Question

If \[f(x) = \sin \left[ {{{\cos }^{ - 1}}\dfrac{{\left( {1 - {2^{2x}}} \right)}}{{\left( {1 + {2^{2x}}} \right)}}} \right]\] and its first derivative with respect to \[x\] is \[\left( {\dfrac{{ - b}}{a}} \right){\log _e}2\] when \[x = 1\], where \[a\] and \[b\] are integers, then the minimum value of \[\left| {{a^2} - {b^2}} \right|\] is

Answer 1

Hint : We are given a function \[f(x) = \sin \left[ {{{\cos }^{ - 1}}\dfrac{{\left( {1 - {2^{2x}}} \right)}}{{\left( {1 + {2^{2x}}} \right)}}} \right]\] and its first derivative is\[\left( {\dfrac{{ - b}}{a}} \right){\log _e}2\]when \[x = 1\]. We have to find the minimum value of \[\left| {{a^2} - {b^2}} \right|\] . We will be first finding the values of \[a\] and \[b\] using below mentioned trigonometric identities and then substitute values of \[a\] and \[b\] in \[\left| {{a^2} - {b^2}} \right|\] to find required answer.

Formula used: Following formulas will be useful for solving this question
\[
  \cos 2\theta = \dfrac{{1 - {{\tan }^2}\theta }}{{1 + {{\tan }^2}\theta }} \\
  \sin 2\theta = \dfrac{{2\tan \theta }}{{1 + {{\tan }^2}\theta }} \\
 \]
\[d(\dfrac{u}{v}) = \dfrac{{u'v - uv'}}{{{v^2}}}\]
\[d({a^x}) = {a^x}\ln a\]

Complete step-by-step Solution :
We are given function \[f(x) = \sin \left[ {{{\cos }^{ - 1}}\dfrac{{\left( {1 - {2^{2x}}} \right)}}{{\left( {1 + {2^{2x}}} \right)}}} \right]\]
Let us assume \[{2^x} = \tan \theta \], then the function becomes
\[f(x) = \sin \left[ {{{\cos }^{ - 1}}\dfrac{{\left( {1 - {{\tan }^2}\theta } \right)}}{{\left( {1 + {{\tan }^2}\theta } \right)}}} \right]\]
We know that \[\dfrac{{1 - {{\tan }^2}\theta }}{{1 + {{\tan }^2}\theta }} = \cos 2\theta \], on substituting this in function, function becomes
\[
  f(x) = \sin \left[ {{{\cos }^{ - 1}}\cos 2\theta } \right] \\
   = \sin 2\theta \\
 \]
We know that \[\sin 2\theta = \dfrac{{2\tan \theta }}{{1 + {{\tan }^2}\theta }}\], so on substituting this in function, we get function as
\[
  f(x) = \sin 2\theta \\
   = \dfrac{{2\tan \theta }}{{1 + {{\tan }^2}\theta }} \\
 \]
On substituting the value of \[\tan \theta \]as \[{2^x}\], we get
\[
  f(x) = \dfrac{{2 \times {2^x}}}{{1 + {2^{2x}}}} \\
   = 2 \times \dfrac{{{2^x}}}{{1 + {4^x}}} \\
 \]
So, we will now be finding the first derivative of function, so on differentiating the function i.e., on using \[d(\dfrac{u}{v}) = \dfrac{{u'v - uv'}}{{{v^2}}}\], we get
\[
  f'(x) = 2 \times \dfrac{{\left( {\dfrac{d}{{dx}}({2^x})(1 + {4^x}) - ({2^x})\dfrac{d}{{dx}}(1 + {4^x})} \right)}}{{{{(1 + {4^x})}^2}}} \\
   = 2 \times \dfrac{{{2^x}(1 + {4^x})\ln 2 - {2^x}({4^x})\ln 4}}{{{{(1 + {4^x})}^2}}} \\
   = 2 \times \dfrac{{{2^x}(1 + {4^x})\ln 2 - {2^x}({4^x})\ln {2^2}}}{{{{(1 + {4^x})}^2}}} \\
   = 2 \times \dfrac{{{2^x}(1 + {4^x})\ln 2 - {2^x} \times 2 \times ({4^x})\ln 2}}{{{{(1 + {4^x})}^2}}} \\
 \]
On substituting \[x = 1\], we get
\[
  f'(1) = 2 \times \dfrac{{{2^1}(1 + {4^1})\ln 2 - {2^1} \times 2 \times ({4^1})\ln 2}}{{{{(1 + {4^1})}^2}}} \\
   = 2 \times \dfrac{{2(5)\ln 2 - 2 \times 2 \times (4)ln2}}{{{{(5)}^2}}} \\
   = \dfrac{{20\ln 2 - 32ln2}}{{25}} \\
   = \dfrac{{ - 12\ln 2}}{{25}} \\
 \]
Now, it is given that the first derivative of function is \[\left( {\dfrac{{ - b}}{a}} \right){\log _e}2\]when \[x = 1\].
So, on comparing \[\left( {\dfrac{{ - b}}{a}} \right){\log _e}2\] with above obtained value of first derivative of function, we get
\[
  a = 25 \\
  b = 12 \\
 \]
Now, on putting this value in \[\left| {{a^2} - {b^2}} \right|\], we get
\[
  {\left| {{a^2} - {b^2}} \right|_{\min }} = \left| {{{25}^2} - {{12}^2}} \right| \\
   = \left| {625 - 144} \right| \\
   = 481 \\
 \]
Therefore the minimum value of \[\left| {{a^2} - {b^2}} \right|\] is \[481\].

Note : Here, students often make mistake while using formulae \[\cos 2\theta = \dfrac{{1 - {{\tan }^2}\theta }}{{1 + {{\tan }^2}\theta }}\], \[\sin 2\theta = \dfrac{{2\tan \theta }}{{1 + {{\tan }^2}\theta }}\]. They often get confused with these formulae and write \[\sin 2\theta = \dfrac{{1 - {{\tan }^2}\theta }}{{1 + {{\tan }^2}\theta }}\],\[\cos 2\theta = \dfrac{{2\tan \theta }}{{1 + {{\tan }^2}\theta }}\]which is wrong. They should use \[\cos 2\theta = \dfrac{{1 - {{\tan }^2}\theta }}{{1 + {{\tan }^2}\theta }}\] , \[\sin 2\theta = \dfrac{{2\tan \theta }}{{1 + {{\tan }^2}\theta }}\] , this formula only.