Question

# The value of $$k$$ for which the number $$3$$ lies between the  roots of the equation $$x^2+(1-2k)x+(k^2-k-2)=0$$ is given by

A
2<k<5
B
k<2
C
2<k<3
D
k>5

Solution

## The correct option is A $$2<k<5$$Let $$\displaystyle f\left( x \right) ={ x }^{ 2 }+\left( 1-2k \right) x+{ k }^{ 2 }-k-2$$ and $$a=1>0$$The number $$3$$ lies between the roots of the given equation, if $$f\left( 3 \right) <0$$Now, $$\displaystyle f\left( 3 \right) =9+\left( 1-2k \right) 3+{ k }^{ 2 }-k-2=10-7k+{ k }^{ 2 }={ k }^{ 2 }-7k+10$$Hence, $$\displaystyle f\left( 3 \right) <0\Rightarrow { k }^{ 2 }-7k+10<0$$$$\displaystyle \Rightarrow \left( k-2 \right) \left( k-5 \right) <0\Rightarrow 2<k<5.$$Mathematics

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