jee-main 2025 Q85

jee-main · India · session2_03apr_shift2 Chain Rule Limit Evaluation Involving Composition or Substitution
Q85. Let $\mathrm { a } > 0$ be a root of the equation $2 x ^ { 2 } + x - 2 = 0$. If $\lim _ { x \rightarrow \frac { 1 } { \mathrm { a } } } \frac { 16 \left( 1 - \cos \left( 2 + x - 2 x ^ { 2 } \right) \right) } { ( 1 - \mathrm { a } x ) ^ { 2 } } = \alpha + \beta \sqrt { 17 }$, where $\alpha , \beta \in Z$, then $\alpha + \beta$ is equal to $\_\_\_\_$ Q86. Let the mean and the standard deviation of the probability distribution
X$\alpha$10- 3
$\mathrm { P } ( \mathrm { X } )$$\frac { 1 } { 3 }$K$\frac { 1 } { 6 }$$\frac { 1 } { 4 }$

be $\mu$ and $\sigma$, respectively. If $\sigma - \mu = 2$, then $\sigma + \mu$ is equal to $\_\_\_\_$
Q85. Let $\mathrm { a } > 0$ be a root of the equation $2 x ^ { 2 } + x - 2 = 0$. If $\lim _ { x \rightarrow \frac { 1 } { \mathrm { a } } } \frac { 16 \left( 1 - \cos \left( 2 + x - 2 x ^ { 2 } \right) \right) } { ( 1 - \mathrm { a } x ) ^ { 2 } } = \alpha + \beta \sqrt { 17 }$, where $\alpha , \beta \in Z$, then $\alpha + \beta$ is equal to $\_\_\_\_$\\
Q86.\\
Let the mean and the standard deviation of the probability distribution

\begin{center}
\begin{tabular}{ | c | c | c | c | c | }
\hline
X & $\alpha$ & 1 & 0 & - 3 \\
\hline
$\mathrm { P } ( \mathrm { X } )$ & $\frac { 1 } { 3 }$ & K & $\frac { 1 } { 6 }$ & $\frac { 1 } { 4 }$ \\
\hline
\end{tabular}
\end{center}

be $\mu$ and $\sigma$, respectively. If $\sigma - \mu = 2$, then $\sigma + \mu$ is equal to $\_\_\_\_$