If the function f defined on -1/3,1/3 by fx= 1/xlog_e(1+3x)/(1-2x), when x≠0 k ,when x=0 , is continuous, then k is equal to

Question

If the function f defined on -1/3,1/3 by fx= 1/xlog_e(1+3x)/(1-2x), when x≠0 k ,when x=0 , is continuous, then k is equal to.

MARKS Content Team · Accepted Answer

Correct answer: 5. limx→0⁡fx=limx→0⁡(ln⁡1+3x)/x-(ln⁡1-2x)/x =limx→0⁡(3ln⁡1+3x)/3x+(2ln1-2x)/(-2x)=3+2=5 limfx→0(ln1+fx)/fx=1 ∴fx, will be continuous if f0=limx→0⁡fx Thus, k=5

If the function $f$ defined on $(- \frac{1}{3}, 1 / 3)$ by $f (x) = \{\begin{matrix} \frac{1}{x} {l o g}_{e} (\frac{1 + 3 x}{1 - 2 x}), & when x \neq 0 \\ k & , when x = 0 \end{matrix}$ , is continuous, then $k$ is equal to.

Solution

Relevant Questions

If the function f defined on -13,1/3 by fx=1xloge1+3x1-2x,when x≠0k,when x=0, is continuous, then k is equal to.

Solution

Relevant Questions

If the function $f$ defined on $(- \frac{1}{3}, 1 / 3)$ by $f (x) = \{\begin{matrix} \frac{1}{x} {l o g}_{e} (\frac{1 + 3 x}{1 - 2 x}), & when x \neq 0 \\ k & , when x = 0 \end{matrix}$ , is continuous, then $k$ is equal to.