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 $\Delta \left( x \right)$ .

& \text{A}\text{. 2} \\\ & \text{B}\text{. 3} \\\ & \text{C}\text{. 0} \\\ & \text{D}\text{. 1} \\\ \end{aligned}$$

Answer 1

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

& 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{aligned}$$ Now let us differentiate equation (1) on both sides, then we get $$\begin{aligned} & \Rightarrow {f}'\left( x \right)=\dfrac{d}{dx}\left( a{{x}^{2}}+bx+c \right) \\\ & \Rightarrow {f}'\left( x \right)=2ax+b.....(4) \\\ \end{aligned}$$ Now let us differentiate equation (2) on both sides, then we get $$\begin{aligned} & \Rightarrow {g}'\left( x \right)=\dfrac{d}{dx}\left( d{{x}^{2}}+ex+f \right) \\\ & \Rightarrow {g}'\left( x \right)=2dx+e.....(5) \\\ \end{aligned}$$ Now let us differentiate equation (3) on both sides, then we get $$\begin{aligned} & \Rightarrow {h}'\left( x \right)=\dfrac{d}{dx}\left( g{{x}^{2}}+hx+i \right) \\\ & \Rightarrow {h}'\left( x \right)=2gx+h....(6) \\\ \end{aligned}$$ Now let us differentiate equation (4) on both sides, then we get $$\begin{aligned} & \Rightarrow {f}''\left( x \right)=\dfrac{d}{dx}\left( 2ax+b \right) \\\ & \Rightarrow {f}''\left( x \right)=2a.....(7) \\\ \end{aligned}$$ Now let us differentiate equation (5) on both sides, then we get $$\begin{aligned} & \Rightarrow {g}''\left( x \right)=\dfrac{d}{dx}\left( 2dx+e \right) \\\ & \Rightarrow {g}''\left( x \right)=2d.....(8) \\\ \end{aligned}$$ Now let us differentiate equation (6) on both sides, then we get $$\begin{aligned} & \Rightarrow {h}''\left( x \right)=\dfrac{d}{dx}\left( 2gx+h \right) \\\ & \Rightarrow {h}''\left( x \right)=2g.....(9) \\\ \end{aligned}$$ Now we have to calculate the value of $$\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{aligned} & \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{aligned}$$ Now let us separate the coefficients of $${{x}^{2}},x$$and constant. Now we will simply each and every term, then we get $$\begin{aligned} & \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{aligned}$$ $$\begin{aligned} & \Rightarrow \Delta \left( x \right)=(0){{x}^{2}}-(0)x+0 \\\ & \Rightarrow \Delta \left( x \right)=0 \\\ \end{aligned}$$ Hence, we can say that the degree of the polynomial $$\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.