Inverse of a Function MCQ Quiz - Objective Question with Answer for Inverse of a Function - Download Free PDF

Explanation:

g(x) is the inverse function of an invertible function f(x)

So, g^-1(x) = f(x)

⇒ x = g{f(x)}

Differentiating both sides

1 = g'{f(x)}f'(x)

⇒ g'{f(x)} = $\frac{1}{f^{\prime }(x)}$

Putting x = -1 we get

g'{f(-1)} = $\frac{1}{{\mathrm{f}}^{\prime }(-1)}$

Option (3) is true.

$g=f^{-1}$

$f(g(x))=x$

Differentiate $w.r.t.x$

$f^{\prime }(g(x))\cdot g^{\prime }(x)=1$

$\therefore {\displaystyle \frac{1}{1+(g(x){)}^4}}\cdot g^{\prime }(x)=1$

$g^{\prime }(x)=1+[g(x){]}^4$

Explanation:

g(x) is the inverse function of an invertible function f(x)

So, g^-1(x) = f(x)

⇒ x = g{f(x)}

Differentiating both sides

1 = g'{f(x)}f'(x)

⇒ g'{f(x)} = $\frac{1}{f^{\prime }(x)}$

Putting x = -1 we get

g'{f(-1)} = $\frac{1}{{\mathrm{f}}^{\prime }(-1)}$

Option (3) is true.

$g=f^{-1}$

$f(g(x))=x$

Differentiate $w.r.t.x$

$f^{\prime }(g(x))\cdot g^{\prime }(x)=1$

$\therefore {\displaystyle \frac{1}{1+(g(x){)}^4}}\cdot g^{\prime }(x)=1$

$g^{\prime }(x)=1+[g(x){]}^4$

Calculation

Given:

$f:R\to R$ is given by $f(x)=\tan x$

We need to find $f^{-1}(1)$.

Let $y=f(x)$, so $y=\tan x$.

⇒ $x={\tan }^{-1}y$

So, $f^{-1}(y)={\tan }^{-1}y$.

⇒ $f^{-1}(1)={\tan }^{-1}(1)$

$f^{-1}(1)=\frac{\pi }{4}$

Hence option 1 is correct

Concept:

For a given function f(x) = y, we say that x = f^-1(y).

Calculation:

Let's say that y = f(x) = 2x - 3.

$\Rightarrow \mathrm{x}={\displaystyle \frac{\mathrm{y}+3}{2}}={\mathrm{f}}^{-1}(\mathrm{y})$

Replacing y by x, we get:

$\therefore {\mathrm{f}}^{-1}(\mathrm{x})={\displaystyle \frac{\mathrm{x}+3}{2}}$.

Concept:

If f(x) = y, we say that x = f^-1(y).

The inverse of the given function will be defined only when the function is bijective.

• Here the base of the function log8 x is greater than 1 so the function will be increasing. So, this will be one - one function.
• Also, the co -domain of the given function will be the same as the range of the function which will be R so this will be onto function.
• So, the function is Bijective.

Calculation:
To find the inverse function, 

Replace x → y

x = f(y) = log₈ y

⇒ x = log8 y

Now, take anti log,

⇒ y = 8^x = f^-1(x)

⇒ f^-1(x) = 8^x.

CONCEPT:

Let f: A → B be one-one and onto (bijective) function. Then f^-1 exists which is a function f^-1: B → A, which maps each element b ∈ B with an element a ∈ A such that f(a) = b is called the inverse function of f: A → B.

Methods to find inverse:

Let f : A → B be a bijective function.

Step – I Put f (x) = y

Step – II Solve the equation y = f (x) to obtain x in terms of y.

Interchange x and y to obtain the inverse of the given function f

CALCULATION:

Given: y = 5^{log x}

Here we have to find the inverse of the given function

By applying log to base 5 on both sides of y = 5^{log x} we get,

⇒ ${\log }_{5}y={\log }_5\left(5^{\log x}\right)$

⇒ ${\log }_{5}y=\log x$

⇒ $\frac{\log y}{\log 5}=\log x$

⇒ $\log \left[y^{\frac{1}{\log 5}}\right]=\log x$

⇒ $x=y^{\frac{1}{\log 5}}$

Hence, option B is the correct answer.

Explanation:

We have, f(x) = 3x - 4 = y (let)

⇒ 3x = y + 4
⇒ x = $\frac{y+4}{3}$

⇒ $f^{-1}(x)=\frac{x+4}{3}$

Concept:

Functions:

If y = f(x), then we say that x = f^-1(y).

Calculation:

Let y = f(x) = ${\displaystyle \frac{1-\mathrm{x}}{1+\mathrm{x}}}$.

⇒ y(1 + x) = 1 - x

⇒ y + xy = 1 - x

⇒ xy + x = 1 - y

⇒ x(1 + y) = 1 - y

⇒ x = ${\displaystyle \frac{1-\mathrm{y}}{1+\mathrm{y}}}$ = f^-1(y)

Replacing y by x, we can say that:

f-1(x) = ${\displaystyle \frac{1-\mathrm{x}}{1+\mathrm{x}}}$.

Let f(x) = y = 10x – 7

So, $x=f^{-1}(y)=\frac{y+7}{10}$

Now, $g(y)=f^{-1}(y)$, {given}

So, $g(x)=\frac{x+7}{10}$

Hence, correct option is (3)

Concept:

• Inverse Function:

Let f: A → B be one-one and onto (bijective) function. Then f-1 exists which is a function f-1: B → A, which maps each element b ∈ B with an element

a ∈ A such that f(a) = b is called the inverse function of f: A → B.

• Methods to find inverse:

Let f : A → B be a bijective function.

Step – I Put f (x) = y

Step – II Solve the equation y = f (x) to obtain x in terms of y.

Interchange x and y to obtain the inverse of the given function f.

Calculation:

Given: f: R → R is a function such that f(x) = 10x + 3

As we can see that, the given function is a bijective function i.e f^-1 exists.

Let y = f(x) = 10x + 3

⇒ x = $\frac{y-3}{10}$

Hence, f^-1 (x) = $\frac{x-3}{10}$

Explanation:

We have f(x) = $\frac{3x+2}{5x-3}$ = y (let)
⇒ 3x + 2 = y(5x - 3)
⇒ 3x + 2 = 5xy - 3y

⇒ 3x - 5xy = -3y - 2

⇒ x(3 - 5y) = -3y - 2

⇒ $x=\frac{3y+2}{5y-3}$

⇒ $f^{-1}(x)=\frac{3x+2}{5x-3}$

∴ f–1(x) = f (x)

Concept:

If g is the inverse function of f then gof(x) = g(f(x)) = x

Chain rule of differentiation: [g(f(x))]' = g'(f(x)).f '(x)

Calculation:

Given, the function f is differentiable at x = c and is one-one in some neighborhood of c, g is the inverse function of f,

⇒ g(f(x)) = x

Differentiating both sides with respect to x,

⇒ g'(f(x)).f '(x) = 1

⇒ g'(f(x)) = $\frac{1}{{\mathrm{f}}^{\prime }(\mathrm{x})}$

So, at x = c,

g'(f(c)) = $\frac{1}{{\mathrm{f}}^{\prime }(\mathrm{c})}$

∴ The correct option is (1).

Concept:

If f(x) = y, we say, y is the image of x under f and x is the pre-image of y under f

Calculation:

Here, f: R→R: f(x) = x2 - 17

⇒f(x) = -1 

⇒ x2 - 17 = -1

⇒ x2 = 16

⇒ x = -4, 4

So, -4, 4 are the pre-images of -1.

Hence, option (3) is correct.