IIT-JEE 2009 · Paper 2 · Mathematics Q37

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MathematicsInteger+4 / -1Question 37 of 57

The smallest value of $k$ , for which both the roots of the equation
$x^2-8kx+16\left(k^2-k+1\right)=0$
are real, distinct and have values at least 4, is

Your answer

Answer and worked solution

Official answer: 2

Conditions for $f(x)=x^2-8kx+16(k^2-k+1)$ :

distinct real roots: $64k^2-64(k^2-k+1)>0\implies k>1$ ;
vertex to the right of 4: $4k>4$ ;
$f(4)\ge0$ : $16k^2-48k+32\ge0\implies(k-1)(k-2)\ge0$ .

Together $k\ge2$ . The smallest value is 2 (roots 4 and 12).

Answer: 2

Topics:Location of roots Nature of roots

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IIT-JEE 2009 · Paper 2, Mathematics question 37