IGDTUW 2025 — Mathematics PYQ

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Question

IGDTUW 2025 — Mathematics PYQ

IGDTUW | Mathematics | 2025

Evaluate: $lim_{x \to 0} \frac{1 - c o s x c o s 2 x c o s 3 x}{s i n ^{2} 2 x}$

Choose the correct answer:

Explanation

Step 1: Simplify the Denominator

\sin^2 2x \approx (2x)^2 = 4x^2

Step 2: Simplify the Numerator

Using the expansion for each cosine term:

\cos x \cos 2x \cos 3x \approx \left(1 - \frac{x^2}{2}\right) \left(1 - \frac{(2x)^2}{2}\right) \left(1 - \frac{(3x)^2}{2}\right)

\approx \left(1 - \frac{x^2}{2}\right) \left(1 - 2x^2\right) \left(1 - \frac{9x^2}{2}\right)

Multiplying these out and ignoring terms with higher powers (like $x^4$ or $x^6$ since they go to zero faster):

\approx 1 - \left(\frac{1}{2} + 2 + \frac{9}{2}\right)x^2

\approx 1 - \left(\frac{1 + 4 + 9}{2}\right)x^2 = 1 - 7x^2

Now, substitute this back into the numerator:

1 - \cos x \cos 2x \cos 3x \approx 1 - (1 - 7x^2) = 7x^2

Step 3: Apply the Limit

\lim_{x \to 0} \frac{7x^2}{4x^2} = \frac{7}{4}

The correct option is (b) $\frac{7}{4}$ .