NIMCET 2015 — Mathematics PYQ

Q: How do I solve NIMCET 2015 Mathematics questions?

Practice NIMCET 2015 Mathematics previous year questions with step-by-step solutions on Mathem Solvex. Each question includes a detailed explanation and video solution to help you master the concept quickly.

Q: What is the difficulty level of NIMCET Mathematics questions?

NIMCET Mathematics questions range from moderate to advanced difficulty. Consistent practice with official previous year papers and topic-wise drills on Mathem Solvex is the most effective preparation strategy.

Q: Are NIMCET previous year papers available for free?

Yes. Mathem Solvex provides free access to NIMCET previous year question papers and topic-wise PDFs. You can download them or attempt them as timed live mock tests directly on the platform.

Question

NIMCET 2015 — Mathematics PYQ

NIMCET | Mathematics | 2015

The value of $k$ for which the equation $(k - 2)x^2 + 8x + k + 4 = 0$ has both real, distinct and negative roots is:

Choose the correct answer:

Explanation

1. Discriminant ( $D$ ) must be greater than zero

For real and distinct roots: $D > 0$

b^2 - 4ac > 0

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

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

Divide by $4$ :

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

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

-k^2 - 2k + 24 > 0 \implies k^2 + 2k - 24 < 0

Factors: $(k + 6)(k - 4) < 0$

So, $-6 < k < 4$ (Condition 1)

2. Sum of roots must be negative

Let roots be $\alpha$ and $\beta$ . If both are negative, $\alpha + \beta < 0$ .

\frac{-b}{a} < 0 \implies \frac{-8}{k - 2} < 0

For this fraction to be negative, the denominator $(k - 2)$ must be positive:

k - 2 > 0 \implies \mathbf{k > 2}

(Condition 2)

3. Product of roots must be positive

If both roots are negative, their product $\alpha\beta > 0$ .

\frac{c}{a} > 0 \implies \frac{k + 4}{k - 2} > 0

From Condition 2, we already know $k - 2 > 0$ . Therefore, for the fraction to be positive:

k + 4 > 0 \implies \mathbf{k > -4}

(Condition 3)

4. Intersection of all conditions

Condition 1: $k \in (-6, 4)$
Condition 2: $k > 2$
Condition 3: $k > -4$

The intersection of these ranges is:

2 < k < 4

Looking at the given options:

(a) $0$ (False)

(b) $2$ (False, as $k$ must be strictly greater than $2$ )

(c) $3$ (True, as $3$ lies between $2$ and $4$ )

(d) $-4$ (False)

Correct Option: (c)