NEB Class 12 Management Business Mathematics Question Paper 2082 (Set A) Nepal

For a non-singular matrix, $A^{-1} = \dfrac{1}{|A|}\,(\text{adj } A)$. Hence the correct option is C.

Expanding along the second row, only the entry $7$ contributes:

Hence the answer is 0 (option A).

The Hawkins-Simon condition (for the matrix $I-A$) requires all leading principal minors to be positive, i.e. the diagonal entries and the determinant must be positive. Only option D qualifies: $\begin{pmatrix} 0.3 & -0.5 \\ -0.2 & 0.7 \end{pmatrix}$ has positive diagonal ($0.3, 0.7$) and determinant $0.3\times 0.7 - (-0.5)(-0.2) = 0.21 - 0.10 = 0.11 > 0$. Hence D.

$f'(x) = 2x - 4$. The function is increasing where $f'(x) > 0$, i.e. $x > 2$. Among the options only $x = 4$ satisfies this. Hence D.

Removing the radical by squaring: $\left(\dfrac{d^4y}{dx^4}\right)^2 = \dfrac{d^3y}{dx^3}$. The degree is the power of the highest-order derivative, which is $2$. Hence B.

Let the numbers be $x, 4x, 12x$. Adding $1$ to the first gives $x+1, 4x, 12x$ in G.P., so $(4x)^2 = (x+1)(12x)$, i.e. $16x^2 = 12x^2 + 12x \Rightarrow 4x^2 - 12x = 0 \Rightarrow x = 3$. The numbers are $3, 12, 36$. Hence D.

Net Present Value (NPV) is the difference between the total present value of future returns and the total investment. Hence C.

Scrap value $= P(1-r)^n = 50000(1-0.10)^3 = 50000 \times 0.729 = \text{Rs. } 36{,}450$. Hence D.

A non-negative variable added to a $\le$ constraint to convert it to an equation is a slack variable. Hence B.

(Note: as printed the equation reads $x+2y+v=0$; the intended form is $x+2y+v=6$, but $v$ remains the slack variable.)

Bowley's coefficient of skewness $= \dfrac{(Q_3 - Q_2) - (Q_2 - Q_1)}{Q_3 - Q_1}$. Since $Q_2 - Q_1 = Q_3 - Q_2$, the numerator is $0$, so skewness $= 0$ (symmetric distribution). Hence B.

For dependent events, the multiplication rule gives $P(X \cap Y) = P(X) \cdot P(Y/X)$. Hence B.

For matrices $A$, $B$, $C$ of the same order, the associative property of addition states:

A square matrix $A$ is called symmetric if it equals its own transpose, i.e. $A^{T} = A$ (equivalently $a_{ij} = a_{ji}$ for all $i, j$).

$A \times (\text{adj } A) = |A|\, I$, where $|A|$ is the determinant of $A$ and $I$ is the identity matrix of the same order.

By Cramer's rule the system cannot be solved (no unique solution) when the determinant of the coefficient matrix is zero, i.e. $|A| = D = 0$.

The Leontief technology (or Leontief) matrix is $I - A$. The system is then written as $(I - A)X = D$, where $D$ is the final demand vector.

The Hawkins-Simon condition requires all leading principal minors of $(I-A)$ to be positive.

• First minor: $0.7 > 0$.
• Second minor (determinant): $|I-A| = (0.7)(2/3) - (-1/3)(-0.4) = 0.4667 - 0.1333 = 0.3333 > 0$.

Both leading principal minors are positive, so the Hawkins-Simon condition is satisfied and the system is viable (economically feasible).

Average revenue is total revenue per unit of output:

Elasticity of supply:

where $Q_s$ is quantity supplied and $P$ is price.

For a function $y = f(x)$ to attain a maximum at a point: the first-order (necessary) condition $f'(x) = 0$ and the second-order (sufficient) condition $f''(x) < 0$.

The break-even point occurs when total revenue equals total cost (profit = 0):

A stationary point occurs where the first derivative is zero:

Example: $\dfrac{dy}{dx} + 2y = 6$ (constant coefficient $2$, constant term $6$).

This is linear of the form $\dfrac{dy}{dx} + Py = Q$ with $P = 2$. Integrating factor $= e^{\int 2\,dx} = e^{2x}$.

where $c$ is an arbitrary constant.

Given $f(x) = \dfrac{1}{x} = x^{-1}$.

First derivative: $f'(x) = -x^{-2} = -\dfrac{1}{x^2}$.

To test whether $f'(x)$ is decreasing, find $f''(x)$: $f''(x) = 2x^{-3} = \dfrac{2}{x^3}$.

For $x \in (-8, 0)$, $x < 0$, so $x^3 < 0$, hence $f''(x) = \dfrac{2}{x^3} < 0$.

Since $f''(x) < 0$ throughout $(-8, 0)$, the slope $f'(x)$ is decreasing on this interval. Hence justified.

Differentiation and integration (anti-derivative) are inverse operations: if we differentiate a function and then integrate the result (or vice versa), we recover the original function (up to a constant).

Example: Let $f(x) = x^3$.

Derivative: $\dfrac{d}{dx}(x^3) = 3x^2$.

Anti-derivative of $3x^2$: $\displaystyle\int 3x^2\,dx = x^3 + c$.

Thus integrating the derivative returns the original function $x^3$ (plus a constant $c$), confirming that derivative and anti-derivative are inverse operations.