How do I solve NIMCET 2012 Mathematics questions?

Practice NIMCET 2012 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.

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.

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.

Tip:A–D to answerE for explanationV for videoS to reveal answer

NIMCET 2012 — Mathematics PYQ

NIMCET | Mathematics | 2012

Normal to the curve $y = x^3 - 3x + 2$ at the point $(2, 4)$ is:

Choose the correct answer:

B.

$x - 9y + 40 = 0$

C.

$x + 9y - 38 = 0$

(Correct Answer)

D.

$-9x + y + 22 = 0$

Explanation

1. Find the derivative (Slope of the tangent):

The equation of the curve is:

y = x^3 - 3x + 2

Differentiating with respect to $x$ :

\frac{dy}{dx} = 3x^2 - 3

2. Calculate the slope of the tangent at $(2, 4)$ :

Substitute $x = 2$ into the derivative:

m_t = \left[ \frac{dy}{dx} \right]_{(2, 4)} = 3(2)^2 - 3

m_t = 3(4) - 3 = 12 - 3 = 9

3. Find the slope of the normal ( $m_n$ ):

Since the normal is perpendicular to the tangent, their slopes satisfy the condition $m_t \cdot m_n = -1$ .

m_n = -\frac{1}{m_t} = -\frac{1}{9}

4. Find the equation of the normal:

Using the point-slope form $y - y_1 = m_n(x - x_1)$ with the point $(2, 4)$ and slope $-\frac{1}{9}$ :

y - 4 = -\frac{1}{9}(x - 2)

Multiply the entire equation by $9$ :

9(y - 4) = -1(x - 2)

9y - 36 = -x + 2

Rearrange the terms into the standard form $Ax + By + C = 0$ :

x + 9y - 36 - 2 = 0

x + 9y - 38 = 0

Conclusion:

The correct equation is $x + 9y - 38 = 0$ .

Correct Option: (c)