NIMCET 2023 Previous Year Questions (PYQs) – Page 1 of 12

Given:

\( \vec{a} = \hat{i} - \hat{k}, \quad \) \(\vec{b} = x\hat{i} + \hat{j} + (1 - x)\hat{k},\) \( \quad \vec{c} = y\hat{i} + x\hat{j} + (1 + x - y)\hat{k} \)

Form the matrix:

\( M = \begin{bmatrix} 1 & x & y \\ 0 & 1 & x \\ -1 & 1 - x & 1 + x - y \end{bmatrix} \)

Find the determinant:

\( \det(M) = \begin{vmatrix} 1 & x & y \\ 0 & 1 & x \\ -1 & 1 - x & 1 + x - y \end{vmatrix} = 1 \)

Since the determinant is constant and non-zero, the vectors are linearly independent.

\( \boxed{\text{The matrix does not depend on } x \text{ or } y} \)

Quick Solution

Given: \( \vec{a}, \vec{b} \) are unit vectors and

\( 2\vec{a} + \vec{b} = 3 \)

Take magnitude on both sides:

\( |2\vec{a} + \vec{b}| = 3 \Rightarrow |2\vec{a} + \vec{b}|^2 = 9 \)

Use identity:

\[ |2\vec{a} + \vec{b}|^2 = 4|\vec{a}|^2 + |\vec{b}|^2 + 4(\vec{a} \cdot \vec{b}) = 4 + 1 + 4(\vec{a} \cdot \vec{b}) = 5 + 4(\vec{a} \cdot \vec{b}) \]

Set equal to 9:

\[ 5 + 4(\vec{a} \cdot \vec{b}) = 9 \Rightarrow \vec{a} \cdot \vec{b} = 1 \Rightarrow \cos\theta = 1 \Rightarrow \theta = 0^\circ \]

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:

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

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