NIMCET 2011 — Mathematics PYQ

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Question

NIMCET 2011 — Mathematics PYQ

NIMCET | Mathematics | 2011

The least integral value of $k$ for which $(k - 2)x^2 + 8x + k + 4 > 0$ for all $x \in R$ , is:

Choose the correct answer:

Explanation

1. First Condition ( $a > 0$ ):

In our equation, $a = k - 2$ .

k - 2 > 0

k > 2

2. Second Condition ( $D < 0$ ):

The discriminant is $D = b^2 - 4ac$ . Here, $a = (k - 2)$ , $b = 8$ , and $c = (k + 4)$ .

8^2 - 4(k - 2)(k + 4) < 0

64 - 4(k^2 + 4k - 2k - 8) < 0

64 - 4(k^2 + 2k - 8) < 0

Divide the entire inequality by 4 to simplify:

16 - (k^2 + 2k - 8) < 0

16 - k^2 - 2k + 8 < 0

-k^2 - 2k + 24 < 0

Multiply by $-1$ (and reverse the inequality sign):

k^2 + 2k - 24 > 0

3. Solve the Quadratic Inequality:

Factorize the expression:

(k + 6)(k - 4) > 0

The critical values are $k = -6$ and $k = 4$ . For the expression to be greater than zero:

k < -6 \quad \text{or} \quad k > 4

4. Combine All Conditions:

We have:

$k > 2$ (from the first condition)
$k < -6$ or $k > 4$ (from the second condition)

The intersection of these conditions is $k > 4$ .

5. Find the Least Integral Value:

Since $k$ must be an integer and $k$ is strictly greater than 4, the smallest integer value is:

k = 5

The correct option is (a) 5.