MCQ
The domain of the function $f(x) = \log (\sqrt {x - 4} + \sqrt {6 - x} )$ is
  • A
    $[4,\infty )$
  • B
    $( - \infty ,\;6]$
  • $[4,\;6]$
  • D
    None of these

Answer

Correct option: C.
$[4,\;6]$
c
(c) $f(x) = \log (\sqrt {x - 4} + \sqrt {6 - x} )$

==> $y = {x^x}\, \Rightarrow \,\,\,\log y = x\log x$ and $6 - x \ge 0$==>$x \ge 4$ and $x \le 6$

$\therefore $ Domain of $f(x)$ = $[4,\,\,6]$.

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

Similar questions

Let distinct lines $L_1,L_2$ belong to family of lines $(x -2y -3) + \lambda (x + 3y + 2) = 0$ and $B_1$ is angle bisector of $L_1$ and $L_2$ which passes through point $A(2,3),$ then equation of other bisector of $L_1$ and $L_2$ is ($\lambda$ is a parameter)
The standard deviation of the observations $6, 5, 9, 13, 12, 8, 10$ is:
If ${z_1},{z_2},{z_3},{z_4}$ are the affixes of four points in the Argand plane and $z$ is the affix of a point such that $|z - {z_1}|\, = \,|z - {z_2}|\, = \,|z - {z_3}|\, = |z - {z_4}|$, then ${z_1},{z_2},{z_3},{z_4}$ are
The position of the point $(4, -3)$ with respect to the ellipse $2{x^2} + 5{y^2} = 20$ is
The ratio of the coefficient of $x^{15}$ to the term independent of $x$ in the expansion of ${\left( {{x^2} + \frac{2}{x}} \right)^{15}}$ is
If $ (1 + \text{x})^\text{ⁿ} = \text{C}_{0} +\text{C}_1\text{x} +\text{C}_2\text{ x}^² + …+ \ ^\text{C}\text{n} \text{ x}ⁿ,$ then the value of$\ ^\text{C}0^² + \ ^\text{C}1^² + \ ^\text{C}2^² + .....+ ^\text{C}\text{n}^\text{ⁿ} =\ ^{2\text{n}}\text{C}_\text{n}$ is:
$\mathop {\lim }\limits_{x \to 1} f(x)$ is equal to, where

$f(x) = \left\{ {\begin{array}{*{20}{c}}
{\frac{{{e^{\frac{1}{{x - 1}}}} - 2}}{{{e^{\frac{1}{{x - 1}}}} + 2}}}&{x \ne 1}\\
{1\,\,\,\,\,\,\,\,\,\,\,\,\,}&{x = 1}
\end{array}} \right.$

Let $f:R \to R$ be a differentiable function having $f(2) = 6,f'(2) = \left( {\frac{1}{{48}}} \right).$ Then $\mathop {\lim }\limits_{x \to 2} \int\limits_6^{f(x)} {\frac{{4{t^3}}}{{x - 2}}} dt$ equals
Choose the correct answer. $\lim\limits_{\text{x} \rightarrow0}\frac{\text{x}^{\text{m}}-1}{\text{x}^{\text{n}}-1}$ is equal to:
If $\cot \theta + \tan \theta = 2{\rm{cosec}}\theta $, the general value of $\theta $ is