \in \left[ 0 , \frac { \pi } { 2 } \right]$. (Justify the inequality, if any used).

Sol. Let $\mathrm { f } ( \mathrm { x } ) = 3 \mathrm { x } ^ { 2 } + ( 3 - 2 \pi ) \mathrm { x } - \pi \sin \mathrm { x }$\\
$\mathrm { f } ( 0 ) = 0 , \mathrm { f } \left( \frac { \pi } { 2 } \right) = - \mathrm { ve }$\\
$\mathrm { f } ^ { \prime } ( \mathrm { x } ) = 6 \mathrm { x } + 3 - 2 \pi - \pi \cos \mathrm { x }$\\
$\mathrm { f } ^ { \prime \prime } ( \mathrm { x } ) = 6 + \pi \sin \mathrm { x } > 0$\\
$\Rightarrow \mathrm { f } ^ { \prime } ( \mathrm { x } )$ is increasing function in $\left[ 0 , \frac { \pi } { 2 } \right]$\\
⇒ there is no local maxima of $\mathrm { f } ( \mathrm { x } )$ in $\left[ 0 , \frac { \pi } { 2 } \right]$\\
⇒ graph of $\mathrm { f } ( \mathrm { x } )$ always lies below the x -axis\\
\includegraphics[max width=\textwidth, alt={}, center]{bf0af8eb-2b9c-49f0-8222-a9220127f02f-6_543_666_285_1017}\\
in $\left[ 0 , \frac { \pi } { 2 } \right]$.\\
$\Rightarrow \mathrm { f } ( \mathrm { x } ) \leq 0$ in $\mathrm { x } \in \left[ 0 , \frac { \pi } { 2 } \right]$.\\
$3 \mathrm { x } ^ { 2 } + 3 \mathrm { x } \leq 2 \pi \mathrm { x } + \pi \sin \mathrm { x } \Rightarrow \sin \mathrm { x } + 2 \mathrm { x } \geq \frac { 3 \mathrm { x } ( \mathrm { x } + 1 ) } { \pi }$.\\

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