Domain of the function f(x) = [ _ 10 ( 5x - x^2 4 ) ]^ 1/2 is - Vidyadip

MCQ

Domain of the function $f(x) = {\left[ {{{\log }_{10}}\left( {\frac{{5x - {x^2}}}{4}} \right)} \right]^{1/2}}$ is

A
$ - \infty < x < \infty $
✓
$1 \le x \le 4$
C
$4 \le x \le 16$
D
$ - 1 \le x \le 1$

✓

Answer

Correct option: B.

$1 \le x \le 4$

b
(b) We have $f(x) = {\left[ {{{\log }_{10}}\left( {\frac{{5x - {x^2}}}{4}} \right)} \right]^{1/2}}$…..(i)

From (i), clearly $f(x)$ is defined for those values of $x$ for which ${\log _{10}}\left[ {\frac{{5x - {x^2}}}{4}} \right] \ge 0$

==> $\left( {\frac{{5x - {x^2}}}{4}} \right) \ge {10^0} \Rightarrow \left( {\frac{{5x - {x^2}}}{4}} \right) \ge 1$

==> ${x^2} - 5x + 4 \le 0$ ==> $(x - 1)(x - 4) \le 0$

Hence domain of the function is $[1, 4].$

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 "MCQ" from 2. Relation and Function→2. Relation and Function — all question types→MATHS STD 11 Science question papers→Gujarat Board STD 11 Science question papers→Gujarat Board question paper generator→

Similar questions

An ellipse having foci at $(3, 3) $ and $(- 4, 4)$ and passing through the origin has eccentricity equal to

The sum of all possible values of $\theta \in[-\pi, 2 \pi]$, for which $\frac{1+i \cos \theta}{1-2 i \cos \theta}$ is purely imaginary, is equal to

Equation of the ellipse in its standard form is $\frac{\text{x}^2}{\text{a}^2}-\frac{\text{y}^2}{\text{b}^2}=1$:

Let ${a_1},{a_2}...,{a_{10}}$ be a $G.P.$ If $\frac{{{a_3}}}{{{a_1}}} = 25,$ then $\frac {{{a_9}}}{{{a_{ 5}}}}$ equal

Let $0 \leq \mathrm{r} \leq \mathrm{n}$. If ${ }^{\mathrm{n}+1} \mathrm{C}_{\mathrm{r}+1}:{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}}:{ }^{\mathrm{n}-1} \mathrm{C}_{\mathrm{r}-1}=55: 35: 21$, then $2 n+5 r$ is equal to:

There are mn letters and n post boxes.The number of ways in which these letters can be posted is:

One of the diameters of the circle $x^2+y^2-12 x+4 y+6=0$ is given by:

If two vertices of a triangle are $(5, -1)$ and $( - 2, 3)$ and its orthocentre is at $(0, 0)$, then the third vertex

For n∈ N,$\big(\frac{1}{5}\big)\text{n}^5+\big(\frac{1}{3}\big)\text{n}^3+\big(\frac{1}{15}\big)$ is:

If a tangent to the ellipse $x^{2}+4 y^{2}=4$ meets the tangents at the extremities of its major axis at $\mathrm{B}$ and $\mathrm{C}$, then the circle with $\mathrm{BC}$ as diameter passes through the point: