Question

$\begin{array}{cccc}& \text{Column I}& & \text{Column II}\\ \text{(A)}& \text{If the roots of the equation}& \text{(P)}& 7\\ & x^3-9x^2+26x-k=0\text{are positive}& & \\ & \text{and in A.P., then}k\text{is equal to}& & \\ \text{(B)}& \text{If the roots of the equation}& \text{(Q)}& 11\\ & x^3-14x^2+kx-64=0\text{are positive}& & \\ & \text{and in G.P., then}k\text{is equal to}& & \\ \text{(C)}& \text{If the roots of the equation}& \text{(R)}& 24\\ & 6x^3-k{x}^2+6x-1=0\text{are positive}& & \\ & \text{and in H.P., then}k\text{is equal to}& & \\ \text{(D)}& \text{The harmonic mean for the roots of}& \text{(S)}& 26\\ & \text{equation}{x}^3-11x^2+3x-26=0\text{is}& & \\ & & \text{(T)}& 56\end{array}$
Which of the following is the only CORRECT combination?

• A. $\left(A\right)\to \left(Q\right),\left(B\right)\to \left(T\right),\left(C\right)\to \left(S\right),\left(D\right)\to \left(R\right)$
• B. $\left(A\right)\to \left(R\right),\left(B\right)\to \left(S\right),\left(C\right)\to \left(Q\right),\left(D\right)\to \left(Q\right)$
• C. $\left(A\right)\to \left(R\right),\left(B\right)\to \left(T\right),\left(C\right)\to \left(Q\right),\left(D\right)\to \left(S\right)$
• D. $\left(A\right)\to \left(Q\right),\left(B\right)\to \left(P\right),\left(C\right)\to \left(R\right),\left(D\right)\to \left(S\right)$

Answer 1

The correct option is C $\left(A\right)\to \left(R\right),\left(B\right)\to \left(T\right),\left(C\right)\to \left(Q\right),\left(D\right)\to \left(S\right)$

$\left(A\right)$
Let the roots be $a-d,a,a+d$
$\Rightarrow a-d+a+a+d=9$
$\Rightarrow a=3$
So $3$ is the root of the equation $x^3-9x^2+26x-k=0$
$\Rightarrow 3^3-9(3{)}^2+26(3)-k=0$
$\Rightarrow k=24$
$\left(A\right)\to \left(R\right)$

$\left(B\right)$
Let the roots be $\frac{a}{r},a,ar$
$\Rightarrow \left(\frac{a}{r}\right)\left(a\right)\left(ar\right)=64$
$\Rightarrow a=4$
So $4$ is the root of the equation $x^3-14x^2+k-64=0$
$\Rightarrow 4^3-14(4{)}^2+k(4)-64=0$
$\Rightarrow k=56$
$\left(B\right)\to \left(T\right)$

$\left(C\right)$
$f\left(x\right)=6x^3-k{x}^2+6x-1=0$, here roots are in H.P.
$f\left(\frac{1}{x}\right)=x^3-6x^2+kx-6=0$, here the roots will be in A.P.
Let the roots be $a-d,a,a+d$
$\Rightarrow a-d+a+a+d=6$
$\Rightarrow a=2$
So $2$ is the root of the equation $x^3-6x^2+kx-6=0$
$\Rightarrow k=11$
​​​​​​​$\left(C\right)\to \left(Q\right)$

$\left(D\right)$
$x^3-11x^2+3x-26=0$
Let the roots be $\alpha ,\beta ,\gamma$
H.M.$=\frac{3}{\frac{1}{\alpha }+\frac{1}{\beta }+\frac{1}{\gamma }}=\frac{3\alpha \beta \gamma }{\alpha \beta +\beta \gamma +\gamma \alpha }=26$
​​​​​​​$\left(D\right)\to \left(S\right)$