Question

The least integral value of $k$ for which $\left( {k - 2} \right){x^2} + 8x + k + 4 > 0$ for all $x \in R$, is
A.5
B.4
C.3
D.None of these

Answer 1

Hint: The equation represents a parabola. If we want the $\left( {k - 2} \right){x^2} + 8x + k + 4 > 0$, then the parabola is an upward facing parabola with the coefficient of the ${x^2}$ greater than 0 with discriminant of the equation less than zero. Hence, find the values of $k$ which satisfies the given conditions.

Complete step-by-step answer:

Since, the given equation is a quadratic equation, then the graph of the equation can be an upward parabola or a downward parabola.

But, we want the case, when the equation is strictly greater than zero, then the graph will be an upward parabola.

As it is known, for an upward parabola, the value of discriminant is always less than 0

We know that if a quadratic equation is $a{x^2} + bx + c = 0$, then the discriminant is given by ${b^2} - 4ac$

We will now calculate the discriminant of the given equation $\left( {k - 2} \right){x^2} + 8x + k + 4 = 0$, where $a = {\left( {k - 2} \right)^2}$, $b = 8$ and $c = k + 4$

Then, we have $D = {\left( 8 \right)^2} - 4{\left( {k - 2} \right)}\left( {k + 4} \right)$

For an upward parabola, the value of discriminant should be equal to 0.

That is,

${\left( 8 \right)^2} - 4\left( {k - 2} \right)\left( {k + 4} \right) < 0$

$\Rightarrow 64 - 4\left( {k - 2} \right)\left( {k + 4} \right) < 0$

$\Rightarrow - 4\left( {k - 2} \right)\left( {k + 4} \right) < \- 64$

We will divide the inequality by $- 4$. Since, we are dividing with the negative quantity, there will be a change in the sign of inequality as,

$\left( {k - 2} \right)\left( {k + 4} \right) > 16$

$\Rightarrow {k^2} + 4k - 2k - 8 - 16 > 0$

$\Rightarrow {k^2} + 2k - 24 > 0$

$\Rightarrow {k^2} + 6k - 4k - 24 > 0$

$\Rightarrow k\left( {k + 6} \right) - 4\left( {k + 6} \right) > 0$

$\Rightarrow \left( {k - 4} \right)\left( {k + 6} \right) > 0$

Now, we will plot the roots on the number line and check the interval for which the above inequality is satisfied.

The roots will be $- 6$ and 4.

Let us take a test value between $\left( { - \infty , - 6} \right)$ let $k = - 7$

Substitute in the inequality $\left( {k - 4} \right)\left( {k + 6} \right) > 0$ to check the sign of inequality

Then, $\left( { - 7 - 4} \right)\left( { - 7 + 6} \right) = \left( { - 11} \right)\left( { - 1} \right) = 11 > 0$

Similarly, take a test value from $\left( { - 6,4} \right)$, say 0

Then, $\left( {0 - 4} \right)\left( {0 + 6} \right) = \left( { - 4} \right)\left( 6 \right) = - 24 < 0$

Let 5 be a value of $k$

Then, $\left( {5 - 4} \right)\left( {5 + 6} \right) = \left( 1 \right)\left( {11} \right) = 11 > 0$

Hence,

The interval satisfying the inequality $\left( {k - 4} \right)\left( {k + 6} \right) > 0$ is $\left( { - \infty , - 6} \right) \cup \left( {4,\infty } \right)$

But, we want strict inequality, that is we want the coefficient of ${x^2}$ should be greater than 0.

Then,

$k - 4 > 0$

$k \in \left( {4,\infty } \right)$

Hence, the given expression is $\left( {k - 2} \right){x^2} + 8x + k + 4 > 0$ is true when $k \in \left( {4,\infty } \right)$

The least integral value will be 5.

Thus, option A is correct.

Note: When the discriminant of the equation is less than 0, then the equation represents an upward facing parabola and when the value of discriminant is greater than 0, then the equation represents the downward facing parabola.