Question

Match the two columns. Column- II gives value of $k$ for polynomials given in Column- I when it is divided by $x–1$.

Column 1	Column 2
$k{x}^2–3x+k$	$-2$
$x^2+x+k$	$\frac{3}{2}$
$2x^2+kx+\sqrt{2}$	$\sqrt{2}-1$
$k{x}^2–\sqrt{2}x+1$	$-\left(2+\sqrt{2}\right)$

Answer 1

Solving the polynomials in Column-I to find the value of $k$:

Explanation for (A):

The polynomial $k{x}^2–3x+k$ is divided by $x–1$

By remainder theorem when a polynomial $a\left(x\right)$ is divided by a linear polynomial $b\left(x\right)$ where $x=p$, then the remainder is $r=a\left(p\right)$

Since, $k{x}^2–3x+k$ is divided by $x–1$ i.e. $x=1$ completely, the remainder $a\left(1\right)=0$

$$\begin{gathered} \therefore a\left(1\right)=k{\left(1\right)}^2–3\left(1\right)+k=0 \\ \Rightarrow k-3+k=0 \\ \Rightarrow 2k=3 \\ \Rightarrow k=\frac{3}{2} \end{gathered}$$

Hence, (A) matches with (II).

Explanation for (B):

The polynomial is $x^2+x+k$ is divided by $x–1$

By remainder theorem when a polynomial $a\left(x\right)$ is divided by a linear polynomial $b\left(x\right)$ where $x=p$, then the remainder is $r=a\left(p\right)$

Since, $x^2+x+k$ is divided by $x–1$ i.e. $x=1$ completely, the remainder $a\left(1\right)=0$

$$\begin{gathered} \therefore a\left(1\right)={\left(1\right)}^2+\left(1\right)+k=0 \\ \Rightarrow 1+1+k=0 \\ \Rightarrow k=-2 \end{gathered}$$

Hence, (B) matches with (I).

Explanation for (C):

The polynomial is $2x^2+kx+\sqrt{2}$ is divided by $x–1$

By remainder theorem when a polynomial $a\left(x\right)$ is divided by a linear polynomial $b\left(x\right)$ where $x=p$, then the remainder is $r=a\left(p\right)$

Since, $2x^2+kx+\sqrt{2}$ is divided by $x–1$ i.e. $x=1$ completely, the remainder $a\left(1\right)=0$

$$\begin{gathered} \therefore a\left(1\right)=2{\left(1\right)}^2+k\left(1\right)+\sqrt{2}=0 \\ \Rightarrow 2+k+\sqrt{2}=0 \\ \Rightarrow k=-2-\sqrt{2} \\ \Rightarrow k=-(2+\sqrt{2}) \end{gathered}$$

Hence, (C) matches with (IV).

Explanation for (D):

The polynomial is $k{x}^2–\sqrt{2}x+1$ is divided by $x–1$

By remainder theorem when a polynomial $a\left(x\right)$ is divided by a linear polynomial $b\left(x\right)$ where $x=p$, then the remainder is $r=a\left(p\right)$

Since, $k{x}^2–\sqrt{2}x+1$ is divided by $x–1$ i.e. $x=1$ completely, the remainder $a\left(1\right)=0$

$$\begin{gathered} \therefore a\left(1\right)=k{\left(1\right)}^2-\sqrt{2}\left(1\right)+1=0 \\ \Rightarrow k-\sqrt{2}+1=0 \\ \Rightarrow k=\sqrt{2}-1 \end{gathered}$$

Hence, (D) matches with (III).

Column-I	Column-II
(A) $k{x}^2–3x+k$	(II) $-2$
(B) $x^2+x+k$	(I) $\frac{3}{2}$
(C) $2x^2+kx+\sqrt{2}$	(IV) $-\left(2+\sqrt{2}\right)$
(D) $k{x}^2–\sqrt{2}x+1$	(III) $\sqrt{2}-1$