Question

If f(x), g(x) and h(x) are three polynomials of degree 2 and \[\Delta \left( x \right)=\left( \begin{matrix}
   f(x) & g(x) & h(x) \\
   {f}'(x) & {g}'(x) & {h}'(x) \\
   {f}''(x) & {g}''(x) & {h}''(x) \\
\end{matrix} \right)\] then the degree of polynomial \[\Delta \left( x \right)\] .
\[\begin{align}
  & \text{A}\text{. 2} \\
 & \text{B}\text{. 3} \\
 & \text{C}\text{. 0} \\
 & \text{D}\text{. 1} \\
\end{align}\]

Answer 1

Hint: Let us assume \[f(x)=a{{x}^{2}}+bx+c,g(x)=d{{x}^{2}}+ex+f,h(x)=g{{x}^{2}}+hx+i\]. Let us consider \[f(x)=a{{x}^{2}}+bx+c,g(x)=d{{x}^{2}}+ex+f,h(x)=g{{x}^{2}}+hx+i\] as equation (1), equation (2) and equation (3) respectively. Now we should find \[\text{{f}'(x)}\], \[\text{{g}'(x)}\] and \[\text{{h}'(x)}\]. Now we should find \[\text{{f}'''(x)}\], \[\text{{g}'''(x)}\] and \[\text{{h}'''(x)}\]. Now we should substitute all the values in \[\Delta \left( x \right)=\left( \begin{matrix}
   f(x) & g(x) & h(x) \\
   {f}'(x) & {g}'(x) & {h}'(x) \\
   {f}''(x) & {g}''(x) & {h}''(x) \\
\end{matrix} \right)\]. Now we will find the determinant of \[\Delta \left( x \right)\].

Complete step-by-step solution
Let us assume \[f(x)=a{{x}^{2}}+bx+c,g(x)=d{{x}^{2}}+ex+f,h(x)=g{{x}^{2}}+hx+i\].
Now let us consider
\[\begin{align}
  & f(x)=a{{x}^{2}}+bx+c......(1) \\
 & g(x)=d{{x}^{2}}+ex+f......(2) \\
 & h(x)=g{{x}^{2}}+hx+i.......(3) \\
\end{align}\]
Now let us differentiate equation (1) on both sides, then we get
\[\begin{align}
  & \Rightarrow {f}'\left( x \right)=\dfrac{d}{dx}\left( a{{x}^{2}}+bx+c \right) \\
 & \Rightarrow {f}'\left( x \right)=2ax+b.....(4) \\
\end{align}\]
Now let us differentiate equation (2) on both sides, then we get
\[\begin{align}
  & \Rightarrow {g}'\left( x \right)=\dfrac{d}{dx}\left( d{{x}^{2}}+ex+f \right) \\
 & \Rightarrow {g}'\left( x \right)=2dx+e.....(5) \\
\end{align}\]
Now let us differentiate equation (3) on both sides, then we get
\[\begin{align}
  & \Rightarrow {h}'\left( x \right)=\dfrac{d}{dx}\left( g{{x}^{2}}+hx+i \right) \\
 & \Rightarrow {h}'\left( x \right)=2gx+h....(6) \\
\end{align}\]
Now let us differentiate equation (4) on both sides, then we get
\[\begin{align}
  & \Rightarrow {f}''\left( x \right)=\dfrac{d}{dx}\left( 2ax+b \right) \\
 & \Rightarrow {f}''\left( x \right)=2a.....(7) \\
\end{align}\]
Now let us differentiate equation (5) on both sides, then we get
\[\begin{align}
  & \Rightarrow {g}''\left( x \right)=\dfrac{d}{dx}\left( 2dx+e \right) \\
 & \Rightarrow {g}''\left( x \right)=2d.....(8) \\
\end{align}\]
Now let us differentiate equation (6) on both sides, then we get
\[\begin{align}
  & \Rightarrow {h}''\left( x \right)=\dfrac{d}{dx}\left( 2gx+h \right) \\
 & \Rightarrow {h}''\left( x \right)=2g.....(9) \\
\end{align}\]
Now we have to calculate the value of \[\Delta \left( x \right)=\left( \begin{matrix}
   f(x) & g(x) & h(x) \\
   {f}'(x) & {g}'(x) & {h}'(x) \\
   {f}''(x) & {g}''(x) & {h}''(x) \\
\end{matrix} \right)\].
We know that \[\left( \begin{matrix}
   a & b & c \\
   d & e & f \\
   g & h & i \\
\end{matrix} \right)=a\left( ei-hf \right)-b\left( di-fg \right)+c\left( dh-eg \right)\]. By using this concept, we should find the value of \[\Delta \left( x \right)\].
So, from this concept we should find the value of
\[\Rightarrow \Delta \left( x \right)=\left( \begin{matrix}
   a{{x}^{2}}+bx+c & d{{x}^{2}}+ex+f & g{{x}^{2}}+hx+i \\
   2ax+b & 2dx+e & 2gx+h \\
   2a & 2d & 2g \\
\end{matrix} \right)\]
\[\begin{align}
  & \Rightarrow \Delta \left( x \right)=\left( a{{x}^{2}}+bx+c \right)\left( \left( 2dx+e \right)\left( 2g \right)-\left( 2gx+h \right)\left( 2d \right) \right)-\left( d{{x}^{2}}+ex+f \right)\left( \left( 2ax+b \right)\left( 2g \right)-\left( 2gx+h \right)\left( 2a \right) \right) \\
 & \text{ }+\left( g{{x}^{2}}+hx+i \right)\left( \left( 2ax+b \right)\left( 2d \right)-\left( 2dx+e \right)\left( 2a \right) \right) \\
 & \Rightarrow \Delta \left( x \right)=\left( a{{x}^{2}}+bx+c \right)\left( \left( 4dgx+2ge \right)-\left( 4dgx+2hd \right) \right)-\left( d{{x}^{2}}+ex+f \right)\left( \left( 4agx+2bg \right)-\left( 4agx+2ah \right) \right) \\
 & \text{ }+\left( g{{x}^{2}}+hx+i \right)\left( \left( 4adx+2bd \right)-\left( 4adx+2ea \right) \right) \\
 & \Rightarrow \Delta \left( x \right)=\left( a{{x}^{2}}+bx+c \right)\left( 2ge-2hd \right)-\left( d{{x}^{2}}+ex+f \right)\left( 2bg-2ah \right)+\left( g{{x}^{2}}+hx+i \right)\left( 2bd-2ea \right) \\
\end{align}\]
Now let us separate the coefficients of \[{{x}^{2}},x\]and constant. Now we will simply each and every term, then we get
\[\begin{align}
  & \Rightarrow \Delta \left( x \right)=\left( 2aeg-2ahd-2bdg+2ahd+2bdg-2aeg \right){{x}^{2}} \\
 & +\left( 2beg-2bdh+2beg-2aeg+2bdh-2aeg \right)x+\left( 2ceg-2cdh-2bfg+2afh-2ibd-2aei \right) \\
\end{align}\]
\[\begin{align}
  & \Rightarrow \Delta \left( x \right)=(0){{x}^{2}}-(0)x+0 \\
 & \Rightarrow \Delta \left( x \right)=0 \\
\end{align}\]
Hence, we can say that the degree of the polynomial \[\Delta \left( x \right)\] is zero.
Hence, option C is correct.

Note: Some students may assume that the degree of the \[\Delta \left( x \right)\] is equal to 3 by simply seeing the equation below.
\[\Delta \left( x \right)=\left( \begin{matrix}
   a{{x}^{2}}+bx+c & d{{x}^{2}}+ex+f & g{{x}^{2}}+hx+i \\
   2ax+b & 2dx+e & 2gx+h \\
   2a & 2d & 2g \\
\end{matrix} \right)\]
But after finding the determinant, we are getting the value of \[\Delta \left( x \right)\] is equal to constant.
So, students should solve the determinant until the last step. Then we will get the correct answer.