Question

State 'T' for true and 'F' for false.
I. If the square of the sum of the zeroes of a quadratic polynomial is 4 times the product of them, then the zeroes are equal in magnitude.
II. A quadratic polynomial, the sum of whose zeroes is 0 and one zero is 4, then the linear term of the polynomial doesn't exist.
III. If the sum of the zeroes of a cubic polynomial is 5 and the sum of the product of two zeroes is 8, then the sum of the square of the zeroes will be 10.
IV. If one of the roots of the polynomial f(x) = $6x^2+13x+k$ is reciprocal of the other, then k is 6.

• A. I - T II - F III - T IV - F
• B. I - F II - T III - F IV - T
• C. I - T II - F III - F IV - T
• D. I - T II - T III - F IV - T

Answer 1

The correct option is D I - T II - T III - F IV - T

I. True
Since, $(\alpha +\beta {)}^2=4\alpha \beta$
$\Rightarrow {\alpha }^2+{\beta }^2+2\alpha \beta -4\alpha \beta =0\Rightarrow {\alpha }^2+{\beta }^2-2\alpha \beta =0\Rightarrow (\alpha -\beta {)}^2=0\Rightarrow \alpha -\beta =0\Rightarrow \alpha =\beta$
II. True
Given, $\alpha +\beta =0$
$4+\beta =0\Rightarrow \beta =-4$
$\therefore$ Coefficient of linear term:
$=\alpha +\beta =0$
$\therefore x^2-(\alpha +\beta )x+\alpha \beta$
$=x^2-16$
III. False
$\alpha +\beta +\gamma =5and\alpha \beta +\beta \gamma +\alpha \gamma =8$
Now,
$(\alpha +\beta +\gamma {)}^2$
$={\alpha }^2+{\beta }^2+{\gamma }^2+$
$2(\alpha \beta +\beta \gamma +\alpha \gamma )\Rightarrow 25={\alpha }^2+{\beta }^2+{\gamma }^2+2\times 8\Rightarrow 9={\alpha }^2+{\beta }^2+{\gamma }^2$
IV. True
$\alpha =\frac{1}{\beta }\therefore Wehave,\alpha \beta =\frac{k}{6}\Rightarrow \frac{1}{\beta }\times \beta =\frac{k}{6}\Rightarrow k=6$