Q 21 :    

Let $f:\mathbf{R}–\left\{0\right\}\to \mathbf{R}$ be a function such that $f\left(x\right)–6f\left(\frac{1}{x}\right)=\frac{35}{3x}–\frac{5}{2}$. If the $\underset{x\to 0}{\lim }(\frac{1}{\alpha x}+f(x\left)\right)=\beta ; \alpha ,\beta \in \mathbf{R}$, than $\alpha +2\beta$ is equal to          [2025]

• 4

   

• 3

   

• 5

   

• 6

   

Q 22 :    

The value of $\underset{n\to \infty }{\lim }\left(\sum _{k=1}^n\frac{k^3+6k^2+11k+5}{(k+3)!}\right)$ is :          [2025]

• 5/3

   

• 4/3

   

• 7/3

   

• 2

   

Q 23 :    

If $\underset{x\to 0}{\lim }{\left(\frac{\tan x}{x}\right)}^{\frac{1}{x^2}}=p$, then $96{\log }_{e}p$ is equal to __________.          [2025]

Q 24 :    

For t > –1, let ${\alpha }_t$ and ${\beta }_t$ be the roots of the equation $({(t+2)}^{1/7}–1)x^2+({(t+2)}^{1/6}–1)x+({(t+2)}^{1/21}–1)=0$. If $\underset{t\to –1^{+}}{\lim }{\alpha }_t=a \mathrm{and} \underset{t\to –1^{+}}{\lim }{\beta }_t=b$, then $72{(a+b)}^2$ is equal to __________.          [2025]

Q 25 :    

Let $f\left(x\right)=\underset{n\to \infty }{\lim }\sum _{r=0}^n\left(\frac{\tan (x/2^{r+1})+{\tan }^3(x/2^{r+1})}{1–{\tan }^2(x/2^{r+1})}\right)$. Then $\underset{x\to 0}{\lim }\frac{e^x–e^{f\left(x\right)}}{(x–f(x\left)\right)}$ is equal to __________.          [2025]

Q 26 :    

Let [t] be the greatest integer less than or equal to t. Then the least value of $p\in \mathbf{N}$ for which $\underset{x\to 0^{+}}{\lim }\left(x\right(\left[\frac{1}{x}\right]+\left[\frac{2}{x}\right]+...+\left[\frac{p}{x}\right])–x^2(\left[\frac{1}{x^2}\right]+\left[\frac{2^2}{x^2}\right]+...+\left[\frac{9^2}{x^2}\right]\left)\right)\ge 1$ is equal to _________.          [2025]