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\(f(x)-6 f\left(\frac{1}{x}\right)=\frac{35}{3 x}-\frac{5}{2} . \lim _{\text {If }}\left(\frac{1}{\alpha x}+f(x)\right)=…

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\(f(x)-6 f\left(\frac{1}{x}\right)=\frac{35}{3 x}-\frac{5}{2} . \lim _{\text {If }}\left(\frac{1}{\alpha x}+f(x)\right)=\beta . \quad \text { find }(\alpha+2 \beta)\). (24 Jan, Shift I, Memory based)

a

1

b

2

c

3

d

4

✓ Correct answer: d)

4

Explanation

\( f(x)-6 f\left(\frac{1}{x}\right)=\frac{35}{3 x}-\frac{5}{2}\)

\( f\left(\frac{1}{x}\right)-6 f(x)=\left(\frac{35 x}{3}-\frac{5}{2}\right)\)
\( 6 f\left(\frac{1}{x}\right)-36 f(x)=\left(\frac{35 x}{3}-\frac{5}{2}\right) \times 6 \)
\( -35 f(x)=\frac{35}{3 x}-\frac{5}{2}+70 x-15 \)
\( -35 f(x)=70 x+\frac{35}{3 x}-\frac{35}{2} \)
\( f(x)=\frac{1}{2}-2 x-\frac{1}{3 x}\)
\( {lim}_{x \rightarrow 0} \frac{1}{\alpha x}+\frac{1}{2}-2 x-\frac{1}{3 x}=\beta \)
\(={lim}_{x \rightarrow 0}\left(\frac{1}{\alpha}-\frac{1}{3}\right)\frac{1}{x}+\frac{1}{2}-2 x=\beta\)
\( \alpha=3 \)
\( \beta=\frac{1}{2}\)

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