Let f: R- 0 R be a function such that f(x)-6 f (1 x )=35 3 x -5 2… - Vidyadip

MCQ

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

A
3
B
5
✓
4
D
6

✓

Answer

Correct option: C.

4

(C)
Sol. $F(x)-6 f(1 / x)=\frac{35}{3 x}-\frac{5}{2}.....(1)$
Replace $\mathrm{x} \rightarrow \frac{1}{\mathrm{x}}$
$F(1 / x)-6(x)=\frac{35 x}{3}-\frac{5}{2}....(2)$
Using (1) & (2)
$f(x)=-2 x-\frac{1}{3 x}+\frac{1}{2}$
$B=\lim _{x \rightarrow 0}\left(\frac{1}{\alpha x}+f(x)\right)$
$=\lim _{x \rightarrow 0}\left(\frac{1}{\alpha \mathrm{x}}-2 \mathrm{x}-\frac{1}{3 \mathrm{x}}+\frac{1}{2}\right)$
$\alpha=3, \quad B=1 / 2$
So, $\alpha+2 \mathrm{~B}=3+1=4$

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Explore more

More "SECTION - A [MATHS - MCQ]" from JEE Main 24-Jan-2025 Paper - Shift 1→JEE Main 24-Jan-2025 Paper - Shift 1 — all question types→JEE STD 11 Science question papers→Gujarat Board STD 11 Science question papers→Gujarat Board question paper generator→

Similar questions

Let $X=\{x \in R: \cos (\sin x)=\sin (\cos x)\} .$ The number of elements in $X$ is

The domain of the function $f(x) = {\sin ^{ - 1}}[{\log _2}(x/2)]$ is

Area bounded by $y = x\sin x$ and $x - $ axis between $x = 0$ and $x = 2\pi ,$ is

$a > 1,\;\mathop \smallint \limits_1^a \left[ x \right]f'\left( x \right)dx = $

If $A = \left[ {\begin{array}{*{20}{c}}2&5\\3&7\end{array}} \right]$and $B = \left[ {\begin{array}{*{20}{c}}0&3\\4&1\end{array}} \right],$then

If $x{\sin ^3}\alpha + y{\cos ^3}\alpha = \sin \alpha \cos \alpha $ and $x\sin \alpha - y\cos \alpha = 0,$ then ${x^2} + {y^2} = $

Product of real roots of the equation ${t^2}{x^2} + |x| + \,9 = 0$

Let $P=\left[\begin{array}{ccc}3 & -1 & -2 \\ 2 & 0 & \alpha \\ 3 & -5 & 0\end{array}\right]$, where $\alpha \in \mathbb{R}$. Suppose $Q=\left[q_{i j}\right]$ is a matrix such that $P Q=k I$, where $k \in \mathbb{R}, k \neq 0$ and $I$ is the identity matrix of order $3$ . If $q_{23}=-\frac{k}{8}$ and $\operatorname{det}(Q)=\frac{k^2}{2}$, then

($A$) $\quad \alpha=0, k=8$

($B$) $4 \alpha-k+8=0$

($C$) $\operatorname{det}(P \operatorname{adj}(Q))=2^9$

($D$) $\operatorname{det}(Q \operatorname{adj}(P))=2^{13}$

In a box, there are $20$ cards, out of which $10$ are lebelled as $\mathrm{A}$ and the remaining $10$ are labelled as $B$. Cards are drawn at random, one after the other and with replacement, till a second $A-$card is obtained. The probability that the second $A-$card appears before the third $B-$card is

$\left[\frac{2^{2020}+1}{2^{2018}+1}\right]+\left[\frac{3^{2020}+1}{3^{2018}+1}\right]+\left[\frac{4^{2020}+1}{4^{2018}+1}\right] +\left[\frac{5^{2020}+1}{5^{2018}+1}\right] + \left[\frac{6^{2020}+1}{6^{2018}+1}\right]$ is