Choose the correct answer from the given four options. Let f : R … - Vidyadip

Question

Choose the correct answer from the given four options.

Let f : R → R be the functions defined by f(x) = x³ + 5. Then f^-1(x) is:

$(\text{x}+5)^\frac{1}{3}$
$(\text{x}-5)^\frac{1}{3}$
$(5-\text{x})^\frac{1}{3}$
$5-\text{x}$

✓

Answer

$(\text{x}-5)^\frac{1}{3}$

Solution:

we are given that, $\text{f}(\text{x})=\text{x}^3 +5$

Let us suppose, $\text{y}=\text{x}^3+5$

$\Rightarrow\ \text{x}^3=\text{y}-5$

$\Rightarrow\text{x}=(\text{y}-5)^{\frac{1}{3}}$

$\begin{bmatrix}\because\text{f}(\text{x})=\text{y}\\\Rightarrow\text{x}=\text{f}^{-1}(\text{y})\end{bmatrix}$

$\Rightarrow\text{f}^{-3}(\text{y})=(\text{y}-5)^{\frac{1}{3}}$

$\Rightarrow\text{f}^{-1}(\text{x})=(\text{x}-5)^{\frac{1}{3}}$

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Explore more

More "M.C.Q (1 Marks)" from RELATIONS AND FUNCTIONS→RELATIONS AND FUNCTIONS — all question types→Maths STD 12 Science question papers→Gujarat Board STD 12 Science question papers→Gujarat Board question paper generator→

Similar questions

The position vectors of points $P$ and $Q$ are $\vec{p}$ and $\vec{q}$ respectively. The point $R$ divides line segment $P Q$ in the ratio $3: 1$ and $S$ is the mid-point of line segment $P R$. The position vector of $S$ is

The graphs of $f (x) = x^2 \,\& \,g(x) = cx^3 \,\, (c > 0)$ intersect at the points $(0, 0) \& \left( {\frac{1}{c},\,\,\frac{1}{{{c^2}}}} \right)$. If the region which lies between these graphs & over the interval $[0, 1/c]$ has the area equal to $2/3$ then the value of $c$ is

If $A = \left[ {\begin{array}{*{20}{c}}1&3\\2&1\end{array}} \right]$, then determinant of ${A^2} - 2A$ is

The value of $\int\limits^\frac{\pi}{2}_0\log\Big(\frac{4+3\sin\text{x}}{4+3\cos\text{x}}\Big)\text{dx}$ is:

2
$\frac{3}{4}$
0
-2

In the given graph, the feasible region for a LPP is shaded. The objective function $Z=2 x-3 y$, will be minimum at

If area bounded by the curves ${y^2} = 4\,ax$ and $y = mx$ is ${a^2}/3,$, then the value of $m$ is

Let $f : R \rightarrow R$ be defined as,

$f(x)=\left\{\begin{array}{ll}-55 x, & \text { if } x<-5 \\ 2 x^{3}-3 x^{2}-120 x, & \text { if }-5 \leq x \leq 4 \\ 2 x^{3}-3 x^{2}-36 x-336, & \text { if } x>4\end{array}\right.$

Let $A=\{ x \in R : f$ is increasing $\} .$ Then $A$ is equal to :

The value of $\int\frac{\cos2\text{x}}{{\cos}{\text{ x}}}\text{dx}$ is equal to:

$2\sin\text{x}-\ell\text{ n }\mid\sec\text{x}+\tan\text{x}\mid+\text{ c}$
$2\sin\text{x}-\ell\text{ n }\mid\sec\text{x}-\tan\text{x}\mid+\text{ c}$
$2\sin\text{x}+\ell\text{ n }\mid\sec\text{x}+\tan\text{x}\mid+\text{ c}$
$3\sin\text{x}-\ell\text{ n }\mid\sec\text{x}+\tan\text{x}\mid+\text{ c}$

The number of distinct real roots of $x ^{4}-4 x +1=0$ is

Given the system of equation $a(x + y + z)=x,b(x + y + z) = y, c(x + y + z) = z$ where $a,b,c$ are non-zero real numbers. If the real numbers $x,y,z$ are such that $xyz \neq 0,$ then $(a + b + c)$ is equal to-