JEE Main 2016 (4)

Algebra Level 5

The smallest integral value of $p$ such that $px^{2}+12x+6>3x^{2}-p$ for all real values of $x$ is

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None of these 8 5 6 7

1 solution

Sahil Bansal
Mar 21, 2016

The inequality can be rewritten as : $(p-3){ x }^{ 2 }+12x+(6+p)>0\quad$

If this is to be true for all real x, then coefficient of $x^{2}$ must be positive and the value of discriminant of the quadratic equation $(p-3){ x }^{ 2 }+12x+(6+p)=0\quad$ must be negative.

$\Rightarrow p-3>0\quad \\ \Rightarrow p>3$

Also, ${ 12 }^{ 2 }-4(p-3)(6+p)<0\\ \Rightarrow 144-4({ p }^{ 2 }+3p-18)<0\\ \Rightarrow { p }^{ 2 }+3p-18>36\\ \Rightarrow { p }^{ 2 }+3p-54>0\\ \Rightarrow (p+9)(p-6)>0\\ \Rightarrow p\epsilon (-\infty ,-9)\sqcup (6,\infty )\\$

Combining both the inequalities : $p>6\\$

Hence smallest integral value of p required is : $\boxed{7}$

Smallest integral value can be any integer less than -9. Therefore answer should be 'none of these'. Where am I wrong??

Rinkon Saha - 5 years, 2 months ago

'p' should be greater than zero, as Sahil Bansal has written in the solution. Hence 'p' can't be negative.

Rajdeep Bharati - 5 years, 2 months ago

$p-3$ must be greater than zero and as well as the value of Discriminant must be less than 0. You are just making the $Discriminant < 0$ and ignoring the condition that coefficient of $x^{2}>0$ . Remember, if coefficient of $x^{2}$ is positive then the parabola opens upward and hence it can be +ve for all real values of 'x' if along with this, the Discriminant is -ve. (As it will always be above the x-axis).

Sahil Bansal - 5 years, 2 months ago

JEE Main 2016 (4)

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