Mathematics (311) | Set A1 • Code 71/SS/311/A1

Ans. (A) \(-1\)
Derivation: Let the intercept on each coordinate axis be \(a\), where \(a > 0\). Using the standard intercept form of a line: \[\frac{x}{a} + \frac{y}{a} = 1 \implies x + y = a \implies y = -x + a\] Comparing with the slope-intercept form \(y = mx + c\), we obtain the slope \(m = -1\).

Ans. (B) \(\left(-\frac{1}{10}, \frac{37}{10}\right)\)
Derivation: The line equation in standard form \(Ax + By + C = 0\) is \(3x - y + 4 = 0\), so \(A = 3, B = -1, C = 4\).
For point \((x_1, y_1) = (2, 3)\), the foot \((h, k)\) satisfies: \[\frac{h - x_1}{A} = \frac{k - y_1}{B} = -\frac{Ax_1 + By_1 + C}{A^2 + B^2}\] \[Ax_1 + By_1 + C = 3(2) - (3) + 4 = 7, \quad A^2 + B^2 = 3^2 + (-1)^2 = 10\] \[\frac{h - 2}{3} = \frac{k - 3}{-1} = -\frac{7}{10} \implies h = 2 - \frac{21}{10} = -\frac{1}{10}, \quad k = 3 + \frac{7}{10} = \frac{37}{10}\]

Derivation: The triangle formed by \(O(0, 0)\), \(A(a, 0)\), and \(B(0, b)\) is a right-angled triangle with \(\angle AOB = 90^\circ\). By Thales' theorem, the hypotenuse segment joining \((a, 0)\) and \((0, b)\) is a diameter of the circumcircle. The centre is the midpoint of the diameter: \[\text{Centre} = \left(\frac{a + 0}{2}, \frac{0 + b}{2}\right) = \left(\frac{a}{2}, \frac{b}{2}\right)\]

Derivation: Divide by 12: \(\frac{x^2}{4} + \frac{y^2}{12} = 1\). Here \(a^2 = 4 \implies a = 2\) and \(b^2 = 12 \implies b = \sqrt{12} = 2\sqrt{3}\).
Since \(b > a\), the major axis is along the \(y\)-axis. The length of latus rectum is: \[\text{L.R.} = \frac{2a^2}{b} = \frac{2(4)}{2\sqrt{3}} = \frac{4}{\sqrt{3}}\]

Derivation: Vertex \((h, k) = (-3, 0)\) and vertical directrix \(x = -5\). Since the directrix is to the left of the vertex, the axis of the parabola is the line \(y = 0\) (\(x\)-axis) and it opens to the right. Distance \(a = |-3 - (-5)| = 2\). \[(y - k)^2 = 4a(x - h) \implies (y - 0)^2 = 4(2)(x - (-3)) \implies y^2 = 8(x + 3)\]

Derivation: Equating corresponding matrix entries: \(y = 4\) (\(a_{11}\)), \(z = 3\) (\(a_{12}\)), and \(x = 1\) (\(a_{21}\)). Thus \(x = 1, y = 4, z = 3\).

Derivation: Reading entries: \(a_{13} = 3\) (row 1, col 3), \(a_{23} = 4\) (row 2, col 3), and \(a_{31} = 2\) (row 3, col 1). \[a_{13} a_{23} + a_{31} = (3)(4) + 2 = 12 + 2 = 14\]

Derivation: The minor \(M_{23}\) is formed by deleting row 2 and column 3: \[M_{23} = \begin{vmatrix} 1 & 0 \\ 0 & 1 \end{vmatrix} = (1)(1) - (0)(0) = 1\] Cofactor \(C_{23} = (-1)^{2+3} M_{23} = -1(1) = -1\).

Derivation:
• One-one check: \(f(1) = 1^4 = 1\) and \(f(-1) = (-1)^4 = 1\). Since \(1 \ne -1\) but \(f(1) = f(-1)\), \(f\) is not one-one.
• Onto check: For any \(x \in \mathbb{R}\), \(x^4 \ge 0\). The range is \([0, \infty) \ne \mathbb{R}\) (codomain). Negative numbers have no pre-image. Thus \(f\) is not onto.

Derivation: \(g(x) = 1 - \frac{1}{1-x} = \frac{1 - x - 1}{1 - x} = \frac{-x}{1 - x}\). \[fog(x) = f(g(x)) = (g(x))^2 + 2 = \left(\frac{-x}{1-x}\right)^2 + 2 = \frac{x^2}{(1-x)^2} + 2\]

Derivation:
• \((1, 1), (2, 2), (3, 3) \in R \implies R\) is reflexive.
• \((1, 2) \in R\) but \((2, 1) \notin R \implies R\) is not symmetric.
• \((1, 2) \in R\) and \((2, 3) \in R\) but \((1, 3) \notin R \implies R\) is not transitive.

Derivation: First, \(2 * 3 = \frac{2 \times 3}{3} = 2\). Next, \((2 * 3) * 5 = 2 * 5 = \frac{2 \times 5}{3} = \frac{10}{3}\).

Derivation: \(\frac{dy}{dx} = \frac{1}{2x+3} \cdot \frac{d}{dx}(2x+3) = \frac{2}{2x+3}\).

Derivation: Using the power chain rule: \(\frac{d}{dx}[(\sin x)^3] = 3\sin^2 x \cdot \frac{d}{dx}(\sin x) = 3\sin^2 x \cos x\).

Derivation: Distance, time, and work done (\(W = \vec{F}\cdot\vec{d}\)) are scalar quantities having only magnitude. Velocity possesses both magnitude and spatial direction; hence it is a vector quantity.

Derivation: Magnitude \(|\vec{a}| = \sqrt{2^2 + 3^2 + 1^2} = \sqrt{4 + 9 + 1} = \sqrt{14}\). \[\hat{a} = \frac{\vec{a}}{|\vec{a}|} = \frac{1}{\sqrt{14}}(2\hat{i} + 3\hat{j} + \hat{k})\]

Derivation: In parallelogram \(OACB\), diagonal \(\vec{OC} = \vec{OA} + \vec{OB} \implies \vec{a} = \vec{OA} + \vec{OB}\) (1).
Vector \(\vec{AB} = \vec{OB} - \vec{OA} \implies \vec{b} = \vec{OB} - \vec{OA}\) (2).
Subtract (2) from (1): \(\vec{a} - \vec{b} = 2\vec{OA} \implies \vec{OA} = \frac{1}{2}(\vec{a} - \vec{b})\).

Derivation: The degree of a differential equation is the power of the highest order derivative when it is expressible as a polynomial in derivatives. Due to the term \(\log\left(\frac{d^2y}{dx^2}\right)\), it cannot be written as a polynomial in derivatives; hence the degree is not defined.

Derivation: Substitute \(t = x^2 + 1 \implies dt = 2x\,dx \implies x\,dx = \frac{dt}{2}\).
When \(x = 0 \to t = 1\); when \(x = 1 \to t = 2\). \[\int_{1}^{2} \frac{dt/2}{t} = \frac{1}{2}[\log t]_1^2 = \frac{1}{2}(\log 2 - \log 1) = \frac{1}{2}\log 2\]

Derivation: For any implication "\(p \implies q\)", its converse is "\(q \implies p\)".
Here \(p\): "a number is divisible by 9" and \(q\): "a number is divisible by 3".
Therefore, the converse is: "If a number is divisible by 3, then it is divisible by 9."

• For (a) \(a_{23}\): Substitute \(i = 2, j = 3 \implies a_{23} = \frac{2 - 3}{2 + 3} = -\frac{1}{5} \implies \mathbf{(Q)}\)
• For (b) \(a_{12}\): Substitute \(i = 1, j = 2 \implies a_{12} = \frac{1 - 2}{1 + 2} = -\frac{1}{3} \implies \mathbf{(S)}\)
Answer: (a) \(\to\) (Q),   (b) \(\to\) (S)

• (a) Normal vector \(\vec{n} = \hat{i} + \hat{j} - \hat{k}\) has magnitude \(|\vec{n}| = \sqrt{1^2 + 1^2 + (-1)^2} = \sqrt{3}\).
  Distance \(D = \frac{|d_1 - d_2|}{|\vec{n}|} = \frac{|4 - 5|}{\sqrt{3}} = \mathbf{\frac{1}{\sqrt{3}}}\text{ units}\).
• (b) Multiplying by \(\frac{dy}{dx}\) gives \(y\frac{dy}{dx} = \left(\frac{dy}{dx}\right)^2 + 1\). The highest order derivative occurring is \(\frac{dy}{dx}\), so the order is \(\mathbf{1}\).

• (a) \(\lim_{x\to 0} f(x) = 1\), but \(f(0) = 2\). Since limit \(\ne\) value at point, \(f\) is discontinuous.  — FALSE
• (b) For any \(c \in \mathbb{R}\), \(\lim_{x\to c} |x| = |c| = f(c)\). Modulus function is continuous everywhere on \(\mathbb{R}\).  — TRUE

• (a) "The capital of Gujarat is not Ahmedabad." (or "It is false that the capital of Gujarat is Ahmedabad.")
• (b) "Not all circles are congruent." (or "There exist circles that are not congruent.")

Writing in matrix form \(AX = B\): \[\begin{bmatrix} 5 & -7 \\ 1 & 3 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 3 \\ 5 \end{bmatrix}\] \(|A| = 5(3) - (-7)(1) = 15 + 7 = 22 \ne 0 \implies A^{-1} \text{ exists}\).
\(\operatorname{adj}(A) = \begin{bmatrix} 3 & 7 \\ -1 & 5 \end{bmatrix} \implies A^{-1} = \frac{1}{22}\begin{bmatrix} 3 & 7 \\ -1 & 5 \end{bmatrix}\).
\[X = A^{-1}B = \frac{1}{22} \begin{bmatrix} 3 & 7 \\ -1 & 5 \end{bmatrix} \begin{bmatrix} 3 \\ 5 \end{bmatrix} = \frac{1}{22} \begin{bmatrix} 9 + 35 \\ -3 + 25 \end{bmatrix} = \frac{1}{22} \begin{bmatrix} 44 \\ 22 \end{bmatrix} = \begin{bmatrix} 2 \\ 1 \end{bmatrix}\] Answer: \(x = 2, \quad y = 1\)

Let \(\theta = \operatorname{cosec}^{-1} x \implies \operatorname{cosec}\theta = x\) for \(|x| \ge 1\).
Using \(\cot^2\theta = \operatorname{cosec}^2\theta - 1 \implies \cot\theta = \pm\sqrt{x^2 - 1}\).
For \(x \ge 1\), \(\cot(\operatorname{cosec}^{-1} x) = \mathbf{\sqrt{x^2 - 1}}\) (in general: \(\frac{\sqrt{x^2 - 1}}{\operatorname{sgn}(x)}\)).

Using \(1 - \cos x = 2\sin^2(x/2)\): \[\lim_{x\to 0} \frac{2\sin^2(x/2)}{x^2} = 2 \lim_{x\to 0} \left(\frac{\sin(x/2)}{x/2}\right)^2 \cdot \frac{1}{4} = 2(1)^2\left(\frac{1}{4}\right) = \mathbf{\frac{1}{2}}\]

Dividing by 36: \(\frac{x^2}{9} + \frac{y^2}{4} = 1 \implies a^2 = 9 \implies a = 3, b^2 = 4 \implies b = 2\) (\(a > b\)).
• Eccentricity: \(e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{4}{9}} = \mathbf{\frac{\sqrt{5}}{3}}\).
• Vertices: \((\pm a, 0) = \mathbf{(\pm 3, 0)}\).
• Foci: \((\pm ae, 0) = \mathbf{(\pm\sqrt{5}, 0)}\).
• Length of Latus Rectum: \(\frac{2b^2}{a} = \frac{2(4)}{3} = \mathbf{\frac{8}{3}}\).

Let \(\Delta = \begin{vmatrix} 0 & a & -b \\ -a & 0 & -c \\ b & c & 0 \end{vmatrix}\). Taking transpose (since \(\det(A) = \det(A^T)\)): \[\Delta = \begin{vmatrix} 0 & -a & b \\ a & 0 & c \\ -b & -c & 0 \end{vmatrix}\] Taking common factor \((-1)\) from each of the three rows: \[\Delta = (-1)^3 \begin{vmatrix} 0 & a & -b \\ -a & 0 & -c \\ b & c & 0 \end{vmatrix} = -\Delta\] \[\Delta = -\Delta \implies 2\Delta = 0 \implies \mathbf{\Delta = 0}. \quad \text{Hence Proved.}\]

Set \(A = IA\): \(\begin{bmatrix} 3 & 5 \\ 4 & 7 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} A\)
1. \(R_2 \to R_2 - R_1\): \(\begin{bmatrix} 3 & 5 \\ 1 & 2 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ -1 & 1 \end{bmatrix} A\)
2. \(R_1 \leftrightarrow R_2\): \(\begin{bmatrix} 1 & 2 \\ 3 & 5 \end{bmatrix} = \begin{bmatrix} -1 & 1 \\ 1 & 0 \end{bmatrix} A\)
3. \(R_2 \to R_2 - 3R_1\): \(\begin{bmatrix} 1 & 2 \\ 0 & -1 \end{bmatrix} = \begin{bmatrix} -1 & 1 \\ 4 & -3 \end{bmatrix} A\)
4. \(R_2 \to -R_2\): \(\begin{bmatrix} 1 & 2 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} -1 & 1 \\ -4 & 3 \end{bmatrix} A\)
5. \(R_1 \to R_1 - 2R_2\): \(\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 7 & -5 \\ -4 & 3 \end{bmatrix} A\)
Answer: \(A^{-1} = \mathbf{\begin{bmatrix} 7 & -5 \\ -4 & 3 \end{bmatrix}}\)

Using \(\int a^x dx = \frac{a^x}{\log a} + C\): \[\int_{0}^{2} 3^x dx = \left[\frac{3^x}{\log 3}\right]_0^2 = \frac{3^2 - 3^0}{\log 3} = \mathbf{\frac{8}{\log 3}}\]

A line \(Ax + By + C = 0\) is parallel to the \(x\)-axis if and only if its slope is \(0\), meaning \(A = 0\) and \(B \ne 0\).
Setting the coefficient of \(x\) to zero: \[k - 3 = 0 \implies \mathbf{k = 3}\] Verification of coefficient of \(y\): \(-(4 - 3^2) = -(4 - 9) = 5 \ne 0\).
The equation simplifies to \(5y - 6 = 0 \implies y = 6/5\), which is parallel to the \(x\)-axis.

The given line has direction ratios \(\langle a, b, c \rangle = \langle 3, -2, 0 \rangle\).
Any parallel line has the same direction ratios \(\langle 3, -2, 0 \rangle\). Passing through \((x_1, y_1, z_1) = (1, -1, 5)\), its Cartesian equation is: \[\mathbf{\frac{x - 1}{3} = \frac{y + 1}{-2}, \quad z = 5} \quad \left(\text{or } \frac{x-1}{3} = \frac{y+1}{-2} = \frac{z-5}{0}\right)\]

Intersection points: \(x^2 = x + 2 \implies x^2 - x - 2 = 0 \implies (x - 2)(x + 1) = 0 \implies x = -1, 2\).
In \([-1, 2]\), line \(y = x + 2\) lies above the parabola \(y = x^2\): \[\text{Area} = \int_{-1}^{2} [(x + 2) - x^2] dx = \left[ \frac{x^2}{2} + 2x - \frac{x^3}{3} \right]_{-1}^{2}\] \[\text{At } x = 2: \quad \frac{4}{2} + 4 - \frac{8}{3} = 6 - \frac{8}{3} = \frac{10}{3}\] \[\text{At } x = -1: \quad \frac{1}{2} - 2 - \left(-\frac{1}{3}\right) = -\frac{3}{2} + \frac{1}{3} = -\frac{7}{6}\] \[\text{Area} = \frac{10}{3} - \left(-\frac{7}{6}\right) = \frac{20 + 7}{6} = \frac{27}{6} = \mathbf{\frac{9}{2} \text{ (or } 4.5)\text{ sq. units}}\]

Let the points be \(A(1, 1, 0)\), \(B(1, 2, 1)\), and \(C(-2, 2, -1)\).
Vectors in the plane: \[\vec{AB} = (1-1)\hat{i} + (2-1)\hat{j} + (1-0)\hat{k} = \hat{j} + \hat{k}\] \[\vec{AC} = (-2-1)\hat{i} + (2-1)\hat{j} + (-1-0)\hat{k} = -3\hat{i} + \hat{j} - \hat{k}\] Normal vector \(\vec{n} = \vec{AB} \times \vec{AC}\): \[\vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 0 & 1 & 1 \\ -3 & 1 & -1 \end{vmatrix} = \hat{i}(-1 - 1) - \hat{j}(0 - (-3)) + \hat{k}(0 - (-3)) = -2\hat{i} - 3\hat{j} + 3\hat{k}\] Multiplying by \(-1\), we take \(\vec{n} = 2\hat{i} + 3\hat{j} - 3\hat{k}\).
The Cartesian equation passing through \((1, 1, 0)\): \[2(x - 1) + 3(y - 1) - 3(z - 0) = 0 \implies 2x + 3y - 3z - 5 = 0 \implies \mathbf{2x + 3y - 3z = 5}\] The Vector equation of the plane is: \[\mathbf{\vec{r} \cdot (2\hat{i} + 3\hat{j} - 3\hat{k}) = 5}\]

Given \(y = (\tan^{-1} x)^2\). Differentiating w.r.t. \(x\): \[y_1 = 2(\tan^{-1} x) \cdot \frac{1}{1 + x^2} \implies (1 + x^2) y_1 = 2\tan^{-1} x\] Differentiating both sides again w.r.t. \(x\) using the product rule: \[(1 + x^2) y_2 + 2x y_1 = 2 \cdot \frac{1}{1 + x^2}\] Multiplying the entire equation by \((1 + x^2)\): \[\mathbf{(x^2 + 1)^2 y_2 + 2x(x^2 + 1)y_1 = 2}. \quad \text{Hence Proved.}\]

Let \(t = x^2 \implies dt = 2x\,dx\). The integral transforms to: \[I = \int \frac{dt}{(t + 1)(t + 3)}\] Using partial fractions decomposition: \[\frac{1}{(t + 1)(t + 3)} = \frac{A}{t + 1} + \frac{B}{t + 3} \implies 1 = A(t + 3) + B(t + 1)\] • For \(t = -1 \implies 1 = A(2) \implies A = 1/2\)
• For \(t = -3 \implies 1 = B(-2) \implies B = -1/2\)
\[I = \frac{1}{2} \int \frac{dt}{t + 1} - \frac{1}{2} \int \frac{dt}{t + 3} = \frac{1}{2} \log|t + 1| - \frac{1}{2}\log|t + 3| + C = \frac{1}{2}\log\left|\frac{t + 1}{t + 3}\right| + C\] Substituting back \(t = x^2\): \[\mathbf{I = \frac{1}{2}\log\left(\frac{x^2 + 1}{x^2 + 3}\right) + C}\]

This is a first-order linear differential equation \(\frac{dy}{dx} + P(x)y = Q(x)\) with \(P(x) = \cot x\) and \(Q(x) = 2\cos x\).
Integrating Factor: \[\text{I.F.} = e^{\int \cot x\,dx} = e^{\log(\sin x)} = \sin x\] General solution: \[y \cdot (\text{I.F.}) = \int Q(x) \cdot (\text{I.F.})\,dx + C\] \[y \sin x = \int 2\cos x \sin x\,dx + C = \int \sin 2x\,dx + C = -\frac{\cos 2x}{2} + C\] \[\mathbf{y\sin x = -\frac{1}{2}\cos 2x + C} \quad \left(\text{or } y = -\frac{\cos 2x}{2\sin x} + C\operatorname{cosec} x\right)\]