NIMCET 2014 — Mathematics PYQ

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Question

NIMCET 2014 — Mathematics PYQ

NIMCET | Mathematics | 2014

If $a = \log_{12} 18$ and $b = \log_{24} 54$ , then the value of $ab + 5(a - b)$ is:

Choose the correct answer:

Explanation

Step 1: Express $a$ and $b$ using change of base rule.

Using the property $\log_m(n) = \frac{\log n}{\log m}$ :

For $a$ :

a = \frac{\log 18}{\log 12} = \frac{\log(3^2 \times 2)}{\log(2^2 \times 3)} = \frac{2\log 3 + \log 2}{2\log 2 + \log 3}

For $b$ :

b = \frac{\log 54}{\log 24} = \frac{\log(3^3 \times 2)}{\log(2^3 \times 3)} = \frac{3\log 3 + \log 2}{3\log 2 + \log 3}

Step 2: Simplify using substitution.

Let $\log 2 = x$ and $\log 3 = y$ .

Then:

a = \frac{2y + x}{2x + y}, \quad b = \frac{3y + x}{3x + y}

Step 3: Substitute $a$ and $b$ into the expression $S = ab + 5(a - b)$ .

S = \left( \frac{2y + x}{2x + y} \right) \left( \frac{3y + x}{3x + y} \right) + 5 \left( \frac{2y + x}{2x + y} - \frac{3y + x}{3x + y} \right)

Step 4: Expand and find a common denominator.

The common denominator for both terms is $(2x + y)(3x + y) = 6x^2 + 5xy + y^2$ .

Expanding the product $ab$ :

ab = \frac{6y^2 + 5xy + x^2}{6x^2 + 5xy + y^2}

Expanding the difference $5(a - b)$ :

5(a - b) = 5 \left( \frac{(2y + x)(3x + y) - (3y + x)(2x + y)}{6x^2 + 5xy + y^2} \right)

5(a - b) = 5 \left( \frac{(6xy + 2y^2 + 3x^2 + xy) - (6xy + 3y^2 + 2x^2 + xy)}{6x^2 + 5xy + y^2} \right)

5(a - b) = 5 \left( \frac{x^2 - y^2}{6x^2 + 5xy + y^2} \right) = \frac{5x^2 - 5y^2}{6x^2 + 5xy + y^2}

Step 5: Combine the terms.

S = \frac{6y^2 + 5xy + x^2 + 5x^2 - 5y^2}{6x^2 + 5xy + y^2}

S = \frac{6x^2 + 5xy + y^2}{6x^2 + 5xy + y^2}

S = 1

Final Answer:

The correct option is 1: $1$ .

Choose the correct answer:

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