If f-Q - Q is defined as fx -x2- then f-19 is equal to

F(x) =x^2 Replace f(x) by y, ⇒ y = x^2 ⇒ x = ±√(y) Replace x by f^ 1(x) and y by x. ⇒ f^ 1(x) = ±√(x) ⇒ f^ 1(9)=±√(9) ⇒ f^ 1 (9) = ± 3 Therefore, option ...

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Standard XII

Mathematics

Higher Order Equations

If f:Q → Q is...

Question

If $f : Q \to Q$ is defined as $f (x) = x^{2}$ , then $f^{- 1} (9)$ is equal to

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Solution

$f (x) = x^{2}$

Replace $f (x)$ by $y$ ,

$\Rightarrow y = x^{2}$

$\Rightarrow x = \pm \sqrt{y}$

Replace $x$ by $f^{- 1} (x)$ and $y$ by $x$ .

$\Rightarrow f^{- 1} (x) = \pm \sqrt{x}$

$\Rightarrow f^{- 1} (9) = \pm \sqrt{9}$

$\Rightarrow f^{- 1} (9) = \pm 3$

Therefore, option (C) is correct.

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If $f : Q \to Q$ is defined as $f (x) = x^{2}$ , then $f^{- 1} (9)$ is equal to

Q. If f : Q → Q is defined as f(x) = x², then f⁻¹ (9) is equal to

(a) 3
(b) −3
(c) {−3, 3}
(d) ϕ

Q. Show that the function f : Q → Q, defined by f(x) = 3x + 5, is invertible. Also, find f⁻¹

Q. If f : Q → Q, g : Q → Q are two functions defined by f(x) = 2 x and g(x) = x + 2, show that f and g are bijective maps. Verify that (gof)⁻¹ = f⁻¹ og ⁻¹.

Q. Classify the following functions as injection, surjection or bijection :
(i) f : N → N given by f(x) = x²
(ii) f : Z → Z given by f(x) = x²
(iii) f : N → N given by f(x) = x³
(iv) f : Z → Z given by f(x) = x³
(v) f : R → R, defined by f(x) = |x|
(vi) f : Z → Z, defined by f(x) = x² + x
(vii) f : Z → Z, defined by f(x) = x − 5
(viii) f : R → R, defined by f(x) = sinx
(ix) f : R → R, defined by f(x) = x³ + 1
(x) f : R → R, defined by f(x) = x³ − x
(xi) f : R → R, defined by f(x) = sin²x + cos²x
(xii) f : Q − {3} → Q, defined by

f (x) = \frac{2 x + 3}{x - 3}

(xiii) f : Q → Q, defined by f(x) = x³ + 1
(xiv) f : R → R, defined by f(x) = 5x³ + 4
(xv) f : R → R, defined by f(x) = 3 − 4x
(xvi) f : R → R, defined by f(x) = 1 + x²
(xvii) f : R → R, defined by f(x) =

\frac{x}{x^{2} + 1}

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Higher Order Equations

Standard XII Mathematics

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If f:Q→Q is defined as f(x)=x2, then f−1(9) is equal to

f(x)=x2 Replace f(x) by y, ⇒y=x2 ⇒x=±√y Replace x by f−1(x) and y by x. ⇒f−1(x)=±√x ⇒f−1(9)=±√9 ⇒f−1(9)=±3 Therefore, option (C) is correct.

If $f : Q \to Q$ is defined as $f (x) = x^{2}$ , then $f^{- 1} (9)$ is equal to

$f (x) = x^{2}$

Replace $f (x)$ by $y$ ,

$\Rightarrow y = x^{2}$

$\Rightarrow x = \pm \sqrt{y}$

Replace $x$ by $f^{- 1} (x)$ and $y$ by $x$ .

$\Rightarrow f^{- 1} (x) = \pm \sqrt{x}$

$\Rightarrow f^{- 1} (9) = \pm \sqrt{9}$

$\Rightarrow f^{- 1} (9) = \pm 3$

Therefore, option (C) is correct.