Question

Topic:- Functions Class XI and XII

Find the domain of the function:

$\sf f(x) = log_{x+ \frac{1}{2}} |x^2-x-6| + {}^{16-x}C_{2x-1} + {}^{20-3x}P_{2x-5}$ ​

Answer 1

Consider,

$\small\text{ \longrightarrow f_1(x)=\log_{x+\frac{1}{2}}\big|x^2-x-6\big| }$

We know $\small\log_ba$ is defined for $\smalla\in(0,\ \infty)$ and $\smallb\in(0,\ \infty)-\{1\}.$ So,

$\small\longrightarrow|x^2-x-6|\in(0,\ \infty)$

$\small\Longrightarrow x^2-x-6\neq0$

$\small\longrightarrow(x-3)(x+2)\neq0$

$\small\Longrightarrow x\in\mathbb{R}-\{-2,\ 3\}$

and,

$\small\longrightarrow x+\dfrac{1}{2}\in(0,\ \infty)-\{1\}$

$\small\text{ \longrightarrow x\in\left(-\dfrac{1}{2},\ \infty\right)-\left\{\dfrac{1}{2}\right\} }$

Then domain of $f_1$ is,

$\small\text{ \longrightarrow x\in\bigg[\mathbb{R}-\{-2,\ 3\}\bigg]\cap\left[\left(-\dfrac{1}{2},\ \infty\right)-\left\{\dfrac{1}{2}\right\}\right] }$

$\small\text{ \longrightarrow x\in\left(-\dfrac{1}{2},\ \infty\right)-\left\{\dfrac{1}{2},\ 3\right\} }$

The conditions to find domain of $\small^n\!C_r$ is $\smalln,\ r,\ n-r\in\mathbb{W}$ so that $\smalln,\ r,\ n-r\in[0,\ \infty).$

Consider,

$\small\text{ \longrightarrow f_2(x)=\,^{16-x}\!C_{_{2x-1}} }$

Then,

$\small\longrightarrow 16-x\in[0,\ \infty)$

$\small\longrightarrow x\in(-\infty,\ 16]$

and,

$\small\longrightarrow 2x-1\in[0,\ \infty)$

$\small\longrightarrow x\in\left[\dfrac{1}{2},\ \infty\right)$

and,

$\small\longrightarrow (16-x)-(2x-1)=17-3x\in[0,\ \infty)$

$\small\longrightarrow x\in\left(-\infty,\ \dfrac{17}{3}\right]$

Then domain of $\smallf_2$ is,

$\small\longrightarrow x\in(-\infty,\ 16]\cap\left[\dfrac{1}{2},\ \infty\right)\cap\left(-\infty,\ \dfrac{17}{3}\right]$

$\small\longrightarrow x\in\left[\dfrac{1}{2},\ \dfrac{17}{3}\right]$

But the domain is actually given as,

$\small\text{ \longrightarrow x\in\left\{x:x=\dfrac{n}{k},\ n\in\mathbb{Z},\ a\leq n\leq b\right\} }$

where $\small[a,\ b]$ is the inequality got as domain of $\smallx$ [Signs in $\smalla\leq n\leq b$ vary according to the brackets of interval] and $\smallk$ is GCD of modulus of coefficients of $\smallx$ in $\smalln,\ r,\ n-r$ of $\small^n\!C_r.$

Here $\small[a,\ b]=\left[\dfrac{1}{2},\ \dfrac{17}{3}\right]$ and $\smallk=\gcd(1,\ 2,\ 3)=1.$ So,

$\small\text{ \longrightarrow x\in\left\{x:x=n,\ n\in\mathbb{Z},\ \dfrac{1}{2}\leq n\leq \dfrac{17}{3}\right\} }$

$\small\longrightarrow x\in\left\{1,\ 2,\ 3,\ 4,\ 5\right\}$

Consider,

$\small\text{ \longrightarrow f_3(x)=\,^{20-3x}\!C_{_{2x-5}} }$

Then,

$\small\longrightarrow 20-3x\in[0,\ \infty)$

$\small\longrightarrow x\in\left(-\infty,\ \dfrac{20}{3}\right]$

and,

$\small\longrightarrow 2x-5\in[0,\ \infty)$

$\small\longrightarrow x\in\left[\dfrac{5}{2},\ \infty\right)$

and,

$\small\longrightarrow (20-3x)-(2x-5)=25-5x\in[0,\ \infty)$

$\small\longrightarrow x\in\left(-\infty,\ 5\right]$

Then domain of $\smallf_3$ is,

$\small\longrightarrow x\in\left(-\infty,\ \dfrac{20}{3}\right]\cap\left[\dfrac{5}{2},\ \infty\right)\cap\left(-\infty,\ 5\right]$

$\small\longrightarrow x\in\left[\dfrac{5}{2},\ 5\right]$

Here $\small[a,\ b]=\left[\dfrac{5}{2},\ 5\right]$ and $\smallk=\gcd(2,\ 3,\ 5)=1.$ So,

$\small\text{ \longrightarrow x\in\left\{x:x=n,\ n\in\mathbb{Z},\ \dfrac{5}{2}\leq n\leq5\right\} }$

$\small\longrightarrow x\in\left\{3,\ 4,\ 5\right\}$

Then domain of $\smallf$ is,

$\small\text{ \longrightarrow x\in\left[\left(-\dfrac{1}{2},\ \infty\right)-\left\{\dfrac{1}{2},\ 3\right\}\right]\cap\{1,\ 2,\ 3,\ 4,\ 5\}\cap\{3,\ 4,\ 5\} }$

$\small\text{ \longrightarrow\underline{\underline{x\in\{4,\ 5\}}} }$

In general $\smallx!$ is defined $\forall x\in[0,\ \infty)$ as factorial is also defined for non - negative real numbers but not integers, in terms of Gamma function.

Then domains of $\smallf_2$ and $\smallf_3$ should be $\small\left[\dfrac{1}{2},\ \dfrac{17}{3}\right]$ and $\small\left[\dfrac{5}{2},\ 5\right]$ respectively, so the domain of $\smallf$ should be,

$\small\text{ \longrightarrow x\in\left[\left(-\dfrac{1}{2},\ \infty\right)-\left\{\dfrac{1}{2},\ 3\right\}\right]\cap\left[\dfrac{1}{2},\ \dfrac{17}{3}\right]\cap\left[\dfrac{5}{2},\ 5\right] }$

$\small\text{ \longrightarrow\underline{\underline{x\in\left[\dfrac{5}{2},\ 5\right]-\left\{3\right\}}} }$

Answer 2

Answer:

Consider,

\small\text{$\longrightarrow f_1(x)=\log_{x+\frac{1}{2}}\big|x^2-x-6\big|$}⟶f1(x)=logx+21∣∣∣x2−x−6∣∣∣

We know \small\text{$\log_ba$}logba is defined for \small\text{$a\in(0,\ \infty)$}a∈(0, ∞) and \small\text{$b\in(0,\ \infty)-\{1\}.$}b∈(0, ∞)−{1}. So,

\small\text{$\longrightarrow|x^2-x-6|\in(0,\ \infty)$}⟶∣x2−x−6∣∈(0, ∞)

\small\text{$\Longrightarrow x^2-x-6\neq0$}⟹x2−x−6=0

\small\text{$\longrightarrow(x-3)(x+2)\neq0$}⟶(x−3)(x+2)=0

\small\text{$\Longrightarrow x\in\mathbb{R}-\{-2,\ 3\}$}⟹x∈R−{−2, 3}

and,

\small\text{$\longrightarrow x+\dfrac{1}{2}\in(0,\ \infty)-\{1\}$}⟶x+21∈(0, ∞)−{1}

\small\text{$\longrightarrow x\in\left(-\dfrac{1}{2},\ \infty\right)-\left\{\dfrac{1}{2}\right\}$}⟶x∈(−21, ∞)−{21}

Then domain of f_1f1 is,

\small\text{$\longrightarrow x\in\bigg[\mathbb{R}-\{-2,\ 3\}\bigg]\cap\left[\left(-\dfrac{1}{2},\ \infty\right)-\left\{\dfrac{1}{2}\right\}\right]$}⟶x∈[R−{−2, 3}]∩[(−21, ∞)−{21}]

\small\text{$\longrightarrow x\in\left(-\dfrac{1}{2},\ \infty\right)-\left\{\dfrac{1}{2},\ 3\right\}$}⟶x∈(−21, ∞)−{21, 3}

The conditions to find domain of \small\text{$^n\!C_r$}nCr is \small\text{$n,\ r,\ n-r\in\mathbb{W}$}n, r, n−r∈W so that \small\text{$n,\ r,\ n-r\in[0,\ \infty).$}n, r, n−r∈[0, ∞).

Consider,

\small\text{$\longrightarrow f_2(x)=\,^{16-x}\!C_{_{2x-1}}$}⟶f2(x)=16−xC2x−1

Then,

\small\text{$\longrightarrow 16-x\in[0,\ \infty)$}⟶16−x∈[0, ∞)

\small\text{$\longrightarrow x\in(-\infty,\ 16]$}⟶x∈(−∞, 16]

and,

\small\text{$\longrightarrow 2x-1\in[0,\ \infty)$}⟶2x−1∈[0, ∞)

\small\text{$\longrightarrow x\in\left[\dfrac{1}{2},\ \infty\right)$}⟶x∈[21, ∞)

and,

\small\text{$\longrightarrow (16-x)-(2x-1)=17-3x\in[0,\ \infty)$}⟶(16−x)−(2x−1)=17−3x∈[0, ∞)

\small\text{$\longrightarrow x\in\left(-\infty,\ \dfrac{17}{3}\right]$}⟶x∈(−∞, 317]

Then domain of \small\text{$f_2$}f2 is,

\small\text{$\longrightarrow x\in(-\infty,\ 16]\cap\left[\dfrac{1}{2},\ \infty\right)\cap\left(-\infty,\ \dfrac{17}{3}\right]$}⟶x∈(−∞, 16]∩[21, ∞)∩(−∞, 317]

\small\text{$\longrightarrow x\in\left[\dfrac{1}{2},\ \dfrac{17}{3}\right]$}⟶x∈[21, 317]

But the domain is actually given as,

\small\text{$\longrightarrow x\in\left\{x:x=\dfrac{n}{k},\ n\in\mathbb{Z},\ a\leq n\leq b\right\}$}⟶x∈{x:x=kn, n∈Z, a≤n≤b}

where \small\text{$[a,\ b]$}[a, b] is the inequality got as domain of \small\text{$x$}x [Signs in \small\text{$a\leq n\leq b$}a≤n≤b vary according to the brackets of interval] and \small\text{$k$}k is GCD of modulus of coefficients of \small\text{$x$}x in \small\text{$n,\ r,\ n-r$}