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Maths NCERT Exemplar Solutions Class 11th Chapter Eleven Relations and Functions Ncert Solutions Maths class 12th Maths Class 12th

Let f : R -> R be defined as f(x) = 2x -1 and g : R - {1} -> R be defined as g(x) = $\frac{x - \frac{1}{2}}{x - 1} .$ Then the composition function f(g(x)) is:

Option 1 -

Neither one-one nor onto

Option 2 -

onto but not one-one

Option 3 -

both one-one and onto

Option 4 -

one-one but not onto

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1 Answer

Answered by

alok kumar singh | Contributor-Level 10

a month ago

Correct Option - 3

Detailed Solution:

$f (x) = 2 x - 1 g : R - {1} \to R$

$g (x) = \frac{x - 1 / 2}{x - 1}$

$f {g (x)} = 2 (\frac{x - 1 / 2}{x - 1}) - 1 = \frac{2 x - 1 - x + 1}{x - 1} = \frac{x}{x - 1}, x \neq 1$

$y = \frac{x}{x - 1} \Rightarrow x = \frac{y}{y - 1} ∴ y \neq 1$

One – one but not onto

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Let S = {1, 2, 3 ,....,20}

R₁ = {(a, b) : a divide b}

R₂ = {(a, b) : a is integral multiple of b} a, b Î s

n(R₁ – R₂) = ?

alok kumar singh

R₁ = { (1, 1) (1, 2), (1, 3)., (1, 20), (2, 2), (2, 4). (2, 20), (3, 3), (3, 6), . (3, 18),
(4, 4), (4, 8), . (4, 20), (5, 5), (5, 10), (5, 15), (5, 20), (6, 6), (6, 12), (6, 18), (7. 7),
(7, 14), (8, 8), (8, 16), (9, 9), (9, 18), (10, 10), (10, 20), (11, 11), (12, 12), . (20, 20)}

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R₂ = {a is integral multiple of b}

So n (R₁ – R₂) = 66 – 20 = 46

as R₁ Ç R₂ = { (a, a) : a Î s} = { (1, 1), (2, 2), ., (20, 20)}

$f (x) = {\begin{matrix} e^{- x}, & x < 0 \\ l n x, & x > 0 \end{matrix}$

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$g o f (x) = {\begin{matrix} f (x), & f (x) < 0 \\ f (x), & f (x) > 0 \end{matrix}$

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If A = {1, 2, 3, … 100}, R = {(x, y) | 2x = 3y, x, y ∈ A} is symmetric relation on A and the number of elements in R is n, the smallest integer value of n is

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$?$ R is symmetric relation

⇒ (y, x) ∈ R V (x, y) ∈ R

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f is increasing function

x < 5x < 7x

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So total no. of functions = 10⁵ × 1 = 10⁵

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Let f : R -> R be defined as f(x) = 2x -1 and g : R - {1} -> R be defined as g(x) = $\frac{x - \frac{1}{2}}{x - 1} .$ Then the composition function f(g(x)) is:

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Let f : R -> R be defined as f(x) = 2x -1 and g : R - {1} -> R be defined as g(x) = x − 1 2 x − 1 . Then the composition function f(g(x)) is:

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Let f : R -> R be defined as f(x) = 2x -1 and g : R - {1} -> R be defined as g(x) = $\frac{x - \frac{1}{2}}{x - 1} .$ Then the composition function f(g(x)) is: