Let $f : R - \{0\} \rightarrow R$ be a function such that $f(x) - 6f\left(\frac{1}{x}\right) = \frac{35}{3x} - \frac{5}{2}$. If $\lim_{x \rightarrow 0} \left(\frac{1}{\alpha x} + f(x)\right) = \beta$,where $\alpha, \beta \in R$,then $\alpha + 2\beta$ is equal to:

  • A
    $3$
  • B
    $5$
  • C
    $4$
  • D
    $6$

Explore More

Similar Questions

If $a$ and $b$ are roots of the equation $px^2 + qx + r = 0$,then $\lim_{x \rightarrow b} \frac{1 - \cos 2(px^2 + qx + r)}{2(px - pb)^2}$ is equal to

For $t > -1$,let $\alpha_t$ and $\beta_t$ be the roots of the equation $\left((t+2)^{\frac{1}{7}}-1\right) x^2+\left((t+2)^{\frac{1}{6}}-1\right) x+\left((t+2)^{\frac{1}{21}}-1\right)=0$. If $\lim _{t \rightarrow -1^{+}} \alpha_t$ and $\lim _{t \rightarrow -1^{+}} \beta_t$ are the roots of the limiting equation,and $a+b$ is the sum of these roots,then $72(a+b)^2$ is equal to . . . . . . .

If $\lim _{x \rightarrow 1} \frac{x^2-ax+b}{x-1}=5$,then $(a+b)$ is equal to

$\lim _{x}$ ${\rightarrow -a} \frac{x^7+a^7}{x+a} = 7$ $\Rightarrow a = ?$

If $\lim _{x \rightarrow 0} \frac{\alpha x e^{x}-\beta \log _{e}(1+x)+\gamma x^{2} e^{-x}}{x \sin ^{2} x}=10$,where $\alpha, \beta, \gamma \in R$,then the value of $\alpha+\beta+\gamma$ is:

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo