Correct Answer - Option 1 : 18
The given equation is:
\(\mathop \smallint \nolimits_6^{{\rm{f}}\left( {\rm{x}} \right)} \left( {4{{\rm{t}}^3}} \right){\rm{dt}} = \left( {{\rm{x}} - 2} \right){\rm{g}}\left( {\rm{x}} \right)\)
Now, g(x),
\(\Rightarrow {\rm{g}}\left( {\rm{x}} \right) = \frac{{\mathop \smallint \nolimits_6^{{\rm{f}}\left( {\rm{x}} \right)} \left( {4{{\rm{t}}^3}} \right){\rm{dt}}}}{{\left( {{\rm{x}} - 2} \right)}}\)
On integrating,
\(\Rightarrow {\rm{g}}\left( {\rm{x}} \right) = \frac{{\left[ {\frac{{4{{\rm{t}}^4}}}{4}} \right]_6^{{\rm{f}}\left( {\rm{x}} \right)}}}{{\left( {{\rm{x}} - 2} \right)}}\)
\(\Rightarrow {\rm{g}}\left( {\rm{x}} \right) = \frac{{\left[ {{{\rm{t}}^4}} \right]_6^{{\rm{f}}\left( {\rm{x}} \right)}}}{{\left( {{\rm{x}} - 2} \right)}}\)
\(\Rightarrow {\rm{g}}\left( {\rm{x}} \right) = \frac{{{{\left( {{\rm{f}}\left( {\rm{x}} \right)} \right)}^4} - {6^4}}}{{\left( {{\rm{x}} - 2} \right)}}\)
Now applying the limit given in question,
\(\Rightarrow \mathop {\lim }\limits_{{\rm{x}} \to 2} {\rm{g}}\left( {\rm{x}} \right) = \mathop {\lim }\limits_{{\rm{x}} \to 2} \frac{{{{\left( {{\rm{f}}\left( {\rm{x}} \right)} \right)}^4} - {6^4}}}{{\left( {{\rm{x}} - 2} \right)}}\)
Which gives \(\left( {\frac{0}{0}} \right)\) form.
On applying L'Hospital rule,
\(\Rightarrow \mathop {\lim }\limits_{{\rm{x}} \to 2} {\rm{g}}\left( {\rm{x}} \right) = \mathop {\lim }\limits_{{\rm{x}} \to 2} \frac{{4{{\left( {{\rm{f}}\left( {\rm{x}} \right)} \right)}^3}.{\rm{f'}}\left( {\rm{x}} \right)}}{1}\)
On direct substitution,
⇒ g(x) = 4(f(2))3.f'(2)
∵ f(2) = 6 (given) and \({\rm{f'}}\left( 2 \right) = \frac{1}{{48}}{\rm{}}\left( {{\rm{given}}} \right)\)
\(\Rightarrow {\rm{g}}\left( {\rm{x}} \right) = 4 \times {\left( 6 \right)^3} \times \frac{1}{{48}}\)
\(\Rightarrow {\rm{g}}\left( {\rm{x}} \right) = 216 \times \frac{1}{{12}}\)
∴ g(x) = 18
