Let fx-ax-bx-2- limx-0fx-2 and limx- fx-1- The value of f-1 is

Given that fx=ax+bx+2,limx→0 nbsp;fx=2 nbsp;and nbsp;limx→∞ nbsp;fx=1.Now, nbsp;limx→0 nbsp;fx=2 nbsp;⇒limx→0 nbsp;ax+bx+2=2⇒b2=2⇒b=4limx→∞fx=1⇒ ...

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Byju's Answer

Standard XI

Mathematics

Substitution Method to Remove Indeterminate Form

Let fx=ax+bx+...

Question

Let $f (x) = \frac{ax + b}{x + 2}, lim x \to 0 f (x) = 2 and lim x \to \infty f (x) = 1 .$ The value of f(-1) is

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Solution

Given that
$f (x) = \frac{a x + b}{x + 2}, lim x \to 0 f (x) = 2 and lim x \to \infty f (x) = 1 . Now, lim x \to 0 f (x) = 2 \Rightarrow lim x \to 0 \frac{a x + b}{x + 2} = 2 \Rightarrow \frac{b}{2} = 2 \Rightarrow b = 4 lim x \to \infty f (x) = 1 \Rightarrow lim x \to \infty \frac{a x + b}{x + 2} = 1 \Rightarrow lim x \to \infty \frac{a + \frac{b}{x}}{1 + \frac{2}{x}} = 1 \Rightarrow a = 1 ∴ f (x) = \frac{x + 4}{x + 2} f (- 1) = \frac{- 1 + 4}{- 1 + 2} = 3$

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Similar questions

Q. Let

f (x) = \frac{a x + b}{x + 1}, l i m_{x \to 0} f (x) = 2

and

l i m_{x \to \infty} f (x) = 1

then

f (- 2) =

f (x) = \frac{a x^{2} + b}{x^{2} + 1}, \lim_{x \to 0} f (x) = 1

and

\lim_{x \to \infty} f (x) = 1,

then prove that f(−2) = f(2) = 1

$f (x) = \frac{a x^{2} + b}{x^{2} + 1}, {lim}_{x \to 0} f (x) = 1 a n d {lim}_{x \to \infty} f (x) = 1, t h e n p r o v e t h a t f (- 2) = f (2) = 1.$

Q. If

f (x) = \frac{(1 + x)^{1 / x} - e}{x}

then

Q. Let

a_{1} > a_{2} > a_{3} > \dots a_{n} > 1;

p_{1} > p_{2} > p_{3} > . . . p_{n} > 0

be such that

p_{1} + p_{2} + p_{3} + \dots + p_{n} = 1.

F (x) = {(p_{1} a_{1}^{x} + p_{2} a_{2}^{x} + \dots + p_{n} a_{n}^{x})}_{1}^{1 / x},

then

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Let f(x)=ax+bx+2, limx→0f(x)=2 and limx→∞ f(x)=1. The value of f(-1) is

Given that f(x)=ax+bx+2,limx→0 f(x)=2 and limx→∞ f(x)=1.Now, limx→0 f(x)=2 ⇒limx→0 ax+bx+2=2⇒b2=2⇒b=4limx→∞f(x)=1⇒limx→∞ ax+bx+2=1⇒limx→∞ a+bx1+2x=1⇒a=1∴f(x)=x+4x+2f(−1)=−1+4−1+2=3

Let $f (x) = \frac{ax + b}{x + 2}, lim x \to 0 f (x) = 2 and lim x \to \infty f (x) = 1 .$ The value of f(-1) is