question_answer

Match the following.        

Column-I	Column - II
(P) If one of the zero of the polynomial \[f(x)=({{k}^{2}}+4){{x}^{2}}\] \[+13x+4\]                           is reciprocal of the other, then k is equal to	(i) 1
(Q) Sum of the zeroes  of the polynomial \[f(x)=2{{x}^{3}}+k{{x}^{2}}+4x+5\]is 3, then A-is equal to	(ii) 0
(R) If the polynomial          \[f(x)=a{{x}^{3}}+bx+c\]is exactly divisible by \[g(x)={{x}^{2}}+bx+c,\]then \[ab\] is equal to	(iii) \[-6\]

           

A)  (P) \[\to \] (iii); (Q) \[\to \] (i); (R) \[\to \] (ii)

B)         (P) \[\to \](ii);  (Q)\[\to \] (iii);  (R)\[\to \]          (i)

C)         (P) \[\to \] (i); (Q) \[\to \] (iii);             (R)\[\to \]          (ii)

D)         (P) \[\to \] (ii); (Q)\[\to \]  (i);   (R)\[\to \]          (iii)

Correct Answer: B

Solution :

\[(P)\,\,f(x)=({{k}^{2}}+4){{x}^{2}}+13x+4\] Let one root be a, then other root must be\[\frac{1}{\alpha }\] \[\therefore \]  Product of roots \[=\frac{1}{\alpha }\times \alpha =1\] \[\therefore \]   \[1=\frac{4}{{{k}^{2}}+4}\,\,\Rightarrow \,\,{{k}^{2}}+4=4\,\,\Rightarrow \,\,{{k}^{2}}=0\,\,\Rightarrow \,k=0\] (Q) \[f(x)=2{{x}^{3}}+k{{x}^{2}}+4x+5\] Sum of zeroes of \[f(x)=-\frac{k}{2}\] According to question, \[-\frac{k}{2}=3\,\,\Rightarrow \,\,\,k=-6\] (R) \[f(x)\] is exactly divisible by \[g(x),\] i.e, when \[f(x)\] is divided by g(x) remainder must be zero. \[\therefore \]  \[(b-ac+a{{b}^{2}})x+(abc-c)=0\] \[\Rightarrow \] \[b-ac+a{{b}^{2}}=0\] and \[abc-c=0\] \[\therefore \]  \[abc-c=0\]  \[\Rightarrow \]  \[ab=1\]