NIMCET Mathematics Previous Year Questions (PYQs) – Page 2 of 95

Quick Solution

Given:

\( \int f(x)\, dx = g(x) \)

Required: \( \int x^5 f(x^3)\, dx \)

Use substitution:

Let \( u = x^3 \Rightarrow du = 3x^2\, dx \Rightarrow dx = \frac{du}{3x^2} \)

Now rewrite the integral:

\[ \int x^5 f(x^3)\, dx = \int x^5 f(u) \cdot \frac{du}{3x^2} = \frac{1}{3} \int x^3 f(u)\, du \]

But \( x^3 = u \), so:

\[ \frac{1}{3} \int u f(u)\, du \]

Now integrate by parts or use the identity:

\[ \int u f(u)\, du = u g(u) - \int g(u)\, du \]

Final answer:

\[ \int x^5 f(x^3)\, dx = \frac{1}{3} \left[ x^3 g(x^3) - \int g(x^3) \cdot 3x^2\, dx \right] = x^3 g(x^3) - \int x^2 g(x^3)\, dx \]

\[ \boxed{ \int x^5 f(x^3)\, dx = x^3 g(x^3) - \int x^2 g(x^3)\, dx } \]

Quick DL Method Solution

Given:

\[ \lim_{x \to 1} \frac{x^4 - 1}{x - 1} = \lim_{x \to k} \frac{x^3 - k^2}{x^2 - k^2} \]

LHS using derivative:

\[ \lim_{x \to 1} \frac{x^4 - 1}{x - 1} = \left.\frac{d}{dx}(x^4)\right|_{x=1} = 4x^3|_{x=1} = 4 \]

RHS using DL logic:

\[ \lim_{x \to k} \frac{x^3 - k^2}{x^2 - k^2} \approx \frac{3k^2(x - k)}{2k(x - k)} = \frac{3k}{2} \]

Equating both sides:

\[ \frac{3k}{2} = 4 \Rightarrow k = \frac{8}{3} \]

\[ \boxed{k = \frac{8}{3}} \]

If A ⊆ B and B ⊆ C, then find the cardinality of A ∪ B ∪ C.

Explanation

• Since A is a subset of B, everything in A is already in B.
• Since B is a subset of C, everything in B (and hence A) is already in C.
• Therefore, A ∪ B ∪ C = C.

✅ Final Answer

|A ∪ B ∪ C| = |C|