JEE Advanced Relations and Functions PYQ – Domain, Codomain & Range Questions with Solutions

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Q 1 :

Range of the function $f (x) = \frac{x^{2} + x + 2}{x^{2} + x + 1}, x \in R$ is [2003]

$(1, \infty)$
$(1, 11 / 7]$
$(1, 7 / 3]$
$(1, 7 / 5]$

(3)

$f (x) = \frac{x^{2} + x + 2}{x^{2} + x + 1} = \frac{(x^{2} + x + 1) + 1}{x^{2} + x + 1}$

$= 1 + \frac{1}{{(x + \frac{1}{2})}^{2} + \frac{3}{4}}$

We can see here that as $x \to \infty, f (x) \to 1$ which is the minimum value of $f (x), i . e . f_{\min} = 1$ .

Also $f (x)$ is maximum when ${(x + \frac{1}{2})}^{2} + \frac{3}{4}$ is minimum, which is so when $x = - \frac{1}{2}$ .

$∴$ $f_{\max} = 1 + \frac{1}{\frac{3}{4}} = \frac{7}{3},$ $∴$ $R_{f} = (1, \frac{7}{3}]$

Q 2 :

Let the function $f : [0, 1] \to ℝ$ be defined by $f (x) = \frac{4^{x}}{4^{x} + 2} .$ Then the value of $f (\frac{1}{40}) + f (\frac{2}{40}) + f (\frac{3}{40}) + \dots + f (\frac{39}{40}) - f (\frac{1}{2})$ is _______ . [2020]

0%

(19)

Since, $f (x) + f (1 - x) = \frac{4^{x}}{4^{x} + 2} + \frac{4^{1 - x}}{4^{1 - x} + 2}$

$= \frac{4^{x}}{4^{x} + 2} + \frac{4 / 4^{x}}{4 / 4^{x} + 2} = \frac{4^{x}}{4^{x} + 2} + \frac{2}{2 + 4^{x}} = 1$

$∴$ $f (\frac{1}{40}) + f (\frac{2}{40}) + \dots + f (\frac{39}{40}) - f (\frac{1}{2})$

$= [f (\frac{1}{40}) + f (\frac{39}{40})] + [f (\frac{2}{40}) + f (\frac{38}{40})] + \dots + f (\frac{20}{40}) - f (\frac{1}{2})$

$= (1 + 1 + \dots 19 times) + f (\frac{1}{2}) - f (\frac{1}{2}) = 19$

Topic Question Set

Q 1 :

Range of the function $f (x) = \frac{x^{2} + x + 2}{x^{2} + x + 1}, x \in R$ is [2003]

Q 2 :

Let the function $f : [0, 1] \to ℝ$ be defined by $f (x) = \frac{4^{x}}{4^{x} + 2} .$ Then the value of $f (\frac{1}{40}) + f (\frac{2}{40}) + f (\frac{3}{40}) + \dots + f (\frac{39}{40}) - f (\frac{1}{2})$ is _______ . [2020]

0%

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Topic Question Set

Q 1 : Range of the function f(x)=x2+x+2x2+x+1, x∈R is [2003]

Q 2 : Let the function f:[0,1]→ℝ be defined by f(x)=4x4x+2. Then the value of f(140)+f(240)+f(340)+⋯+f(3940)-f(12) is _______ . [2020]

0%

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Q 1 :

Range of the function $f (x) = \frac{x^{2} + x + 2}{x^{2} + x + 1}, x \in R$ is [2003]

Q 2 :

Let the function $f : [0, 1] \to ℝ$ be defined by $f (x) = \frac{4^{x}}{4^{x} + 2} .$ Then the value of $f (\frac{1}{40}) + f (\frac{2}{40}) + f (\frac{3}{40}) + \dots + f (\frac{39}{40}) - f (\frac{1}{2})$ is _______ . [2020]