CBSE Class 12 Applied Mathematics — NCERT Solutions

Given: Two numbers 132 and 121; modulus = 23.

Step 1: Find 132 mod 23.
$$132 = 23 \times 5 + 17 \implies 132 \equiv 17 \pmod{23}$$

Step 2: Find 121 mod 23.
$$121 = 23 \times 5 + 6 \implies 121 \equiv 6 \pmod{23}$$

Step 3: Add the residues.
$$17 + 6 = 23 \equiv 0 \pmod{23}$$

Answer: $(132 + 121) \bmod 23 = \mathbf{0}$

Given: $7^6 \pmod{3}$.

Step 1: Find $7 \bmod 3$.
$$7 = 3 \times 2 + 1 \implies 7 \equiv 1 \pmod{3}$$

Step 2: Apply the power.
$$7^6 \equiv 1^6 = 1 \pmod{3}$$

Answer: $7^6 \bmod 3 = \mathbf{1}$

We identify each matrix by examining its rows, columns, and element structure.

i) $A = [2\quad 3]$ has 1 row and 2 columns. Type: Row matrix. Order: $1 \times 2$.

ii) $B = \begin{bmatrix} 3 \\ -2 \\ 1 \end{bmatrix}$ has 3 rows and 1 column. Type: Column matrix. Order: $3 \times 1$.

iii) $C = \begin{bmatrix} 2 & 5 \\ -1 & 2 \\ 4 & 0 \end{bmatrix}$ has 3 rows and 2 columns; number of rows $\neq$ number of columns. Type: Rectangular matrix. Order: $3 \times 2$.

iv) $D = [1\quad 3\quad -4]$ has 1 row and 3 columns. Type: Row matrix. Order: $1 \times 3$.

v) $E = \begin{bmatrix} 3 & 4 \\ 4 & 6 \end{bmatrix}$ has 2 rows and 2 columns (square). Type: Square matrix. Order: $2 \times 2$.

vi) $P = \begin{bmatrix} 0 & -4 & 3 \\ 1 & 0 & -7 \\ 2 & 2 & 0 \end{bmatrix}$ has 3 rows and 3 columns (square). Type: Square matrix. Order: $3 \times 3$.

vii) $Q = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$ — all elements are zero. Type: Zero matrix. Order: $2 \times 2$.

viii) $R = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}$ — all elements are zero, rows $\neq$ columns. Type: Zero matrix. Order: $2 \times 3$.

ix) $X = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$ — square matrix with all diagonal elements = 1 and all off-diagonal elements = 0. Type: Identity matrix. Order: $3 \times 3$.

x) $Z = \begin{bmatrix} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{bmatrix}$ — square matrix with all diagonal elements equal (= 2) and all off-diagonal elements = 0. Type: Scalar matrix. Order: $3 \times 3$.

The element $a_{ij}$ denotes the element in the $i$th row and $j$th column.

i) $a_{12}$ is the element in row 1, column 2 of matrix $A$.
$$a_{12} = -4$$

ii) $b_{22}$ is the element in row 2, column 2 of matrix $B$: $b_{22} = 0$.
$b_{32}$ is the element in row 3, column 2 of matrix $B$: $b_{32} = 2$.
$$b_{22} + b_{32} = 0 + 2 = \boxed{2}$$

iii) From matrix $C$:

• $c_{21}$ = element in row 2, column 1 $= -1$
• $c_{32}$ = element in row 3, column 2 $= 2$
• $c_{13}$ = element in row 1, column 3 $= -3$

$$c_{21} + c_{32} - c_{13} = -1 + 2 - (-3) = -1 + 2 + 3 = \boxed{4}$$

Given: $x^3 + y^3 = 3axy$

Concept: Implicit differentiation — differentiate both sides with respect to $x$.

Working:

Differentiating both sides w.r.t. $x$:
$$3x^2 + 3y^2\frac{dy}{dx} = 3a\left(y + x\frac{dy}{dx}\right)$$

$$3x^2 + 3y^2\frac{dy}{dx} = 3ay + 3ax\frac{dy}{dx}$$

$$3y^2\frac{dy}{dx} - 3ax\frac{dy}{dx} = 3ay - 3x^2$$

$$\frac{dy}{dx}(3y^2 - 3ax) = 3ay - 3x^2$$

$$\boxed{\frac{dy}{dx} = \frac{ay - x^2}{y^2 - ax}}$$

Given: $e^{xy} - axy = a$

Concept: Implicit differentiation using chain rule and product rule.

Working:

Differentiating both sides w.r.t. $x$:
$$e^{xy}\cdot\frac{d}{dx}(xy) - a\frac{d}{dx}(xy) = 0$$

$$e^{xy}\left(y + x\frac{dy}{dx}\right) - a\left(y + x\frac{dy}{dx}\right) = 0$$

$$(e^{xy} - a)\left(y + x\frac{dy}{dx}\right) = 0$$

Since $e^{xy} - a \neq 0$ in general:
$$y + x\frac{dy}{dx} = 0$$

$$\boxed{\frac{dy}{dx} = -\frac{y}{x}}$$

Given: $\int (x^2+1)(x-2)\,dx$

Step 1 – Expand the integrand:
$$(x^2+1)(x-2)=x^3-2x^2+x-2$$

Step 2 – Integrate term by term using $\int x^n\,dx=\dfrac{x^{n+1}}{n+1}+C$:
$$\int(x^3-2x^2+x-2)\,dx=\frac{x^4}{4}-\frac{2x^3}{3}+\frac{x^2}{2}-2x+C$$

Answer: $\dfrac{x^4}{4}-\dfrac{2x^3}{3}+\dfrac{x^2}{2}-2x+C$

Given: $x \dfrac{dy}{dx} + 2y = x^2$

Concept: The order of a differential equation is the order of the highest-order derivative present. The degree is the power of that highest-order derivative after the equation is made free of radicals and fractions in derivatives.

Working:

• The highest-order derivative present is $\dfrac{dy}{dx}$ (first derivative).
• It appears with power 1.

Answer: Order = 1, Degree = 1

For a valid probability distribution two conditions must hold:
(i) Each $P(X) \geq 0$
(ii) $\sum P(X) = 1$

(a)
$$\sum P(X) = 0.1 + 0.8 + 0.001 + 0.2 = 1.101 \neq 1$$
Since the sum of probabilities is not equal to 1, this is NOT a valid probability distribution.

(b)
$P(X = 4) = -0.2 < 0$
Since one of the probabilities is negative, this is NOT a valid probability distribution.

(c)
$$\sum P(X) = 0.5 + 0.2 + 0.3 = 1.0$$
All probabilities are non-negative and their sum equals 1.
Hence, this IS a valid probability distribution.

Given: A survey about daily mobile use by students uses random selection of twenty students from a school.

Concept: A sampling method is unbiased if every member of the population has an equal chance of being selected. It is biased if certain members are more likely to be selected than others.

Analysis:
In the given statement, students are selected randomly from the school. Random selection ensures that every student has an equal probability of being chosen, so no particular group is favoured or excluded.

Conclusion: The statement represents an Unbiased sampling method, because random sampling gives every student an equal chance of selection, thereby eliminating systematic bias.

i) Correct. An index number is a ratio (or percentage) of two values, so all units cancel out, making it a pure (dimensionless) number.

ii) Correct. Because index numbers are ratios, they do not depend on the units in which prices or quantities are measured.

iii) Incorrect. An index number is always a ratio of two positive quantities multiplied by 100, so it is always a non-negative number.

iv) Correct. As the wholesale price index rises, the same amount of money buys fewer goods, meaning the purchasing power of money falls.

Given:

• Periodic payment, R = ₹ 80
• Interest rate = 4% compounded semi-annually
• Semi-annual interest rate, i = 4%/2 = 2% = 0.02
• Type: Perpetuity (payments continue forever)

Formula for Present Value of a Perpetuity (ordinary):
$$PV = \frac{R}{i}$$

Calculation:
$$PV = \frac{80}{0.02} = ₹\, 4000$$

The present value is ₹ 4,000.

Given:
Let $x$ = quantity (in lbs) of Food I purchased per day, and $y$ = quantity (in lbs) of Food II purchased per day.

Objective Function (Cost to be minimized):
$$Z = 0.60x + 1.00y$$

Constraints:

• Calcium requirement: $10x + 4y \geq 20$
• Protein requirement: $5x + 5y \geq 20$
• Calorie requirement: $2x + 6y \geq 13$
• Non-negativity: $x \geq 0,\ y \geq 0$

LPP Formulation:
$$\text{Minimize } Z = 0.60x + 1.00y$$
$$\text{Subject to:}$$
$$10x + 4y \geq 20$$
$$5x + 5y \geq 20$$
$$2x + 6y \geq 13$$
$$x \geq 0,\ y \geq 0$$