Let f be a real function defined as fx-2x-12x-1- The number of integers which are not in the range of f is

F(x)=2^x+12^x 1 Domain of f is ℝ {0} Let y=2^x+12^x 1 ⇒ y(2^x 1)=2^x+1 ⇒ 2^x(y 1)=y+1 ⇒ nbsp; 2^x=y+1y 1 nbsp; Since 2^x gt;0, y+1y 1 gt;0 ⇒ y ∈( ∞, 1) ...

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Standard XII

Mathematics

Sign of Quadratic Expression

Let f be a re...

Question

Let $f$ be a real function defined as $f (x) = \frac{2^{x} + 1}{2^{x} - 1} .$ The number of integer(s) which are not in the range of $f$ is

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Solution

$f (x) = \frac{2^{x} + 1}{2^{x} - 1}$
Domain of $f$ is $R - {0}$
Let $y = \frac{2^{x} + 1}{2^{x} - 1}$
$\Rightarrow y (2^{x} - 1) = 2^{x} + 1$
$\Rightarrow 2^{x} (y - 1) = y + 1$
$\Rightarrow 2^{x} = \frac{y + 1}{y - 1}$
Since $2^{x} > 0,$
$\frac{y + 1}{y - 1} > 0$
$\Rightarrow y \in (- \infty, - 1) \cup (1, \infty)$
Range of $f$ is $(- \infty, - 1) \cup (1, \infty)$
Integers which are not there in range are $- 1, 0, 1$
Hence, required answer is $3$

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Let f be a real function defined as f(x)=2x+12x−1. The number of integer(s) which are not in the range of f is

f(x)=2x+12x−1 Domain of f is R−{0} Let y=2x+12x−1 ⇒y(2x−1)=2x+1 ⇒2x(y−1)=y+1 ⇒2x=y+1y−1 Since 2x>0, y+1y−1>0 ⇒y∈(−∞,−1)∪(1,∞) Range of f is (−∞,−1)∪(1,∞) Integers which are not there in range are −1,0,1 Hence, required answer is 3

Let $f$ be a real function defined as $f (x) = \frac{2^{x} + 1}{2^{x} - 1} .$ The number of integer(s) which are not in the range of $f$ is