| Directions: Each of these questions contains two statements: Assertion [A] and Reason [R]. Each of these questions also has four alternative choices, any one of which is the correct answer. You have to select one of the codes [a], [b], [c] and [d] given below. |
| Assertion [A] If m and n are the zeroes of the polynomial \[3{{x}^{2}}+11x-4\], then \[12\left( {{m}^{2}}+{{n}^{2}} \right)+145mn=0\] |
| Reason [R] If \[\alpha \] and \[\beta \]are the zeroes of the polynomial\[a{{x}^{2}}+bx+c\], then |
| \[\frac{\alpha }{\beta }+\frac{\beta }{\alpha }=\frac{{{b}^{2}}}{ac}+2\] |
A) A is true, R is true; R is a correct explanation for A.
B) A is true, R is true; R is not a correct explanation for A.
C) A is true; R is False.
D) A is false; R is true.
Correct Answer: C
Solution :
| Let\[f\left( x \right)=3{{x}^{2}}+11x-4\] |
| \[=3{{x}^{2}}+12x-x-4\] |
| \[=3x\left( x+4 \right)-1\left( x+4 \right)\] |
| \[=\left( 3x-1 \right)\left( x+4 \right)\] |
| Let \[m=\frac{1}{3}\] |
| And \[n=-4\] |
| \[12\left( {{m}^{2}}+{{n}^{2}} \right)+145\,mn\] |
| \[=12\left( \frac{1}{9}+16 \right)+145\times \left( \frac{1}{3} \right)\times \left( -4 \right)\] |
| \[=12\left( \frac{1+144}{9} \right)-\frac{145\times 4}{3}\] |
| \[=\frac{12\times 145}{9}-\frac{145\times 4}{3}=0\] |
| Given, a and p are roots of \[a{{x}^{2}}+by+c\] |
| \[\therefore \,\,\alpha +\beta =\frac{-b}{a},\,\alpha \beta =\frac{c}{a}\] |
| \[\therefore \,\,\frac{\alpha }{\beta }+\frac{\beta }{\alpha }=\frac{{{\alpha }^{2}}+{{\beta }^{2}}}{\alpha \beta }=\frac{{{\left( \alpha +\beta \right)}^{2}}-2\alpha \beta }{\alpha \beta }\] |
| \[=\frac{{{\left( \alpha +\beta \right)}^{2}}}{\alpha \beta }-2=\frac{{{b}^{2}}}{ac}-2\] |
| Hence, Assertion is true but Reason is false. |
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