Question

If the fourth term in binomial expansion of ${{\left( \sqrt{{{x}^{\dfrac{1}{1+{{\log }_{10}}x}}}}+{{x}^{\dfrac{1}{12}}} \right)}^{6}}$ is equals to 200, and x > 1, then the value of x is:
(a)${{10}^{3}}$
(b) $100$
(c) ${{10}^{4}}$
(d) 10

Answer 1

To solve this question, first we will expand the given expression by using binomial expansion, then, we will find the fourth term of expansion by using the ${{n}^{th}}$ term formula. After that, we will put that term equal to 200. And then using properties of the logarithmic and exponential function, we will solve for the value of x such that x > 1.

Complete step-by-step solution
Now, in question it is given that fourth term in binomial expansion of ${{\left( \sqrt{{{x}^{\dfrac{1}{1+{{\log }_{10}}x}}}}+{{x}^{\dfrac{1}{12}}} \right)}^{6}}$ is equals to 200.
Let, the fourth term of binomial expansion be denoted by ${{T}_{4}}$.
Using concept of binomial expansion of ${{(a+b)}^{n}}{{=}^{n}}{{C}_{0}}{{a}^{n}}{{b}^{0}}{{+}^{n}}{{C}_{1}}{{a}^{n-1}}{{b}^{1}}{{+}^{n}}{{C}_{2}}{{a}^{n-2}}{{b}^{2}}+......{{+}^{n}}{{C}_{n}}{{a}^{0}}{{b}^{n}}$ , we get
Now, ${{\left( \sqrt{{{x}^{\dfrac{1}{1+{{\log }_{10}}x}}}}+{{x}^{\dfrac{1}{12}}} \right)}^{6}}{{=}^{6}}{{C}_{0}}{{\left( \sqrt{{{x}^{\dfrac{1}{1+{{\log }_{10}}x}}}} \right)}^{6}}{{\left( {{x}^{\dfrac{1}{12}}} \right)}^{0}}{{+}^{6}}{{C}_{1}}{{\left( \sqrt{{{x}^{\dfrac{1}{1+{{\log }_{10}}x}}}} \right)}^{5}}{{\left( {{x}^{\dfrac{1}{12}}} \right)}^{1}}+......+{{\,}^{6}}{{C}_{6}}{{\left( \sqrt{{{x}^{\dfrac{1}{1+{{\log }_{10}}x}}}} \right)}^{0}}{{\left( {{x}^{\dfrac{1}{12}}} \right)}^{6}}$
On simplifying, we get
${{\left( \sqrt{{{x}^{\dfrac{1}{1+{{\log }_{10}}x}}}}+{{x}^{\dfrac{1}{12}}} \right)}^{6}}={{\left( \sqrt{{{x}^{\dfrac{1}{1+{{\log }_{10}}x}}}} \right)}^{6}}+6{{\left( \sqrt{{{x}^{\dfrac{1}{1+{{\log }_{10}}x}}}} \right)}^{5}}\left( {{x}^{\dfrac{1}{12}}} \right)+......+\,{{\left( {{x}^{\dfrac{1}{12}}} \right)}^{6}}$, as $^{n}{{C}_{0}}=1$, $^{n}{{C}_{1}}=n$ and $^{n}{{C}_{n}}=1$ where $^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$, and $n!=n(n-1)(n-2).......3.2.1$ .
Also, we know that ${{n}^{th}}$ term of expansion ${{(a+b)}^{n}}$, is given as ${{T}_{n}}{{=}^{n}}{{C}_{r}}{{a}^{n-r}}{{b}^{r}}$, where r + 1 = n.
So, we can say that fourth term on expansion of ${{\left( \sqrt{{{x}^{\dfrac{1}{1+{{\log }_{10}}x}}}}+{{x}^{\dfrac{1}{12}}} \right)}^{6}}$ will be,${{T}_{4}}{{=}^{6}}{{C}_{3}}{{\left( {{x}^{\dfrac{1}{1+{{\log }_{10}}x}}} \right)}^{\dfrac{3}{2}}}{{\left( {{x}^{\dfrac{1}{12}}} \right)}^{3}}$
As given that, ${{T}_{4}}=200$
So, we can say that
$^{6}{{C}_{3}}{{\left( {{x}^{\dfrac{1}{1+{{\log }_{10}}x}}} \right)}^{\dfrac{3}{2}}}{{\left( {{x}^{\dfrac{1}{12}}} \right)}^{3}}=200$
Now, on simplifying the terms, we get
$\dfrac{6!}{3!\left( 6-3 \right)!}{{\left( {{x}^{\dfrac{1}{1+{{\log }_{10}}x}}} \right)}^{\dfrac{3}{2}}}\left( {{x}^{\dfrac{1}{4}}} \right)=200$, as ${{({{x}^{a}})}^{b}}={{x}^{a}}^{b}$
On simplifying, factorials, we get
$\dfrac{6.5.4.3!}{3.2\left( 3 \right)!}{{\left( {{x}^{\dfrac{1}{1+{{\log }_{10}}x}}} \right)}^{\dfrac{3}{2}}}\left( {{x}^{\dfrac{1}{4}}} \right)=200$
$\Rightarrow 20{{\left( {{x}^{\dfrac{1}{1+{{\log }_{10}}x}}} \right)}^{\dfrac{3}{2}}}\left( {{x}^{\dfrac{1}{4}}} \right)=200$
On solving, we get
${{\left( {{x}^{\dfrac{1}{1+{{\log }_{10}}x}}} \right)}^{\dfrac{3}{2}}}\left( {{x}^{\dfrac{1}{4}}} \right)=10$
Also, we know that ${{x}^{a}}{{x}^{b}}={{x}^{a}}^{+b}$
So, we have
$\left( {{x}^{\dfrac{3}{2}\left( \dfrac{1}{(1+{{\log }_{10}}x)} \right)}} \right)\left( {{x}^{\dfrac{1}{4}}} \right)=10$
$\Rightarrow {{x}^{\dfrac{1}{4}+\dfrac{3}{2}\left( \dfrac{1}{(1+{{\log }_{10}}x)} \right)}}=10$
Now, taking log base 10 on both side, that is ${{\log }_{10}}$ on both side, we get
${{\log }_{10}}\left( {{x}^{\dfrac{1}{4}+\dfrac{3}{2}\left( \dfrac{1}{(1+{{\log }_{10}}x)} \right)}} \right)={{\log }_{10}}(10)$
We know that, ${{\log }_{b}}(b)=1$,
So, we get
${{\log }_{10}}\left( {{x}^{\dfrac{1}{4}+\dfrac{3}{2}\left( \dfrac{1}{(1+{{\log }_{10}}x)} \right)}} \right)=1$
Also, we know that, ${{\log }_{a}}{{b}^{c}}=c{{\log }_{a}}b$
So, we get
$\left( \dfrac{1}{4}+\dfrac{3}{2}\left( \dfrac{1}{(1+{{\log }_{10}}x)} \right) \right){{\log }_{10}}\left( x \right)=1$
Let, ${{\log }_{10}}\left( x \right)=t$, we get
$\left( \dfrac{1}{4}+\dfrac{3}{2}\left( \dfrac{1}{(1+t)} \right) \right)t=1$
On simplifying, we get
$\dfrac{t}{4}+\dfrac{3}{2}\left( \dfrac{t}{(1+t)} \right)=1$
$\Rightarrow {{t}^{2}}+7t=4+4t$
Or, ${{t}^{2}}+3t-4=0$
Now, for quadratic equation, $a{{x}^{2}}+bx+c=0$ value of $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ .
So, for ${{t}^{2}}+3t-4=0$
$t=\dfrac{-(3)\pm \sqrt{{{(3)}^{2}}-4(1)(-4)}}{2(1)}$
$t=1,-4$
As, ${{\log }_{10}}\left( x \right)=t$
So, ${{\log }_{10}}\left( x \right)=1,-4$
As, we know that, ${{\log }_{a}}\left( b \right)=c$ then, $b={{a}^{c}}$
So, $x=10,{{10}^{-4}}$
As I question it is given that x > 1
So, x = 10
Hence, option ( d ) is correct.

Note: To solve such question, one must know basic properties of logarithmic and exponential functions such as ${{\log }_{b}}(b)=1$, ${{\log }_{a}}\left( b \right)=c$ then, $b={{a}^{c}}$, ${{\log }_{a}}{{b}^{c}}=c{{\log }_{a}}b$ and ${{({{x}^{a}})}^{b}}={{x}^{a}}^{b}$, ${{x}^{a}}{{x}^{b}}={{x}^{a}}^{+b}$. Always remember that ${{(a+b)}^{n}}{{=}^{n}}{{C}_{0}}{{a}^{n}}{{b}^{0}}{{+}^{n}}{{C}_{1}}{{a}^{n-1}}{{b}^{1}}{{+}^{n}}{{C}_{2}}{{a}^{n-2}}{{b}^{2}}+......{{+}^{n}}{{C}_{n}}{{a}^{0}}{{b}^{n}}$ whose ${{n}^{th}}$ term of expansion ${{(a+b)}^{n}}$, is given as ${{T}_{n}}{{=}^{n}}{{C}_{r}}{{a}^{n-r}}{{b}^{r}}$, where r + 1 = n. Try to solve these questions without any calculation error, else the answer obtained will be wrong or the solution gets more complex.