Advertisements
Advertisements
प्रश्न
If \[f : R \to \left( - 1, 1 \right)\] is defined by
\[f\left( x \right) = \frac{- x|x|}{1 + x^2}, \text{ then } f^{- 1} \left( x \right)\] equals
पर्याय
\[\sqrt{\frac{\left| x \right|}{1 - \left| x \right|}}\]
\[\text{ Sgn } \left( x \right) \sqrt{\frac{\left| x \right|}{1 - \left| x \right|}}\]
\[- \sqrt{\frac{x}{1 - x}}\]
None of these
उत्तर
(b) \[- Sgn \left( x \right) \sqrt{\frac{\left| x \right|}{1 - \left| x \right|}}\]
\[\text{We have}, f\left( x \right) = \frac{- x|x|}{1 + x^2} x \in \left( - 1, 1 \right)\]
\[\text{Case} - \left( I \right)\]
\[\text{When}, x < 0, \]
\[\text{Then}, \left| x \right| = - x\]
\[\text{And} f\left( x \right) > 0\]
\[\text{Now}, \]
\[f\left( x \right) = \frac{- x\left( - x \right)}{1 + x^2}\]
\[ \Rightarrow y = \frac{x^2}{1 + x^2}\]
\[ \Rightarrow \frac{y}{1} = \frac{x^2}{1 + x^2}\]
\[ \Rightarrow \frac{y + 1}{y - 1} = \frac{x^2 + 1 + x^2}{x^2 - 1 - x^2} \left[ \text { Using Componendo and dividendo } \right]\]
\[ \Rightarrow \frac{y + 1}{y - 1} = \frac{2 x^2 + 1}{- 1}\]
\[ \Rightarrow - \frac{y + 1}{y - 1} = 2 x^2 + 1\]
\[ \Rightarrow \frac{2y}{1 - y} = 2 x^2 \]
\[ \Rightarrow \frac{y}{1 - y} = x^2 \]
\[ \Rightarrow x = - \sqrt{\frac{y}{1 - y}} \left( \text{ As } x < 0 \right)\]
\[ \Rightarrow x = - \sqrt{\frac{\left| y \right|}{1 - \left| y \right|}} \]
\[ \left[ \text{ As } y > 0 \right]\]
\[\text{To find the inverse interchanging x and y we get}, \]
\[ f^{- 1} \left( x \right) = - \sqrt{\frac{\left| x \right|}{1 - \left| x \right|}} . . . \left( i \right)\]
\[\text{Case} - \left( II \right)\]
\[\text{When}, x > 0, \]
\[\text{Then}, \left| x \right| = x\]
\[\text{And} f\left( x \right) < 0\]
\[\text{Now}, \]
\[f\left( x \right) = \frac{- x\left( x \right)}{1 + x^2}\]
\[ \Rightarrow y = \frac{- x^2}{1 + x^2}\]
\[ \Rightarrow \frac{y}{1} = \frac{- x^2}{1 + x^2}\]
\[ \Rightarrow \frac{y + 1}{y - 1} = \frac{- x^2 + 1 + x^2}{- x^2 - 1 - x^2} \left[ \text{Using Componendo and dividendo} \right]\]
\[ \Rightarrow \frac{y + 1}{y - 1} = \frac{1}{- 2 x^2 - 1}\]
\[ \Rightarrow \frac{1 + y}{1 - y} = \frac{1}{2 x^2 + 1}\]
\[ \Rightarrow \frac{1 - y}{1 + y} = 2 x^2 + 1\]
\[ \Rightarrow \frac{- 2y}{1 + y} = 2 x^2 \]
\[ \Rightarrow x^2 = \frac{- y}{1 + y}\]
\[ \Rightarrow x = \sqrt{\frac{- y}{1 + y}} \left( \text{As} x > 0 \right)\]
\[ \Rightarrow x = \sqrt{\frac{\left| y \right|}{1 - \left| y \right|}} \]
\[ \left[ \text{ As } y < 0 \right]\]
\[\text{To find the inverse interchanging x and y we get}, \]
\[ f^{- 1} \left( x \right) = \sqrt{\frac{\left| x \right|}{1 - \left| x \right|}} . . . \left( ii \right)\]
\[\text{Case} - \left( III \right)\]
\[\text{When}, x = 0, \]
\[\text{Then}, f\left( x \right) = 0\]
\[\text{Hence}, f^{- 1} \left( x \right) = 0 . . . \left( iii \right)\]
\[\text{Combinig equation} \left( i \right) , \left( ii \right) \text{and} \left( iii \right) \text{we get}, \]
\[ f^{- 1} \left( x \right) = - Sgn\left( x \right)\sqrt{\frac{\left| x \right|}{1 - \left| x \right|}}\]
APPEARS IN
संबंधित प्रश्न
Give examples of two functions f: N → Z and g: Z → Z such that g o f is injective but gis not injective.
(Hint: Consider f(x) = x and g(x) =|x|)
Find the number of all onto functions from the set {1, 2, 3, …, n} to itself.
Give an example of a function which is one-one but not onto ?
Prove that the function f : N → N, defined by f(x) = x2 + x + 1, is one-one but not onto
Classify the following function as injection, surjection or bijection :
f : R → R, defined by f(x) = sinx
Classify the following function as injection, surjection or bijection :
f : R → R, defined by f(x) = 3 − 4x
Let A = [-1, 1]. Then, discuss whether the following functions from A to itself is one-one, onto or bijective : h(x) = x2
Show that the exponential function f : R → R, given by f(x) = ex, is one-one but not onto. What happens if the co-domain is replaced by`R0^+` (set of all positive real numbers)?
Show that f : R→ R, given by f(x) = x — [x], is neither one-one nor onto.
Let f : N → N be defined by
`f(n) = { (n+ 1, if n is odd),( n-1 , if n is even):}`
Show that f is a bijection.
[CBSE 2012, NCERT]
Let f : R → R and g : R → R be defined by f(x) = x2 and g(x) = x + 1. Show that fog ≠ gof.
Verify associativity for the following three mappings : f : N → Z0 (the set of non-zero integers), g : Z0 → Q and h : Q → R given by f(x) = 2x, g(x) = 1/x and h(x) = ex.
Give examples of two functions f : N → Z and g : Z → Z, such that gof is injective but gis not injective.
If f : A → B and g : B → C are onto functions, show that gof is a onto function.
Find fog and gof if : f (x) = ex g(x) = loge x .
Find fog and gof if : f (x) = x2 g(x) = cos x .
if `f (x) = sqrt(1-x)` and g(x) = `log_e` x are two real functions, then describe functions fog and gof.
State with reason whether the following functions have inverse :
f : {1, 2, 3, 4} → {10} with f = {(1, 10), (2, 10), (3, 10), (4, 10)}
Consider f : R → R given by f(x) = 4x + 3. Show that f is invertible. Find the inverse of f.
Let f be a function from C (set of all complex numbers) to itself given by f(x) = x3. Write f−1 (−1).
Let f : R → R be defined as `f (x) = (2x - 3)/4.` write fo f-1 (1) .
Let f be an invertible real function. Write ( f-1 of ) (1) + ( f-1 of ) (2) +..... +( f-1 of ) (100 )
Let M be the set of all 2 × 2 matrices with entries from the set R of real numbers. Then, the function f : M→ R defined by f(A) = |A| for every A ∈ M, is
If a function\[f : [2, \infty )\text{ to B defined by f}\left( x \right) = x^2 - 4x + 5\] is a bijection, then B =
The function f : [-1/2, 1/2, 1/2] → [-π /2,π/2], defined by f (x) = `sin^-1` (3x - `4x^3`), is
Let
\[f : R - \left\{ n \right\} \to R\]
If the function
\[f : R \to R\] be such that
\[f\left( x \right) = x - \left[ x \right]\] where [x] denotes the greatest integer less than or equal to x, then \[f^{- 1} \left( x \right)\]
Let C be the set of complex numbers. Prove that the mapping f: C → R given by f(z) = |z|, ∀ z ∈ C, is neither one-one nor onto.
Which of the following functions from Z into Z are bijections?
Which of the following functions from Z into Z is bijective?
Let X = {-1, 0, 1}, Y = {0, 2} and a function f : X → Y defiend by y = 2x4, is ____________.
If N be the set of all-natural numbers, consider f: N → N such that f(x) = 2x, ∀ x ∈ N, then f is ____________.
Let f : R → R, g : R → R be two functions such that f(x) = 2x – 3, g(x) = x3 + 5. The function (fog)-1 (x) is equal to ____________.
Raji visited the Exhibition along with her family. The Exhibition had a huge swing, which attracted many children. Raji found that the swing traced the path of a Parabola as given by y = x2.
Answer the following questions using the above information.
- Let : N → R be defined by f(x) = x2. Range of the function among the following is ____________.
Raji visited the Exhibition along with her family. The Exhibition had a huge swing, which attracted many children. Raji found that the swing traced the path of a Parabola as given by y = x2.
Answer the following questions using the above information.
- The function f: Z → Z defined by f(x) = x2 is ____________.
Let n(A) = 4 and n(B) = 6, Then the number of one – one functions from 'A' to 'B' is:
Let f(1, 3) `rightarrow` R be a function defined by f(x) = `(x[x])/(1 + x^2)`, where [x] denotes the greatest integer ≤ x, Then the range of f is ______.
Let f(x) be a polynomial function of degree 6 such that `d/dx (f(x))` = (x – 1)3 (x – 3)2, then
Assertion (A): f(x) has a minimum at x = 1.
Reason (R): When `d/dx (f(x)) < 0, ∀ x ∈ (a - h, a)` and `d/dx (f(x)) > 0, ∀ x ∈ (a, a + h)`; where 'h' is an infinitesimally small positive quantity, then f(x) has a minimum at x = a, provided f(x) is continuous at x = a.
ASSERTION (A): The relation f : {1, 2, 3, 4} `rightarrow` {x, y, z, p} defined by f = {(1, x), (2, y), (3, z)} is a bijective function.
REASON (R): The function f : {1, 2, 3} `rightarrow` {x, y, z, p} such that f = {(1, x), (2, y), (3, z)} is one-one.
If f : R `rightarrow` R is defined by `f(x) = (2x - 7)/4`, show that f(x) is one-one and onto.
