NIMCET 2010 — Mathematics PYQ

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Question

NIMCET 2010 — Mathematics PYQ

NIMCET | Mathematics | 2010

If $y = a \log x + bx^2 + x$ has its extremum value at $x = -1$ and $x = 2$ , then:

Choose the correct answer:

Explanation

Step 1: Find the first derivative of the function.

Given:

y = a \log x + bx^2 + x

Differentiating with respect to $x$ :

\frac{dy}{dx} = \frac{a}{x} + 2bx + 1

Step 2: Apply the condition for extremum values.

Extremum values occur where the derivative is zero ( $\frac{dy}{dx} = 0$ ).

At $x = 2$ :

$\frac{a}{2} + 2b(2) + 1 = 0$

$\frac{a}{2} + 4b + 1 = 0 \implies a + 8b = -2 \quad \dots \text{(Equation 1)}$
At $x = -1$ (Assuming the formal derivative must vanish):

$\frac{a}{-1} + 2b(-1) + 1 = 0$

$-a - 2b + 1 = 0 \implies a + 2b = 1 \quad \dots \text{(Equation 2)}$

Step 3: Solve the system of linear equations.

Subtract Equation 2 from Equation 1:

(a + 8b) - (a + 2b) = -2 - 1

6b = -3

b = -\frac{3}{6} = -\frac{1}{2}

Now, substitute $b = -\frac{1}{2}$ into Equation 2:

a + 2\left(-\frac{1}{2}\right) = 1

a - 1 = 1

a = 2

Step 4: Conclusion.

The values are $a = 2$ and $b = -\frac{1}{2}$ .