Question

If \[f\left( x \right){\rm{ = }}\left| {\begin{array}{*{20}{c}}{2\cos x}&1&0\\{x - \dfrac{\pi }{2}}&{2\cos x}&1\\0&1&{2\cos x}\end{array}} \right|\]then \[\dfrac{{df}}{{dx}}\] at \[{\rm{x = }}\dfrac{\pi }{2}\] is
\[\begin{array}{l}A.\,\,2\\B.\,\,\dfrac{\pi }{2}\\C.\,\,1\\D.\,\,8\end{array}\]

Answer 1

Hint: First expand the determinant in a way that is easy. After expanding you’ll get a function of x. Differentiate both sides with respect to x. After differentiating, pluck the value of x as \[\dfrac{\pi }{2}.\] Solve the equation completely. Then the result of that simplification is the required result.

Complete step by step solution:
Given function in the question is as follows:
\[f\left( x \right){\rm{ = }}\left| {\begin{array}{*{20}{c}}{2\cos x}&1&0\\{x - \dfrac{\pi }{2}}&{2\cos x}&1\\0&1&{2\cos x}\end{array}} \right|\]
By expanding determinant by using first row, we get it as:
\[f\left( x \right){\rm{ = 2cosx}}\left| {\begin{array}{*{20}{c}}{2\cos x}&1\\1&{2\cos x}\end{array}} \right| - 1\left| {\begin{array}{*{20}{c}}{x - \dfrac{\pi }{2}}&1\\0&{2\cos x}\end{array}} \right|\]
By expanding determinants in above equation, we get it as:
\[f\left( x \right){\rm{ = 2cosx }}\left( {{{\left( {2cosx} \right)}^2} - {{\left( 1 \right)}^2}} \right) - 1\left( {\left( {x - \dfrac{\pi }{2}} \right)\left( {2cosx} \right) - 0} \right)\]
By simplifying both the terms, we get it as:
\[f\left( x \right){\rm{ = }}\left( {{{\left( {2cosx} \right)}^3} - 2cosx} \right) - \left( {2xcosx - \pi cosx} \right)\]
By simplifying the terms and removing the bracket, we get:
\[f\left( x \right){\rm{ = 8co}}{{\rm{s}}^3}x - 2cosx - 2xcosx + \pi cosx\]
By differentiating with respect to x on both sides, we get:
\[\dfrac{d}{{dx}}{\rm{f}}\left( x \right) = \dfrac{d}{{dx}}\left( {8co{s^3}x - 2cosx - 2xcosx + \pi cosx} \right)\]
By separating the terms of above equation, we get it as:
\[{f^1}\left( x \right){\rm{ = 8}}\dfrac{d}{{dx}}\left( {co{s^3}x} \right) - 2\dfrac{d}{{dx}}cosx - 2\dfrac{d}{{dx}}\left( {xcosx} \right) + \pi \dfrac{d}{{dx}}\left( {cosx} \right)\]
By basic knowledge of differentiation, we know
\[\dfrac{d}{{dx}}{x^n}{\rm{ = n}}{{\rm{x}}^{{\rm{n - 1}}}}{\rm{ }}\dfrac{d}{{dx}}\left( {cosx} \right){\rm{ }} = {\rm{ }} - sinx\]
By substituting this in our equation, we get it as:
\[\begin{array}{l}{f^1}\left( x \right){\rm{ = 83co}}{{\rm{s}}^{\rm{2}}}{\rm{x }}\left( { - sinx} \right) - 2\left( { - sinx} \right) - 2\dfrac{d}{{dx}}\left( {xcosx} \right) + \left( { - sinx} \right)\\\end{array}\]
By simplifying the above equation, we can write it as:
\[{f^1}\left( x \right){\rm{ = - 24sinx co}}{{\rm{s}}^{\rm{2}}}{\rm{x + 2sinx - }}\pi {\rm{sinx - 2}}\dfrac{d}{{dx}}\left( {xcosx} \right)\]
By knowledge of trigonometry we know u.v rule
\[d\left( {u.v} \right){\rm{ = vdu + udv}}\]
By substituting this we can solve the last term, in form of:
\[{{\rm{f}}^{\rm{1}}}\left( {\rm{x}} \right){\rm{ = - 24sinx co}}{{\rm{s}}^{\rm{2}}}{\rm{x + 2sinx - }}\pi {\rm{sinx - 2x}}\dfrac{d}{{dx}}cosx{\rm{ - 2cosx}}\dfrac{d}{{dx}}x\]
By substituting \[x{\rm{ = }}\dfrac{\pi }{2}\], we know \[sin\dfrac{\pi }{2}{\rm{ = 1 cos}}\dfrac{\pi }{2} = 0,\]we get
\[{f^1}\left( x \right){\rm{ = 0 + 2 - }}\pi {\rm{ + 2}}{\rm{. }}\dfrac{\pi }{2} - 0\]
By cancelling the common terms, we get the value as:
\[{f^1}\left( {x = \dfrac{\pi }{2}} \right) = 2.\]
Therefore, option (a) is correct.

Note: Be careful with “-“sign at differentiation of \[{\rm{cosx}}\]. Generally students forget that sign and end up getting the wrong answer. While applying the u.v rule also be careful that you consider both the differentiations. In hurry students forget to consider the second term of u.v rule.