Let f-6- be defined by fx-5x-36x-Then the value of - for which fofx - x- for all x-6- is-

For nbsp; f(f(x))=x f(x)=f^ 1(x) Finding f^ 1(x) y=5x+36x α⇒ x=3+α y6y 5 ⇒ f^ 1(x)=3+α x6x 5 ∴ f^ 1(x)=f(x) gives 3+α x6x 5=5x+36x α ⇒ (30 6α )x^2+(α ^2 25)x+(3 ...

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Byju's Answer

Standard XII

Mathematics

Discriminant

Let f:ℝ-α6→ℝ ...

Question

Let $f : R - {\frac{α}{6}} \to R$ be defined by $f (x) = \frac{5 x + 3}{6 x - α} .$
Then the value of $α$ for which $(f o f) (x) = x,$ for all
$x \in R - {\frac{α}{6}},$ is:

5

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8

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No such

α

exists

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6

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Solution

The correct option is A $5$
For $f (f (x)) = x$
$f (x) = f^{- 1} (x)$
Finding $f^{- 1} (x)$
$y = \frac{5 x + 3}{6 x - α} \Rightarrow x = \frac{3 + α y}{6 y - 5}$
$\Rightarrow f^{- 1} (x) = \frac{3 + α x}{6 x - 5}$
$∴ f^{- 1} (x) = f (x) gives$
$\frac{3 + α x}{6 x - 5} = \frac{5 x + 3}{6 x - α}$
$\Rightarrow (30 - 6 α) x^{2} + (α^{2} - 25) x + (3 α - 15) = 0$
$∴ α = 5$

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Let f:R−{α6}→R be defined by f(x)=5x+36x−α. Then the value of α for which (fof)(x)=x, for all x∈R−{α6}, is:

The correct option is A 5For f(f(x))=x f(x)=f−1(x) Finding f−1(x) y=5x+36x−α⇒x=3+αy6y−5 ⇒f−1(x)=3+αx6x−5 ∴f−1(x)=f(x) gives 3+αx6x−5=5x+36x−α ⇒(30−6α)x2+(α2−25)x+(3α−15)=0 ∴α=5

Let $f : R - {\frac{α}{6}} \to R$ be defined by $f (x) = \frac{5 x + 3}{6 x - α} .$
Then the value of $α$ for which $(f o f) (x) = x,$ for all
$x \in R - {\frac{α}{6}},$ is: