1e:I[28602,["/_next/static/immutable/chunks/3je-fy9vz51nm.js","/_next/static/immutable/chunks/0ibk90-bju47u.js","/_next/static/immutable/chunks/3t3krnuroclxx.js","/_next/static/immutable/chunks/1iauvmdwoizmy.js","/_next/static/immutable/chunks/0_xxby9y0y8_o.js","/_next/static/immutable/chunks/2lhk81v3-jfqd.js","/_next/static/immutable/chunks/2nit73v3qsojn.js"],"default"] 4f:I[27201,["/_next/static/immutable/chunks/3je-fy9vz51nm.js","/_next/static/immutable/chunks/0ibk90-bju47u.js","/_next/static/immutable/chunks/3t3krnuroclxx.js","/_next/static/immutable/chunks/1iauvmdwoizmy.js"],"IconMark"] 1f:T87c,To determine which of the given statements is true, we first analyze the function $g(x)$ and the given condition. We are given that $x_0$ is the real root of the equation: $$e^{x_0} + x_0 = 0 \implies e^{x_0} = -x_0$$ The function $g(x)$ is defined as: $$g(x) = \frac{3xe^x + 3x - \alpha e^x - \alpha x}{3(e^x + 1)}$$ We can rewrite the numerator by factoring terms: $$3xe^x + 3x - \alpha e^x - \alpha x = 3x(e^x + 1) - \alpha(e^x + x)$$ Thus, $g(x)$ simplifies to: $$g(x) = \frac{3x(e^x + 1) - \alpha(e^x + x)}{3(e^x + 1)} = x - \frac{\alpha}{3}\left(\frac{e^x + x}{e^x + 1}\right)$$ Evaluating $g(x)$ at $x = x_0$: $$g(x_0) = x_0 - \frac{\alpha}{3}\left(\frac{e^{x_0} + x_0}{e^{x_0} + 1}\right)$$ Since $e^{x_0} + x_0 = 0$, we have: $$g(x_0) = x_0 - \frac{\alpha}{3}(0) = x_0$$ Since $x_0 = -e^{x_0}$, we have $g(x_0) = -e^{x_0}$, which gives: $$g(x_0) + e^{x_0} = 0 \implies e^{x_0} = -g(x_0)$$ Now, the given limit can be recognized as the absolute value of the derivative of $g(x)$ at $x = x_0$: $$\lim_{x \to x_0} \left| \frac{g(x) + e^{x_0}}{x - x_0} \right| = \lim_{x \to x_0} \left| \frac{g(x) - g(x_0)}{x - x_0} \right| = |g'(x_0)|$$ Next, we differentiate $g(x)$ with respect to $x$: $$g'(x) = 1 - \frac{\alpha}{3} \cdot \frac{d}{dx}\left(\frac{e^x + x}{e^x + 1}\right)$$ Using the quotient rule: $$\frac{d}{dx}\left(\frac{e^x + x}{e^x + 1}\right) = \frac{(e^x + 1)\frac{d}{dx}(e^x + x) - (e^x + x)\frac{d}{dx}(e^x + 1)}{(e^x + 1)^2}$$ $$= \frac{(e^x + 1)(e^x + 1) - (e^x + x)e^x}{(e^x + 1)^2}$$ Evaluating this derivative at $x = x_0$ using $e^{x_0} + x_0 = 0$: $$\left. \frac{d}{dx}\left(\frac{e^x + x}{e^x + 1}\right) \right|_{x = x_0} = \frac{(e^{x_0} + 1)^2 - 0 \cdot e^{x_0}}{(e^{x_0} + 1)^2} = 1$$ Therefore: $$g'(x_0) = 1 - \frac{\alpha}{3}(1) = 1 - \frac{\alpha}{3}$$ Hence, the limit is: $$\lim_{x \to x_0} \left| \frac{g(x) + e^{x_0}}{x - x_0} \right| = \left| 1 - \frac{\alpha}{3} \right|$$ Evaluating for the values of $\alpha$ given in the options: 1. For $\alpha = 2$: $$\left| 1 - \frac{2}{3} \right| = \frac{1}{3}$$ 2. For $\alpha = 3$: $$\left| 1 - \frac{3}{3} \right| = 0$$ Therefore, the statement in option (C) is TRUE.20:Ta21,To find the area of the region defined by $$ \mathcal{R} = \left\{ (x, y) \in \mathbb{R} \times \mathbb{R} : x > 0, \, y > \frac{1}{x}, \, 5x - 4y - 1 > 0, \, 4x + 4y - 17 < 0 \right\} $$ we first rewrite the inequalities for $y$ in terms of $x$: 1. $y > \frac{1}{x}$ (lower boundary) 2. $y < \frac{5x - 1}{4}$ (upper boundary for a certain range of $x$) 3. $y < \frac{17 - 4x}{4}$ (upper boundary for another range of $x$) ### Step 1: Finding the Points of Intersection 1. **Intersection of $y = \frac{5x - 1}{4}$ and $y = \frac{17 - 4x}{4}$:** $$ \frac{5x - 1}{4} = \frac{17 - 4x}{4} \implies 5x - 1 = 17 - 4x \implies 9x = 18 \implies x = 2 $$ At $x = 2$, the value of $y$ is: $$ y = \frac{5(2) - 1}{4} = \frac{9}{4} $$ 2. **Intersection of $y = \frac{1}{x}$ and $y = \frac{5x - 1}{4}$:** $$ \frac{1}{x} = \frac{5x - 1}{4} \implies 5x^2 - x - 4 = 0 \implies (5x + 4)(x - 1) = 0 $$ Since $x > 0$, we have $x = 1$. 3. **Intersection of $y = \frac{1}{x}$ and $y = \frac{17 - 4x}{4}$:** $$ \frac{1}{x} = \frac{17 - 4x}{4} \implies 4x^2 - 17x + 4 = 0 \implies (4x - 1)(x - 4) = 0 $$ The region of interest has the boundary at $x = 4$. ### Step 2: Setting up the Integrals The upper boundary of the region is: $$ y_{\text{upper}} = \begin{cases} \frac{5x - 1}{4}, & \text{for } 1 \le x \le 2 \\ \frac{17 - 4x}{4}, & \text{for } 2 \le x \le 4 \end{cases} $$ The lower boundary of the region across $x \in [1, 4]$ is: $$ y_{\text{lower}} = \frac{1}{x} $$ Thus, the total area $A$ is given by: $$ A = \int_{1}^{2} \left( \frac{5x - 1}{4} - \frac{1}{x} \right) dx + \int_{2}^{4} \left( \frac{17 - 4x}{4} - \frac{1}{x} \right) dx $$ This can be written as: $$ A = \int_{1}^{2} \frac{5x - 1}{4} \, dx + \int_{2}^{4} \frac{17 - 4x}{4} \, dx - \int_{1}^{4} \frac{1}{x} \, dx $$ ### Step 3: Evaluating the Integrals 1. **First Integral:** $$ \int_{1}^{2} \frac{5x - 1}{4} \, dx = \left[ \frac{5x^2}{8} - \frac{x}{4} \right]_1^2 = \left( \frac{20}{8} - \frac{2}{4} \right) - \left( \frac{5}{8} - \frac{1}{4} \right) = 2 - \frac{3}{8} = \frac{13}{8} $$ 2. **Second Integral:** $$ \int_{2}^{4} \left( \frac{17}{4} - x \right) dx = \left[ \frac{17x}{4} - \frac{x^2}{2} \right]_2^4 = (17 - 8) - \left( \frac{17}{2} - 2 \right) = 9 - \frac{13}{2} = \frac{5}{2} = \frac{20}{8} $$ 3. **Third Integral:** $$ \int_{1}^{4} \frac{1}{x} \, dx = \left[ \log_e x \right]_1^4 = \log_e 4 $$ ### Step 4: Summing the Results $$ A = \frac{13}{8} + \frac{20}{8} - \log_e 4 = \frac{33}{8} - \log_e 4 $$ Thus, the correct option is **(B)**.21:T8f1,To find the total number of real solutions to the equation $$\theta = \tan^{-1}(2\tan\theta) - \frac{1}{2}\sin^{-1}\left(\frac{6\tan\theta}{9+\tan^2\theta}\right)$$ Let $t = \tan\theta$. Note that the argument of the inverse sine function can be rewritten as: $$\frac{6\tan\theta}{9+\tan^2\theta} = \frac{2\left(\frac{t}{3}\right)}{1 + \left(\frac{t}{3}\right)^2}$$ Using the identity for $\sin^{-1}\left(\frac{2x}{1+x^2}\right)$: 1. **Case 1: $|t| \le 3$ (i.e., $\left|\frac{t}{3}\right| \le 1$)** In this interval, $\sin^{-1}\left(\frac{2(t/3)}{1+(t/3)^2}\right) = 2\tan^{-1}\left(\frac{t}{3}\right)$. Substituting this into the given equation yields: $$\theta = \tan^{-1}(2t) - \tan^{-1}\left(\frac{t}{3}\right)$$ Taking the tangent of both sides: $$\tan\theta = \tan\left(\tan^{-1}(2t) - \tan^{-1}\left(\frac{t}{3}\right)\right)$$ $$t = \frac{2t - \frac{t}{3}}{1 + 2t\left(\frac{t}{3}\right)} = \frac{5t}{3 + 2t^2}$$ Rearranging the terms: $$t(3 + 2t^2) - 5t = 0 \implies t(2t^2 - 2) = 0 \implies 2t(t - 1)(t + 1) = 0$$ This gives the roots $t = 0, 1, -1$, all of which lie within the interval $[-3, 3]$: - For $t = 0$: $\theta = 0$, which gives $\text{LHS} = 0 = \text{RHS}$. - For $t = 1$: $\theta = \frac{\pi}{4}$, which gives $\text{LHS} = \frac{\pi}{4}$ and $\text{RHS} = \tan^{-1}(2) - \tan^{-1}\left(\frac{1}{3}\right) = \tan^{-1}(1) = \frac{\pi}{4}$. - For $t = -1$: $\theta = -\frac{\pi}{4}$, which gives $\text{LHS} = -\frac{\pi}{4} = \text{RHS}$. Any other values of $\theta$ satisfying $\tan\theta = 0, \pm 1$ (where $\theta = k\pi$ or $\theta = k\pi \pm \frac{\pi}{4}$ for $k \in \mathbb{Z} \setminus \{0\}$) do not satisfy the equation because the right-hand side is bounded strictly within $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$. 2. **Case 2: $|t| > 3$** - For $t > 3$: $\frac{1}{2}\sin^{-1}\left(\frac{6t}{9+t^2}\right) = \frac{\pi}{2} - \tan^{-1}\left(\frac{t}{3}\right)$, leading to $\text{RHS} = \tan^{-1}(2t) + \tan^{-1}\left(\frac{t}{3}\right) - \frac{\pi}{2} < 0$, while $\theta > \tan^{-1}(3) > 0$ for principal values, yielding no solutions. - For $t < -3$: by symmetry, no solutions exist. Therefore, the only real solutions are $\theta = -\frac{\pi}{4}, 0, \frac{\pi}{4}$. The total number of real solutions is $3$. **Correct Option: (C)**22:T617,To find the locus $S$, we consider the given system of lines for $\alpha \neq 0$: $$4x - 3y = 12\alpha \quad \text{--- (1)}$$ $$4\alpha x + 3\alpha y = 12 \implies 4x + 3y = \frac{12}{\alpha} \quad \text{--- (2)}$$ Multiplying equation (1) and equation (2), we eliminate the parameter $\alpha$: $$(4x - 3y)(4x + 3y) = (12\alpha) \left(\frac{12}{\alpha}\right)$$ $$16x^2 - 9y^2 = 144$$ Dividing both sides by $144$, we obtain the standard equation of a hyperbola: $$\frac{x^2}{9} - \frac{y^2}{16} = 1$$ Here, $a^2 = 9$ and $b^2 = 16$. The tangent line $T$ is parallel to the line $4x - \frac{3}{\sqrt{2}}y = 0$, so the slope $m$ of the tangent is: $$m = \frac{4}{\frac{3}{\sqrt{2}}} = \frac{4\sqrt{2}}{3}$$ The equation of a tangent to the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ in slope form is: $$y = mx \pm \sqrt{a^2m^2 - b^2}$$ Substituting $a^2 = 9$, $b^2 = 16$, and $m = \frac{4\sqrt{2}}{3}$: $$a^2m^2 - b^2 = 9\left(\frac{4\sqrt{2}}{3}\right)^2 - 16 = 9\left(\frac{32}{9}\right) - 16 = 32 - 16 = 16$$ Thus, the tangent lines are: $$y = \frac{4\sqrt{2}}{3}x \pm 4$$ Since the tangent line $T$ passes through $(0, q)$ with $q > 0$, the $y$-intercept must be positive: $$y = \frac{4\sqrt{2}}{3}x + 4$$ From this equation: 1. The $y$-intercept $(0, q)$ gives: $$q = 4$$ 2. The $x$-intercept $(p, 0)$ gives: $$0 = \frac{4\sqrt{2}}{3}p + 4 \implies p = -\frac{3}{\sqrt{2}}$$ Therefore, the value of the product $pq$ is: $$pq = \left(-\frac{3}{\sqrt{2}}\right)(4) = -\frac{12}{\sqrt{2}} = -6\sqrt{2}$$ Hence, the correct option is **(A)**.23:T995,To determine which of the statements are true, we use the given matrix equation: $$QR = RP$$ Let the matrix $R$ be represented as: $$R = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$$ where $a, b, c, d$ are non-zero real numbers. Given matrices are: $$P = \begin{pmatrix} 2 & 0 \\ 0 & 3 \end{pmatrix}, \quad Q = \begin{pmatrix} x & y \\ z & 4 \end{pmatrix}$$ Compute $QR$: $$QR = \begin{pmatrix} x & y \\ z & 4 \end{pmatrix} \begin{pmatrix} a & b \\ c & d \end{pmatrix} = \begin{pmatrix} ax + cy & bx + dy \\ az + 4c & bz + 4d \end{pmatrix}$$ Compute $RP$: $$RP = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \begin{pmatrix} 2 & 0 \\ 0 & 3 \end{pmatrix} = \begin{pmatrix} 2a & 3b \\ 2c & 3d \end{pmatrix}$$ Equating corresponding entries of $QR$ and $RP$: 1. First column equations: $$ax + cy = 2a \implies (x - 2)a + cy = 0$$ $$az + 4c = 2c \implies az + 2c = 0 \implies \frac{c}{a} = -\frac{z}{2}$$ Substituting $\frac{c}{a} = -\frac{z}{2}$ into the first equation: $$(x - 2) + \left(-\frac{z}{2}\right)y = 0 \implies x - 2 = \frac{yz}{2} \implies yz = 2x - 4 \quad \text{--- (1)}$$ 2. Second column equations: $$bx + dy = 3b \implies (x - 3)b + dy = 0$$ $$bz + 4d = 3d \implies bz + d = 0 \implies \frac{d}{b} = -z$$ Substituting $\frac{d}{b} = -z$ into the first equation: $$(x - 3) + (-z)y = 0 \implies x - 3 = yz \quad \text{--- (2)}$$ Equating the two expressions for $yz$ from equations (1) and (2): $$2x - 4 = x - 3 \implies x = 1$$ Substituting $x = 1$ back to find $yz$: $$yz = 1 - 3 = -2$$ Thus, the matrix $Q$ is: $$Q = \begin{pmatrix} 1 & y \\ z & 4 \end{pmatrix} \quad \text{with } yz = -2$$ Now, we evaluate each of the given options: - **Option (A):** $$Q - 2I = \begin{pmatrix} 1 - 2 & y \\ z & 4 - 2 \end{pmatrix} = \begin{pmatrix} -1 & y \\ z & 2 \end{pmatrix}$$ $$\det(Q - 2I) = (-1)(2) - yz = -2 - (-2) = 0$$ Hence, statement (A) is **TRUE**. - **Option (B):** $$Q - 6I = \begin{pmatrix} 1 - 6 & y \\ z & 4 - 6 \end{pmatrix} = \begin{pmatrix} -5 & y \\ z & -2 \end{pmatrix}$$ $$\det(Q - 6I) = (-5)(-2) - yz = 10 - (-2) = 12$$ Hence, statement (B) is **TRUE**. - **Option (C):** $$Q - 3I = \begin{pmatrix} 1 - 3 & y \\ z & 4 - 3 \end{pmatrix} = \begin{pmatrix} -2 & y \\ z & 1 \end{pmatrix}$$ $$\det(Q - 3I) = (-2)(1) - yz = -2 - (-2) = 0 \neq 15$$ Hence, statement (C) is **FALSE**. - **Option (D):** $$yz = -2 \neq 2$$ Hence, statement (D) is **FALSE**. Therefore, the correct statements are **(A)** and **(B)**.24:T98a,To determine the correct statements, we first derive the equation of the locus $S$ of the mid-points of the chords. Let the endpoints of a chord of the parabola $y^2 = x$ be $P(y_1^2, y_1)$ and $Q(y_2^2, y_2)$. The area of the region enclosed between the parabola $y^2 = x$ and the chord $PQ$ is given by the standard formula: $$\text{Area} = \frac{|y_1 - y_2|^3}{6}$$ Given that this area is $\frac{4}{3}$, we have: $$\frac{|y_1 - y_2|^3}{6} = \frac{4}{3} \implies |y_1 - y_2|^3 = 8 \implies |y_1 - y_2| = 2$$ Let $(h, k)$ be the mid-point of the chord $PQ$. Then: $$k = \frac{y_1 + y_2}{2} \implies y_1 + y_2 = 2k$$ $$h = \frac{x_1 + x_2}{2} = \frac{y_1^2 + y_2^2}{2}$$ Using the algebraic identity $(y_1 - y_2)^2 = (y_1 + y_2)^2 - 4y_1 y_2$, we substitute $|y_1 - y_2| = 2$ and $y_1 + y_2 = 2k$: $$2^2 = (2k)^2 - 4y_1 y_2 \implies 4 = 4k^2 - 4y_1 y_2 \implies y_1 y_2 = k^2 - 1$$ Now, expressing $h$ in terms of $k$: $$h = \frac{(y_1 + y_2)^2 - 2y_1 y_2}{2} = \frac{4k^2 - 2(k^2 - 1)}{2} = k^2 + 1$$ Replacing $(h, k)$ with $(x, y)$, the equation of the locus $S$ is: $$x = y^2 + 1 \implies y^2 = x - 1$$ --- ### Step 1: Verification of Options (A) and (B) - For Option (A): Point $(4, \sqrt{3})$ $$(\sqrt{3})^2 = 3 \quad \text{and} \quad 4 - 1 = 3$$ Since $3 = 3$, $(4, \sqrt{3}) \in S$. Thus, **Option (A) is TRUE**. - For Option (B): Point $(5, \sqrt{2})$ $$(\sqrt{2})^2 = 2 \quad \text{and} \quad 5 - 1 = 4 \neq 2$$ Thus, **Option (B) is FALSE**. --- ### Step 2: Calculation of the Area of Region $\mathcal{R}$ The region $\mathcal{R}$ lies in the first quadrant and is bounded above by $y = \sqrt{x}$ (from $y^2 = x$), below by $y = \sqrt{x-1}$ (from the locus $S$), and by the vertical lines $x = 1$ and $x = 4$. The area of $\mathcal{R}$ is given by: $$\text{Area}(\mathcal{R}) = \int_{1}^{4} \left( \sqrt{x} - \sqrt{x-1} \right) dx$$ Evaluating the integral step-by-step: $$\int_{1}^{4} \sqrt{x} \, dx = \left[ \frac{2}{3} x^{3/2} \right]_{1}^{4} = \frac{2}{3} \left( 4^{3/2} - 1^{3/2} \right) = \frac{2}{3} (8 - 1) = \frac{14}{3}$$ $$\int_{1}^{4} \sqrt{x-1} \, dx = \left[ \frac{2}{3} (x-1)^{3/2} \right]_{1}^{4} = \frac{2}{3} \left( 3^{3/2} - 0 \right) = \frac{2}{3} (3\sqrt{3}) = 2\sqrt{3}$$ Subtracting the two integrals: $$\text{Area}(\mathcal{R}) = \frac{14}{3} - 2\sqrt{3}$$ Thus, **Option (C) is TRUE** and **Option (D) is FALSE**. --- ### Conclusion The correct statements are **A** and **C**.25:Tb3a,To determine which statements are true, we start by finding the coordinates of the points on the circle $C$ and the ellipse. ### **1. Coordinates of Points $R$ and $S$ on Circle $C$** The circle equation is given by: $$C: x^2 + y^2 = 9$$ This circle has radius $r = 3$ and centered at the origin $O(0, 0)$. The point $M$ is $(3, 0)$, lying on the positive $x$-axis. Since $R$ lies on the circle $C$ with $y_R > 0$ and $\angle ROM = \frac{\pi}{6}$: $$R = \left(3 \cos \frac{\pi}{6}, 3 \sin \frac{\pi}{6}\right) = \left(3 \cdot \frac{\sqrt{3}}{2}, 3 \cdot \frac{1}{2}\right) = \left(\frac{3\sqrt{3}}{2}, \frac{3}{2}\right)$$ Since $S$ lies on the circle $C$ with $y_S > 0$ and $\angle SOM = \frac{\pi}{3}$: $$S = \left(3 \cos \frac{\pi}{3}, 3 \sin \frac{\pi}{3}\right) = \left(3 \cdot \frac{1}{2}, 3 \cdot \frac{\sqrt{3}}{2}\right) = \left(\frac{3}{2}, \frac{3\sqrt{3}}{2}\right)$$ --- ### **2. Coordinates of Points $P$ and $Q$ on the Ellipse** The ellipse equation is: $$\frac{x^2}{9} + \frac{y^2}{4} = 1$$ The vertical line $x = x_1 = \frac{3\sqrt{3}}{2}$ intersects the ellipse at $P(x_1, y_1)$ with $y_1 > 0$: $$\frac{\left(\frac{3\sqrt{3}}{2}\right)^2}{9} + \frac{y_1^2}{4} = 1 \implies \frac{27/4}{9} + \frac{y_1^2}{4} = 1 \implies \frac{3}{4} + \frac{y_1^2}{4} = 1 \implies y_1 = 1$$ Thus, $P = \left(\frac{3\sqrt{3}}{2}, 1\right)$. The vertical line $x = x_2 = \frac{3}{2}$ intersects the ellipse at $Q(x_2, y_2)$ with $y_2 > 0$: $$\frac{\left(\frac{3}{2}\right)^2}{9} + \frac{y_2^2}{4} = 1 \implies \frac{9/4}{9} + \frac{y_2^2}{4} = 1 \implies \frac{1}{4} + \frac{y_2^2}{4} = 1 \implies y_2 = \sqrt{3}$$ Thus, $Q = \left(\frac{3}{2}, \sqrt{3}\right)$. --- ### **3. Equation of the Line Joining $P$ and $Q$** The slope $m$ of the line segment $PQ$ is: $$m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{\sqrt{3} - 1}{\frac{3}{2} - \frac{3\sqrt{3}}{2}} = \frac{2(\sqrt{3} - 1)}{3(1 - \sqrt{3})} = -\frac{2}{3}$$ Using point-slope form with point $P\left(\frac{3\sqrt{3}}{2}, 1\right)$: $$y - 1 = -\frac{2}{3} \left(x - \frac{3\sqrt{3}}{2}\right)$$ $$3(y - 1) = -2x + 3\sqrt{3}$$ $$2x + 3y = 3 + 3\sqrt{3} = 3(1 + \sqrt{3})$$ This matches **Option A** (and refutes **Option B**). --- ### **4. Checking Options C and D** For point $N_2 = (x_2, 0) = \left(\frac{3}{2}, 0\right)$: - $|N_2 Q| = y_2 = \sqrt{3}$ - $|N_2 S| = y_S = \frac{3\sqrt{3}}{2}$ Evaluating the ratio for **Option C**: $$3 |N_2 Q| = 3\sqrt{3}$$ $$2 |N_2 S| = 2 \left(\frac{3\sqrt{3}}{2}\right) = 3\sqrt{3}$$ Since $3 |N_2 Q| = 2 |N_2 S|$, **Option C** is TRUE. For point $N_1 = (x_1, 0) = \left(\frac{3\sqrt{3}}{2}, 0\right)$: - $|N_1 P| = y_1 = 1$ - $|N_1 R| = y_R = \frac{3}{2}$ Evaluating the ratio for **Option D**: $$9 |N_1 P| = 9(1) = 9$$ $$4 |N_1 R| = 4\left(\frac{3}{2}\right) = 6 \neq 9$$ Thus, **Option D** is FALSE. --- ### **Conclusion** The correct statements are **A** and **C**.26:T606,To analyze the function $$ f(x) = \begin{cases} \frac{6x + \sin x}{2x + \sin x} & \text{if } x \neq 0, \\[6pt] \frac{7}{3} & \text{if } x = 0 \end{cases} $$ we evaluate its behavior at $x = 0$ and its critical points. First, using Taylor expansion near $x = 0$, we have $\sin x \approx x - \frac{x^3}{6}$, which gives: $$f(x) = \frac{7x - \frac{x^3}{6}}{3x - \frac{x^3}{6}} = \frac{7 - \frac{x^2}{6}}{3 - \frac{x^2}{6}} \approx \frac{7}{3}\left(1 + \frac{2x^2}{63}\right) > \frac{7}{3} = f(0)$$ Since $f(x) > f(0)$ in a neighborhood around $x = 0$, $x = 0$ is a point of local minima, making statement (B) TRUE and statement (A) FALSE. For $x \neq 0$, the derivative of $f(x)$ is given by: $$f'(x) = \frac{4(\sin x - x \cos x)}{(2x + \sin x)^2}$$ The sign of $f'(x)$ is determined by $g(x) = \sin x - x \cos x$, with $g'(x) = x \sin x$. - In intervals $((2k-1)\pi, 2k\pi)$ where $\sin x < 0$, $g'(x) < 0$, so $g(x)$ decreases from $(2k-1)\pi > 0$ to $-2k\pi < 0$, changing sign from positive to negative. This yields points of local maxima in $(\pi, 2\pi)$, $(3\pi, 4\pi)$, and $(5\pi, 6\pi)$. Thus, the number of local maxima in $[\pi, 6\pi]$ is $3$, so statement (C) is TRUE. - In intervals $(2k\pi, (2k+1)\pi)$ where $\sin x > 0$, $g'(x) > 0$, so $g(x)$ increases from $-2k\pi < 0$ to $(2k+1)\pi > 0$, changing sign from negative to positive. This yields a point of local minima in $(2\pi, 3\pi)$. Thus, the number of local minima in $[2\pi, 4\pi]$ is $1$, so statement (D) is TRUE. Hence, the correct options are (B), (C), and (D).27:T718,To solve the given differential equation $$x^2 \frac{dy}{dx} + xy = x^2 + y^2, \quad x > \frac{1}{e}$$ we can rewrite it by dividing the entire equation by $x^2$: $$\frac{dy}{dx} + \frac{y}{x} = 1 + \left(\frac{y}{x}\right)^2$$ This is a homogeneous first-order differential equation. Let $y = vx$, where $v$ is a function of $x$. Differentiating with respect to $x$, we get: $$\frac{dy}{dx} = v + x \frac{dv}{dx}$$ Substituting $y = vx$ and $\frac{dy}{dx}$ into the differential equation gives: $$v + x \frac{dv}{dx} + v = 1 + v^2$$ $$x \frac{dv}{dx} = v^2 - 2v + 1$$ $$x \frac{dv}{dx} = (v - 1)^2$$ Separating the variables, we have: $$\frac{dv}{(v - 1)^2} = \frac{dx}{x}$$ Integrating both sides: $$\int \frac{dv}{(v - 1)^2} = \int \frac{dx}{x}$$ $$-\frac{1}{v - 1} = \ln x + C$$ Since $v = \frac{y}{x}$, this becomes: $$-\frac{1}{\frac{y}{x} - 1} = \ln x + C$$ $$\frac{x}{x - y} = \ln x + C$$ We are given the initial condition $y(1) = 0$. Substituting $x = 1$ and $y = 0$: $$\frac{1}{1 - 0} = \ln 1 + C \implies C = 1$$ Thus, the particular solution is: $$\frac{x}{x - y} = 1 + \ln x$$ Rearranging for $x - y$: $$x - y = \frac{x}{1 + \ln x}$$ $$y(x) = x - \frac{x}{1 + \ln x} = \frac{x \ln x}{1 + \ln x}$$ Now, we evaluate $y(e)$ and $y(e^2)$: 1. For $x = e$: $$y(e) = \frac{e \ln e}{1 + \ln e} = \frac{e \cdot 1}{1 + 1} = \frac{e}{2}$$ 2. For $x = e^2$: $$y(e^2) = \frac{e^2 \ln(e^2)}{1 + \ln(e^2)} = \frac{e^2 \cdot 2}{1 + 2} = \frac{2e^2}{3}$$ Finally, we evaluate the required expression $2 \frac{(y(e))^2}{y(e^2)}$: $$2 \frac{(y(e))^2}{y(e^2)} = 2 \cdot \frac{\left(\frac{e}{2}\right)^2}{\frac{2e^2}{3}} = 2 \cdot \frac{\frac{e^2}{4}}{\frac{2e^2}{3}} = 2 \cdot \left(\frac{e^2}{4} \times \frac{3}{2e^2}\right) = 2 \cdot \frac{3}{8} = \frac{3}{4} = 0.75$$ The value of the expression is $0.75$.28:T662,To find the value of $r$ for which $a_r$ is the largest among the coefficients $a_0, a_1, \dots, a_{23}$, we start by writing the general term of the binomial expansion of $\left(1 + \frac{2}{5}x\right)^{23}$. By the Binomial Theorem: $$\left(1 + \frac{2}{5}x\right)^{23} = \sum_{i=0}^{23} \binom{23}{i} \left(\frac{2}{5}\right)^i x^i$$ Comparing this with the given expansion $\sum_{i=0}^{23} a_i x^i$, the coefficient $a_i$ is given by: $$a_i = \binom{23}{i} \left(\frac{2}{5}\right)^i$$ To determine the term with the maximum value, we examine the ratio of consecutive coefficients $\frac{a_k}{a_{k-1}}$: $$\frac{a_k}{a_{k-1}} = \frac{\binom{23}{k} \left(\frac{2}{5}\right)^k}{\binom{23}{k-1} \left(\frac{2}{5}\right)^{k-1}} = \frac{\frac{23!}{k!(23-k)!}}{\frac{23!}{(k-1)!(24-k)!}} \cdot \frac{2}{5} = \frac{24-k}{k} \cdot \frac{2}{5}$$ We set $\frac{a_k}{a_{k-1}} \ge 1$ to find where the sequence of coefficients is non-decreasing: $$\frac{24-k}{k} \cdot \frac{2}{5} \ge 1$$ $$2(24 - k) \ge 5k$$ $$48 - 2k \ge 5k$$ $$7k \le 48$$ $$k \le \frac{48}{7} \approx 6.857$$ Since $k$ must be an integer, $a_k > a_{k-1}$ for $k = 1, 2, 3, 4, 5, 6$. Specifically, for $k = 6$: $$\frac{a_6}{a_5} = \frac{24 - 6}{6} \cdot \frac{2}{5} = \frac{18}{6} \cdot \frac{2}{5} = \frac{6}{5} > 1 \implies a_6 > a_5$$ For $k = 7$: $$\frac{a_7}{a_6} = \frac{24 - 7}{7} \cdot \frac{2}{5} = \frac{17}{7} \cdot \frac{2}{5} = \frac{34}{35} < 1 \implies a_7 < a_6$$ Therefore, the terms follow the order: $$a_0 < a_1 < a_2 < a_3 < a_4 < a_5 < a_6 > a_7 > a_8 > \dots > a_{23}$$ Thus, $a_6$ is the largest coefficient, and the value of $r$ is $6$.29:T861,Let $M_1, M_2$, and $M_3$ denote the events that a randomly chosen bulb was manufactured by unit $M_1, M_2$, and $M_3$, respectively. Let $D$ denote the event that a randomly chosen bulb is defective. From the given proportions of production $2 : 2 : 1$, the probabilities of selecting a bulb from each unit are: $$P(M_1) = \frac{2}{2 + 2 + 1} = \frac{2}{5}$$ $$P(M_2) = \frac{2}{2 + 2 + 1} = \frac{2}{5}$$ $$P(M_3) = \frac{1}{2 + 2 + 1} = \frac{1}{5}$$ We are given the overall probability of a defective bulb in the factory: $$P(D) = 20\% = \frac{1}{5} = 0.20$$ We are also given the conditional probability of a defective bulb given it is from $M_1$: $$P(D \mid M_1) = 15\% = \frac{15}{100} = \frac{3}{20}$$ It is given that if a bulb chosen at random is defective, the probability that it was produced by $M_2$ is $\frac{2}{5}$: $$P(M_2 \mid D) = \frac{2}{5}$$ Using Bayes' Theorem: $$P(M_2 \mid D) = \frac{P(D \mid M_2) \cdot P(M_2)}{P(D)}$$ Substitute the known values into the equation: $$\frac{2}{5} = \frac{P(D \mid M_2) \cdot \frac{2}{5}}{\frac{1}{5}}$$ Simplifying the right-hand side: $$\frac{2}{5} = 2 \cdot P(D \mid M_2) \implies P(D \mid M_2) = \frac{1}{5} = 0.20$$ Now, applying the Law of Total Probability for $P(D)$: $$P(D) = P(D \mid M_1) \cdot P(M_1) + P(D \mid M_2) \cdot P(M_2) + P(D \mid M_3) \cdot P(M_3)$$ Substituting the known probabilities: $$\frac{1}{5} = \left(\frac{3}{20}\right) \cdot \left(\frac{2}{5}\right) + \left(\frac{1}{5}\right) \cdot \left(\frac{2}{5}\right) + P(D \mid M_3) \cdot \left(\frac{1}{5}\right)$$ $$\frac{1}{5} = \frac{6}{100} + \frac{2}{25} + \frac{1}{5} P(D \mid M_3)$$ Convert all fractions on the right side to a common denominator of $100$: $$\frac{20}{100} = \frac{6}{100} + \frac{8}{100} + \frac{1}{5} P(D \mid M_3)$$ $$\frac{20}{100} = \frac{14}{100} + \frac{1}{5} P(D \mid M_3)$$ Subtracting $\frac{14}{100}$ from both sides: $$\frac{6}{100} = \frac{1}{5} P(D \mid M_3)$$ Multiply both sides by $5$: $$P(D \mid M_3) = \frac{30}{100} = 0.3$$ Thus, if a bulb is chosen randomly from the bulbs produced by $M_3$, the probability that it is defective is $0.3$ (or $0.30$).2a:Tb76,To find the value of $\alpha + \beta - 3$, we analyze the condition for the vectors $\vec{X}, \vec{Y},$ and $\vec{Z}$ to be coplanar. ### Step 1: Check the linear independence of $\vec{x}, \vec{y},$ and $\vec{z}$ The scalar triple product of the vectors $\vec{x}, \vec{y},$ and $\vec{z}$ is given by the determinant of their components: $$[\vec{x}, \vec{y}, \vec{z}] = \begin{vmatrix} 1 & 2 & 3 \\ 2 & 3 & 1 \\ 3 & 1 & 2 \end{vmatrix}$$ Evaluating this determinant: $$[\vec{x}, \vec{y}, \vec{z}] = 1(6 - 1) - 2(4 - 3) + 3(2 - 9) = 5 - 2 - 21 = -18$$ Since $[\vec{x}, \vec{y}, \vec{z}] \neq 0$, the vectors $\vec{x}, \vec{y},$ and $\vec{z}$ are non-coplanar and form a basis in three-dimensional space. --- ### Step 2: Set up the coplanarity condition for $\vec{X}, \vec{Y},$ and $\vec{Z}$ The vectors $\vec{X}, \vec{Y},$ and $\vec{Z}$ are expressed as linear combinations of $\vec{x}, \vec{y},$ and $\vec{z}$: $$\vec{X} = \alpha \vec{x} + \beta \vec{y} - \vec{z}$$ $$\vec{Y} = -\vec{x} + \alpha \vec{y} + \beta \vec{z}$$ $$\vec{Z} = \beta \vec{x} - \vec{y} + \alpha \vec{z}$$ The scalar triple product $[\vec{X}, \vec{Y}, \vec{Z}]$ is related to $[\vec{x}, \vec{y}, \vec{z}]$ by: $$[\vec{X}, \vec{Y}, \vec{Z}] = \begin{vmatrix} \alpha & \beta & -1 \\ -1 & \alpha & \beta \\ \beta & -1 & \alpha \end{vmatrix} [\vec{x}, \vec{y}, \vec{z}]$$ For $\vec{X}, \vec{Y},$ and $\vec{Z}$ to lie in a plane (i.e., to be coplanar), their scalar triple product must be zero: $$[\vec{X}, \vec{Y}, \vec{Z}] = 0$$ Since $[\vec{x}, \vec{y}, \vec{z}] = -18 \neq 0$, we must have: $$\begin{vmatrix} \alpha & \beta & -1 \\ -1 & \alpha & \beta \\ \beta & -1 & \alpha \end{vmatrix} = 0$$ --- ### Step 3: Evaluate the determinant Expanding the determinant along the first row: $$\alpha(\alpha^2 + \beta) - \beta(-\alpha - \beta^2) - 1(1 - \alpha\beta) = 0$$ $$\alpha^3 + \alpha\beta + \alpha\beta + \beta^3 - 1 + \alpha\beta = 0$$ $$\alpha^3 + \beta^3 - 1 + 3\alpha\beta = 0$$ Using the algebraic identity $a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca)$ with $a = \alpha, b = \beta,$ and $c = -1$: $$\alpha^3 + \beta^3 + (-1)^3 - 3\alpha\beta(-1) = (\alpha + \beta - 1)\left(\alpha^2 + \beta^2 + 1 - \alpha\beta + \alpha + \beta\right) = 0$$ Rewriting the second factor: $$\alpha^2 + \beta^2 + 1 - \alpha\beta + \alpha + \beta = \frac{1}{2} \left[ (\alpha - \beta)^2 + (\alpha + 1)^2 + (\beta + 1)^2 \right]$$ Since $\alpha$ and $\beta$ are distinct positive real numbers ($\alpha, \beta > 0$ and $\alpha \neq \beta$), the terms $(\alpha - \beta)^2$, $(\alpha + 1)^2$, and $(\beta + 1)^2$ are strictly positive. Therefore: $$\alpha^2 + \beta^2 + 1 - \alpha\beta + \alpha + \beta > 0$$ Thus, the only solution to the equation is: $$\alpha + \beta - 1 = 0 \implies \alpha + \beta = 1$$ --- ### Step 4: Calculate the required expression We need to find the value of $\alpha + \beta - 3$: $$\alpha + \beta - 3 = 1 - 3 = -2$$2b:T670,To find the value of $\frac{3\alpha}{\pi}$, we begin by determining the complex number $\omega$. Since $\omega$ is a non-real cube root of unity with argument $0 < \text{arg}(\omega) < \pi$, we have: $$\omega = e^{i\frac{2\pi}{3}} = -\frac{1}{2} + i\frac{\sqrt{3}}{2}$$ Let $r = -\omega$. We can write $r$ in polar form as: $$r = -\omega = -e^{i\frac{2\pi}{3}} = e^{i\left(\frac{2\pi}{3} - \pi\right)} = e^{-i\frac{\pi}{3}} = \frac{1}{2} - i\frac{\sqrt{3}}{2}$$ Notice that $r^6 = \left(e^{-i\frac{\pi}{3}}\right)^6 = e^{-i2\pi} = 1$. Therefore, the terms $r^n = (-\omega)^n$ are periodic with period $6$, and the sum of any $6$ consecutive powers of $r$ is zero: $$\sum_{k=1}^{6} r^k = r + r^2 + r^3 + r^4 + r^5 + r^6 = r \left(\frac{1 - r^6}{1 - r}\right) = 0$$ Since $2025 = 6 \times 337 + 3$, the given sum simplifies to the sum of the first $3$ terms: $$S = \sum_{n=1}^{2025} (-\omega)^n = \sum_{n=1}^{3} r^n = r + r^2 + r^3$$ Now, calculating each term: 1. $r = -\omega = \frac{1}{2} - i\frac{\sqrt{3}}{2}$ 2. $r^2 = (-\omega)^2 = \omega^2 = e^{i\frac{4\pi}{3}} = -\frac{1}{2} - i\frac{\sqrt{3}}{2}$ 3. $r^3 = (-\omega)^3 = -\omega^3 = -1$ Summing these up: $$S = \left(\frac{1}{2} - i\frac{\sqrt{3}}{2}\right) + \left(-\frac{1}{2} - i\frac{\sqrt{3}}{2}\right) + (-1) = -1 - i\sqrt{3}$$ The principal argument $\alpha = \text{arg}(S)$ of $S = -1 - i\sqrt{3}$ lies in the third quadrant ($-\pi < \text{arg}(z) \le \pi$): $$\alpha = \text{arg}\left(2 e^{-i\frac{2\pi}{3}}\right) = -\frac{2\pi}{3}$$ Finally, we substitute $\alpha$ into the required expression: $$\frac{3\alpha}{\pi} = \frac{3}{\pi} \left( -\frac{2\pi}{3} \right) = -2$$2c:T603,To find the derivative of the composite function $(f \circ g^{-1})(x) = f(g^{-1}(x))$ at $x = 2$, we use the chain rule: $$\left(f \circ g^{-1}\right)'(x) = f'\left(g^{-1}(x)\right) \cdot \left(g^{-1}\right)'(x)$$ Using the formula for the derivative of an inverse function, $\left(g^{-1}\right)'(x) = \frac{1}{g'\left(g^{-1}(x)\right)}$, we get: $$\left(f \circ g^{-1}\right)'(2) = f'\left(g^{-1}(2)\right) \cdot \frac{1}{g'\left(g^{-1}(2)\right)}$$ **Step 1: Find $g^{-1}(2)$** Let $y = g^{-1}(2)$, which means $g(y) = 2$. Given $g(x) = \frac{4}{1 + e^{-2x}}$: $$\frac{4}{1 + e^{-2y}} = 2 \implies 1 + e^{-2y} = 2 \implies e^{-2y} = 1 \implies y = 0$$ Thus, $g^{-1}(2) = 0$. **Step 2: Calculate $f'(0)$** Given $f(x) = \log_e(x^2 + 2x + 4)$, its derivative with respect to $x$ is: $$f'(x) = \frac{2x + 2}{x^2 + 2x + 4}$$ Evaluating $f'(x)$ at $x = 0$: $$f'(0) = \frac{2(0) + 2}{0^2 + 2(0) + 4} = \frac{2}{4} = \frac{1}{2}$$ **Step 3: Calculate $g'(0)$** Given $g(x) = \frac{4}{1 + e^{-2x}}$, its derivative with respect to $x$ is: $$g'(x) = 4 \cdot (-1)(1 + e^{-2x})^{-2} \cdot (-2e^{-2x}) = \frac{8e^{-2x}}{(1 + e^{-2x})^2}$$ Evaluating $g'(x)$ at $x = 0$: $$g'(0) = \frac{8e^0}{(1 + e^0)^2} = \frac{8}{(1 + 1)^2} = \frac{8}{4} = 2$$ **Step 4: Compute the final value** Substituting $g^{-1}(2) = 0$, $f'(0) = \frac{1}{2}$, and $g'(0) = 2$ into our chain rule expression: $$\left(f \circ g^{-1}\right)'(2) = \frac{f'(0)}{g'(0)} = \frac{1/2}{2} = \frac{1}{4} = 0.25$$ The value of the derivative at $x = 2$ is $0.25$.2d:T474,To evaluate the given sum, we multiply both sides of the expression for $\alpha$ by $\sin 1^\circ$: $$\sin 1^\circ \alpha = \frac{\sin(61^\circ - 60^\circ)}{\sin 60^\circ \sin 61^\circ} + \frac{\sin(63^\circ - 62^\circ)}{\sin 62^\circ \sin 63^\circ} + \dots + \frac{\sin(119^\circ - 118^\circ)}{\sin 118^\circ \sin 119^\circ}$$ Using the identity $\frac{\sin(y-x)}{\sin x \sin y} = \cot x - \cot y$, we rewrite the series as: $$\sin 1^\circ \alpha = (\cot 60^\circ - \cot 61^\circ) + (\cot 62^\circ - \cot 63^\circ) + \dots + (\cot 118^\circ - \cot 119^\circ)$$ Grouping the positive and negative terms separately: $$\sin 1^\circ \alpha = \left( \sum_{k=30}^{59} \cot(2k^\circ) \right) - \left( \sum_{k=30}^{59} \cot((2k+1)^\circ) \right)$$ By symmetric cancellation using $\cot(180^\circ - \theta) = -\cot \theta$, all pairs in the second sum cancel to zero, and in the first sum all terms cancel except for $\cot 60^\circ$: $$\sin 1^\circ \alpha = \cot 60^\circ = \frac{1}{\sqrt{3}}$$ Rearranging gives: $$\frac{\csc 1^\circ}{\alpha} = \sqrt{3}$$ Squaring both sides yields: $$\left( \frac{\csc 1^\circ}{\alpha} \right)^2 = 3$$2e:Td42,To find the value of $\sqrt{7} \tan\left( \frac{2\alpha \sqrt{7}}{\pi} \right)$, we begin by simplifying the given definite integral: $$\alpha = \int_{1/2}^2 \frac{\tan^{-1} x}{2x^2 - 3x + 2} \, dx \quad \text{--- (1)}$$ Using the substitution $x = \frac{1}{t}$, we have $dx = -\frac{1}{t^2} \, dt$. The limits of integration change as follows: - When $x = \frac{1}{2}$, $t = 2$. - When $x = 2$, $t = \frac{1}{2}$. Substituting these into equation (1): $$\alpha = \int_{2}^{1/2} \frac{\tan^{-1}\left(\frac{1}{t}\right)}{2\left(\frac{1}{t}\right)^2 - 3\left(\frac{1}{t}\right) + 2} \left( -\frac{1}{t^2} \right) dt$$ $$\alpha = \int_{1/2}^{2} \frac{\tan^{-1}\left(\frac{1}{t}\right)}{\frac{2 - 3t + 2t^2}{t^2}} \cdot \frac{1}{t^2} \, dt$$ $$\alpha = \int_{1/2}^{2} \frac{\tan^{-1}\left(\frac{1}{t}\right)}{2t^2 - 3t + 2} \, dt$$ Replacing the dummy variable $t$ back with $x$: $$\alpha = \int_{1/2}^{2} \frac{\tan^{-1}\left(\frac{1}{x}\right)}{2x^2 - 3x + 2} \, dx \quad \text{--- (2)}$$ Adding equation (1) and equation (2): $$2\alpha = \int_{1/2}^{2} \frac{\tan^{-1} x + \tan^{-1}\left(\frac{1}{x}\right)}{2x^2 - 3x + 2} \, dx$$ Since $x > 0$ for $x \in \left[\frac{1}{2}, 2\right]$, we use the identity $\tan^{-1} x + \tan^{-1}\left(\frac{1}{x}\right) = \frac{\pi}{2}$: $$2\alpha = \int_{1/2}^{2} \frac{\frac{\pi}{2}}{2x^2 - 3x + 2} \, dx$$ $$\alpha = \frac{\pi}{4} \int_{1/2}^{2} \frac{dx}{2x^2 - 3x + 2}$$ Now, we evaluate the remaining integral $I$: $$I = \int_{1/2}^{2} \frac{dx}{2x^2 - 3x + 2}$$ Completing the square in the denominator: $$2x^2 - 3x + 2 = 2\left(x^2 - \frac{3}{2}x + 1\right) = 2\left[\left(x - \frac{3}{4}\right)^2 + 1 - \frac{9}{16}\right] = 2\left[\left(x - \frac{3}{4}\right)^2 + \frac{7}{16}\right]$$ Thus, $$I = \frac{1}{2} \int_{1/2}^{2} \frac{dx}{\left(x - \frac{3}{4}\right)^2 + \left(\frac{\sqrt{7}}{4}\right)^2}$$ Using the standard integral formula $\int \frac{dx}{u^2 + a^2} = \frac{1}{a} \tan^{-1}\left(\frac{u}{a}\right)$: $$I = \frac{1}{2} \cdot \frac{1}{\frac{\sqrt{7}}{4}} \left[ \tan^{-1}\left(\frac{x - \frac{3}{4}}{\frac{\sqrt{7}}{4}}\right) \right]_{1/2}^2$$ $$I = \frac{2}{\sqrt{7}} \left[ \tan^{-1}\left(\frac{4x - 3}{\sqrt{7}}\right) \right]_{1/2}^2$$ Evaluating at the limits: $$I = \frac{2}{\sqrt{7}} \left[ \tan^{-1}\left(\frac{5}{\sqrt{7}}\right) - \tan^{-1}\left(\frac{-1}{\sqrt{7}}\right) \right]$$ $$I = \frac{2}{\sqrt{7}} \left[ \tan^{-1}\left(\frac{5}{\sqrt{7}}\right) + \tan^{-1}\left(\frac{1}{\sqrt{7}}\right) \right]$$ Using the identity $\tan^{-1} A + \tan^{-1} B = \tan^{-1}\left(\frac{A + B}{1 - AB}\right)$: $$\tan^{-1}\left(\frac{5}{\sqrt{7}}\right) + \tan^{-1}\left(\frac{1}{\sqrt{7}}\right) = \tan^{-1}\left( \frac{\frac{5}{\sqrt{7}} + \frac{1}{\sqrt{7}}}{1 - \frac{5}{7}} \right) = \tan^{-1}\left( \frac{\frac{6}{\sqrt{7}}}{\frac{2}{7}} \right) = \tan^{-1}(3\sqrt{7})$$ Hence, $$I = \frac{2}{\sqrt{7}} \tan^{-1}(3\sqrt{7})$$ Substitute $I$ back into the expression for $\alpha$: $$\alpha = \frac{\pi}{4} \cdot \frac{2}{\sqrt{7}} \tan^{-1}(3\sqrt{7}) = \frac{\pi}{2\sqrt{7}} \tan^{-1}(3\sqrt{7})$$ Rearranging terms: $$\frac{2\alpha \sqrt{7}}{\pi} = \tan^{-1}(3\sqrt{7})$$ Taking the tangent on both sides: $$\tan\left( \frac{2\alpha \sqrt{7}}{\pi} \right) = 3\sqrt{7}$$ Multiplying both sides by $\sqrt{7}$: $$\sqrt{7} \tan\left( \frac{2\alpha \sqrt{7}}{\pi} \right) = \sqrt{7} \times 3\sqrt{7} = 21$$2f:T8a1,To find the dimensional formula of the thermoelectric figure of merit quantity $Z = \frac{S^2 \sigma}{\kappa}$, we first determine the dimensions of each constituent physical quantity: $S$, $\sigma$, and $\kappa$. 1. **Dimensions of Seebeck Coefficient ($S$):** The Seebeck coefficient $S$ is defined as the electromotive force (e.m.f.) produced per unit temperature difference: $$S = \frac{V}{\Delta T}$$ The dimension of electric potential (or e.m.f.) $V$ is given by: $$[V] = \frac{[\text{Work}]}{[\text{Charge}]} = \frac{[M L^2 T^{-2}]}{[I T]} = [M L^2 T^{-3} I^{-1}]$$ Therefore, the dimension of $S$ is: $$[S] = \frac{[M L^2 T^{-3} I^{-1}]}{[K]} = [M L^2 T^{-3} I^{-1} K^{-1}]$$ 2. **Dimensions of Electrical Conductivity ($\sigma$):** Electrical conductivity is the reciprocal of resistivity ($\rho$): $$\sigma = \frac{l}{R A}$$ The dimension of electrical resistance $R$ is: $$[R] = \frac{[V]}{[I]} = \frac{[M L^2 T^{-3} I^{-1}]}{[I]} = [M L^2 T^{-3} I^{-2}]$$ Therefore, the dimension of electrical conductivity $\sigma$ is: $$[\sigma] = \frac{[L]}{[M L^2 T^{-3} I^{-2}] \cdot [L^2]} = [M^{-1} L^{-3} T^3 I^2]$$ 3. **Dimensions of Thermal Conductivity ($\kappa$):** Thermal conductivity is defined through the heat transfer equation: $$\frac{dQ}{dt} = \kappa A \frac{\Delta T}{l}$$ Here, $\frac{dQ}{dt}$ represents power, with dimensions $[M L^2 T^{-3}]$. Thus, the dimension of thermal conductivity $\kappa$ is: $$[\kappa] = \frac{\left[\frac{dQ}{dt}\right] [L]}{[A] [\Delta T]} = \frac{[M L^2 T^{-3}] [L]}{[L^2] [K]} = [M L T^{-3} K^{-1}]$$ 4. **Dimensional Formula of $Z$:** Now, substituting the dimensions into $Z = \frac{S^2 \sigma}{\kappa}$: $$[S^2] = \left([M L^2 T^{-3} I^{-1} K^{-1}]\right)^2 = [M^2 L^4 T^{-6} I^{-2} K^{-2}]$$ Combining $[S^2]$ and $[\sigma]$: $$[S^2 \sigma] = [M^2 L^4 T^{-6} I^{-2} K^{-2}] \times [M^{-1} L^{-3} T^3 I^2] = [M^1 L^1 T^{-3} I^0 K^{-2}]$$ Finally, dividing by $[\kappa]$: $$[Z] = \frac{[M^1 L^1 T^{-3} K^{-2}]}{[M^1 L^1 T^{-3} K^{-1}]} = [M^0 L^0 T^0 I^0 K^{-1}]$$ Thus, the dimensional formula of the quantity $Z$ is $[M^0 L^0 T^0 I^0 K^{-1}]$, which corresponds to **Option B**.30:Tc77,To find the flux of the electric field through the imaginary plane, we proceed step-by-step using Gauss's Law and surface integration. ### Step 1: Electric Field in the Dielectric Region The system consists of two coaxial conducting cylinders of length $\ell$. The inner cylinder has a radius $r_1 = \sqrt{2}R$ and carries a charge $Q$. The outer cylinder has a radius $r_2 = 2R$ and is grounded. The region between $r_1$ and $r_2$ is filled with a dielectric material of relative permittivity $\kappa = 5$. By Gauss's law for the electric displacement field $\vec{D}$ in cylindrical coordinates: $$\oint \vec{D} \cdot d\vec{A} = Q_{\text{free}}$$ $$D(2\pi r \ell) = Q \implies D(r) = \frac{Q}{2\pi r \ell}$$ Since $\vec{D} = \kappa \epsilon_0 \vec{E}$, the electric field in the annular dielectric region ($\sqrt{2}R \le r \le 2R$) is: $$\vec{E}(r) = \frac{Q}{2\pi \kappa \epsilon_0 \ell r} \hat{r}$$ For $r < \sqrt{2}R$ (inside the inner conductor) and $r > 2R$ (outside the grounded outer cylinder), the electric field $\vec{E} = 0$. ### Step 2: Parametrizing the Plane Set up a Cartesian coordinate system with the origin at the center of the cross-section along the common axis: - The common axis is along the $z$-axis ($0 \le z \le \ell$). - The plane is parallel to the axis at a distance $R$ from it, so its equation is $x = R$. The distance $r$ from any point $(R, y, z)$ on the plane to the axis is: $$r = \sqrt{x^2 + y^2} = \sqrt{R^2 + y^2}$$ The normal to the plane is in the direction of $\hat{i}$, so the area element is $d\vec{A} = dy \, dz \, \hat{i}$. The component of the electric field normal to the plane is: $$E_x = \vec{E} \cdot \hat{i} = E(r) \cos\theta = E(r) \frac{x}{r} = \frac{Q}{2\pi \kappa \epsilon_0 \ell r} \frac{R}{r} = \frac{Q R}{2\pi \kappa \epsilon_0 \ell (R^2 + y^2)}$$ ### Step 3: Determining the Limits of Integration The electric field is non-zero only within the dielectric region where $\sqrt{2}R \le r \le 2R$: $$(\sqrt{2}R)^2 \le R^2 + y^2 \le (2R)^2$$ $$2R^2 \le R^2 + y^2 \le 4R^2$$ $$R^2 \le y^2 \le 3R^2$$ This yields two symmetric segments for $y$: $$y \in [-\sqrt{3}R, -R] \quad \text{and} \quad y \in [R, \sqrt{3}R]$$ ### Step 4: Calculating the Total Flux The total flux $\Phi$ through the plane is given by the integral of $E_x$ over the area of the plane where $\vec{E} \neq 0$: $$\Phi = \int_{z=0}^{\ell} dz \int_{y \in \text{region}} E_x \, dy$$ $$\Phi = \ell \cdot \frac{Q R}{2\pi \kappa \epsilon_0 \ell} \left[ 2 \int_{R}^{\sqrt{3}R} \frac{dy}{R^2 + y^2} \right]$$ Evaluating the definite integral: $$\int_{R}^{\sqrt{3}R} \frac{dy}{R^2 + y^2} = \left[ \frac{1}{R} \arctan\left(\frac{y}{R}\right) \right]_{R}^{\sqrt{3}R} = \frac{1}{R} \left( \arctan(\sqrt{3}) - \arctan(1) \right)$$ $$\int_{R}^{\sqrt{3}R} \frac{dy}{R^2 + y^2} = \frac{1}{R} \left( \frac{\pi}{3} - \frac{\pi}{4} \right) = \frac{\pi}{12R}$$ Substituting this back into the flux equation: $$\Phi = \frac{Q R}{2\pi \kappa \epsilon_0} \cdot 2 \cdot \left( \frac{\pi}{12R} \right) = \frac{Q}{12 \kappa \epsilon_0}$$ Given $\kappa = 5$: $$\Phi = \frac{Q}{12 \times 5 \epsilon_0} = \frac{Q}{60\epsilon_0}$$ **Correct Answer:** **C** ($\frac{Q}{60\epsilon_0}$)31:Tb19,To find the ratio of the frequencies of small angular oscillations $\frac{f_1}{f_2}$, we analyze the restoring torques acting on the rod for both configurations. Let: - $m$ be the mass of the uniform rod $OO'$, - $l$ be the length of the rod, - $k$ be the spring constant of each spring, - $\theta$ be a small angular displacement of the rod from its vertical equilibrium position. The moment of inertia of the uniform rod about the hinge $O$ at its end is: $$I = \frac{1}{3} m l^2$$ --- ### **Case 1: Figure 1** When the rod is rotated clockwise by a small angle $\theta$: 1. **Spring at midpoint ($y = \frac{l}{2}$):** - The linear displacement of the midpoint is $x_1 = \frac{l}{2} \theta$. - The spring connected to the left wall is stretched, exerting a leftward restoring force $F_1 = k \left(\frac{l}{2} \theta\right)$. - The restoring torque about $O$ is: $$\tau_{1a} = F_1 \cdot \frac{l}{2} = k \left(\frac{l}{2} \theta\right) \left(\frac{l}{2}\right) = \frac{1}{4} k l^2 \theta$$ 2. **Spring at top-end $O'$ ($y = l$):** - The linear displacement of the top-end is $x_2 = l \theta$. - The spring connected to the right wall is compressed, exerting a leftward restoring force $F_2 = k (l \theta)$. - The restoring torque about $O$ is: $$\tau_{1b} = F_2 \cdot l = k (l \theta) \cdot l = k l^2 \theta$$ The total restoring torque for Fig. 1 is: $$\tau_1 = \tau_{1a} + \tau_{1b} = \frac{1}{4} k l^2 \theta + k l^2 \theta = \frac{5}{4} k l^2 \theta$$ Using the rotational equation of motion $I \alpha_1 = -\tau_1$: $$\omega_1^2 = \frac{\tau_1 / \theta}{I} = \frac{\frac{5}{4} k l^2}{\frac{1}{3} m l^2} = \frac{15 k}{4 m}$$ Hence, the frequency of oscillation is: $$f_1 = \frac{1}{2\pi} \sqrt{\frac{15 k}{4 m}}$$ --- ### **Case 2: Figure 2** In this configuration, both springs are attached at the midpoint ($y = \frac{l}{2}$): 1. Upon displacement $\theta$, the midpoint moves by $x = \frac{l}{2} \theta$. 2. The left spring is stretched by $x$, and the right spring is compressed by $x$. Both springs apply restoring forces in the same direction (towards the left). 3. The restoring force from each spring is $F = k \left(\frac{l}{2} \theta\right)$. The total restoring torque about $O$ for Fig. 2 is: $$\tau_2 = 2 \times \left( k \left(\frac{l}{2} \theta\right) \cdot \frac{l}{2} \right) = \frac{1}{2} k l^2 \theta$$ Using the equation of motion $I \alpha_2 = -\tau_2$: $$\omega_2^2 = \frac{\tau_2 / \theta}{I} = \frac{\frac{1}{2} k l^2}{\frac{1}{3} m l^2} = \frac{3 k}{2 m}$$ Hence, the frequency of oscillation is: $$f_2 = \frac{1}{2\pi} \sqrt{\frac{3 k}{2 m}}$$ --- ### **Ratio of Frequencies $\frac{f_1}{f_2}$** $$\frac{f_1}{f_2} = \frac{\omega_1}{\omega_2} = \sqrt{\frac{15 / 4}{3 / 2}} = \sqrt{\frac{15}{4} \times \frac{2}{3}} = \sqrt{\frac{5}{2}}$$ Thus, the correct option is **C**.32:Tc59,To determine the relative rate of change of the separation $r$ between the two stars, we analyze the dynamics of mass transfer in a binary star system. --- ### **1. Binary System Framework** Consider two stars of masses $m_1$ and $m_2$ separated by a distance $r$, where $m_1 \gg m_2$. The total mass of the system is: $$ M = m_1 + m_2 $$ Mass is transferred slowly from the lighter star $m_2$ to the heavier star $m_1$ at a constant rate $\gamma \text{ kg/s}$, with no total mass loss from the system: $$ \frac{dm_1}{dt} = \gamma, \quad \frac{dm_2}{dt} = -\gamma, \quad \frac{dM}{dt} = 0 $$ For a circular orbit, the orbital angular frequency $\omega$ is governed by Kepler's Third Law: $$ \omega = \sqrt{\frac{GM}{r^3}} $$ The reduced mass of the system is: $$ \mu = \frac{m_1 m_2}{m_1 + m_2} = \frac{m_1 m_2}{M} $$ The total orbital angular momentum $J$ of the binary system is: $$ J = \mu r^2 \omega = \frac{m_1 m_2}{M} r^2 \sqrt{\frac{GM}{r^3}} = \frac{m_1 m_2}{\sqrt{M}} \sqrt{Gr} $$ --- ### **2. Case A: Conservative Mass Transfer (Total Angular Momentum Conserved)** Assuming that total orbital angular momentum $J$ is conserved during the mass transfer process ($\frac{dJ}{dt} = 0$), we take the natural logarithm of $J$: $$ \ln J = \ln m_1 + \ln m_2 - \frac{1}{2}\ln M + \frac{1}{2}\ln G + \frac{1}{2}\ln r $$ Differentiating with respect to time $t$: $$ \frac{1}{J}\frac{dJ}{dt} = \frac{1}{m_1}\frac{dm_1}{dt} + \frac{1}{m_2}\frac{dm_2}{dt} - \frac{1}{2M}\frac{dM}{dt} + \frac{1}{2r}\frac{dr}{dt} $$ Substituting $\frac{dJ}{dt} = 0$, $\frac{dM}{dt} = 0$, $\frac{dm_1}{dt} = \gamma$, and $\frac{dm_2}{dt} = -\gamma$: $$ 0 = \frac{\gamma}{m_1} - \frac{\gamma}{m_2} + \frac{1}{2r}\frac{dr}{dt} $$ Rearranging to solve for the relative rate of change of separation $\frac{1}{r}\frac{dr}{dt}$: $$ \frac{1}{r}\frac{dr}{dt} = 2\gamma \left( \frac{1}{m_2} - \frac{1}{m_1} \right) $$ Since $m_1 \gg m_2$, we have $\frac{1}{m_1} \ll \frac{1}{m_2}$, yielding: $$ \frac{1}{r}\frac{dr}{dt} \approx +\frac{2\gamma}{m_2} $$ --- ### **3. Case B: Specific Angular Momentum Loss Model** If the transferred mass leaves star $m_2$ carrying the specific orbital angular momentum of star $m_2$, the rate of change of orbital angular momentum is: $$ \frac{dJ}{dt} = \frac{dm_2}{dt} (r_2 v_2) = -\gamma \sqrt{G m_1 r} $$ Since $J \approx m_2 \sqrt{G m_1 r}$, differentiating $J$ gives: $$ \frac{dJ}{dt} = \frac{dm_2}{dt}\sqrt{G m_1 r} + m_2 \frac{d}{dt}\left(\sqrt{G m_1 r}\right) $$ $$ -\gamma \sqrt{G m_1 r} = -\gamma \sqrt{G m_1 r} + m_2 \sqrt{G m_1 r} \left( \frac{1}{2m_1}\frac{dm_1}{dt} + \frac{1}{2r}\frac{dr}{dt} \right) $$ Simplifying: $$ 0 = \frac{\gamma}{2m_1} + \frac{1}{2r}\frac{dr}{dt} \implies \frac{1}{r}\frac{dr}{dt} = -\frac{\gamma}{m_1} $$ --- ### **Conclusion** Because the mechanism for angular momentum loss/transfer is not uniquely specified in the problem statement, different standard models yield different expressions (e.g., $+\frac{2\gamma}{m_2}$ vs. $-\frac{\gamma}{m_1}$). None of the options unambiguously capture all assumptions without ambiguity. Therefore, the official decision for this question is **MARKS TO ALL**.33:Tdbc,To determine which statements are correct, we analyze the electrostatic properties of the system at each step: ### **1. Potential of the sphere before grounding (Option A)** Before grounding, the conducting sphere is neutral, meaning its net charge is $Q_1 = 0$. Since the sphere is a conductor, it is an equipotential body, and its potential is equal to the electrostatic potential at its center $C$. The total potential at the center $C$ is the sum of the potential due to the point charge $q = 10^{-8}\text{ C}$ located at distance $d_1 = 20\text{ cm} = 0.2\text{ m}$ and the potential due to the induced surface charges $q_{\text{ind}}$: $$V_{\text{sphere}} = V_C = \frac{1}{4\pi\epsilon_0} \frac{q}{d_1} + V_{\text{induced charges at } C}$$ Since all induced surface charges lie at a distance equal to the radius $R = 10\text{ cm} = 0.1\text{ m}$ from the center, their net contribution to the potential at $C$ is: $$V_{\text{induced charges at } C} = \frac{1}{4\pi\epsilon_0} \frac{\int \text{d}q}{R} = \frac{1}{4\pi\epsilon_0} \frac{Q_{\text{net, induced}}}{R} = 0 \quad (\text{since } Q_{\text{net, induced}} = 0)$$ Thus: $$V_{\text{sphere}} = \frac{1}{4\pi\epsilon_0} \frac{q}{d_1} = \left(9 \times 10^9\text{ N m}^2/\text{C}^2\right) \times \frac{10^{-8}\text{ C}}{0.2\text{ m}} = 450\text{ V}$$ **Statement (A) is correct.** --- ### **2. Charge on the sphere after grounding and charge flow (Options B and C)** When the sphere is grounded, its electrostatic potential becomes zero ($V_{\text{sphere}}' = 0$). Let $q'$ be the net charge on the sphere after grounding. The potential at the center of the sphere is now: $$V_C' = \frac{1}{4\pi\epsilon_0} \frac{q}{d_1} + \frac{1}{4\pi\epsilon_0} \frac{q'}{R} = 0$$ Substitute the given values: $$\frac{q}{d_1} + \frac{q'}{R} = 0 \implies q' = -q \left(\frac{R}{d_1}\right)$$ $$q' = -\left(10^{-8}\text{ C}\right) \times \left(\frac{0.1\text{ m}}{0.2\text{ m}}\right) = -5 \times 10^{-9}\text{ C}$$ * **Charge flowing to the ground:** The change in charge on the sphere is: $$\Delta q = q' - 0 = -5 \times 10^{-9}\text{ C}$$ Therefore, a positive charge of $5 \times 10^{-9}\text{ C}$ flows from the sphere to the ground (or $5 \times 10^{-9}\text{ C}$ of negative charge flows from the ground to the sphere). **Statement (B) is correct.** * **Charge remaining on the sphere:** After the grounding wire is disconnected, the charge remaining on the isolated sphere is $q' = -5 \times 10^{-9}\text{ C}$. **Statement (C) is correct.** --- ### **3. Final electrostatic potential of the sphere (Option D)** After removing the ground connection, the point charge $q = 10^{-8}\text{ C}$ is moved by an additional distance of $10\text{ cm}$ radially away from the center. The new distance from the center is: $$d_2 = 20\text{ cm} + 10\text{ cm} = 30\text{ cm} = 0.3\text{ m}$$ The charge on the sphere remains constant at $q' = -5 \times 10^{-9}\text{ C}$. The new electrostatic potential of the sphere is calculated at its center: $$V_{\text{final}} = \frac{1}{4\pi\epsilon_0} \left( \frac{q}{d_2} + \frac{q'}{R} \right)$$ $$V_{\text{final}} = \left(9 \times 10^9\right) \left( \frac{10^{-8}}{0.3} + \frac{-5 \times 10^{-9}}{0.1} \right)$$ $$V_{\text{final}} = 90 \left( \frac{10}{3} - 5 \right) = 90 \left( -\frac{5}{3} \right) = -150\text{ V}$$ Thus, the final electrostatic potential of the sphere is $-150\text{ V}$, not $300\text{ V}$. **Statement (D) is incorrect.** --- ### **Conclusion** The correct choices are **A, B, and C**.34:Ta67,To determine the distance $d$ between the two concave mirrors such that the final image is formed on the source $S$ itself, we analyze the path of light rays from $S$ through refraction at the slab's surface and reflection from the mirrors. ### 1. Apparent Position of the Source $S$ The monochromatic point light source $S$ is embedded at the center of the glass slab of refractive index $n_0$ and thickness $t$. Therefore, $S$ is at a real distance of $\frac{t}{2}$ from either face of the slab. When light rays travel from inside the glass medium ($n_1 = n_0$) into air ($n_2 = 1$), refraction occurs at the plane interface. Using the plane refraction formula: $$\frac{n_2}{v} = \frac{n_1}{u}$$ Taking the real depth as $u = -\frac{t}{2}$: $$\frac{1}{v} = \frac{n_0}{-\frac{t}{2}} \implies v = -\frac{t}{2n_0}$$ Thus, to an observer in air (or looking from the concave mirror), the source appears to be at a virtual position $S'$ located at a distance of $\frac{t}{2n_0}$ behind the slab face. ### 2. Distance of Virtual Source $S'$ from the Mirror Let $d$ be the total distance between the two concave mirrors. Since the slab is placed symmetrically between them, the distance $x$ from either mirror to the nearest surface of the slab is: $$x = \frac{d - t}{2}$$ The distance $u$ of the virtual source $S'$ from either mirror is: $$u = x + \frac{t}{2n_0} = \frac{d - t}{2} + \frac{t}{2n_0} = \frac{d}{2} - \frac{t}{2}\left(1 - \frac{1}{n_0}\right)$$ --- ### 3. Conditions for Image Formation at $S$ #### **Case I: Rays retrace their path (Reflection at Center of Curvature)** If the virtual source $S'$ is located at the center of curvature of the concave mirror, light rays strike the mirror normally and reflect back along their original paths, re-focusing at $S$. Setting $u = 2f$: $$\frac{d}{2} - \frac{t}{2}\left(1 - \frac{1}{n_0}\right) = 2f$$ $$d = 4f + \left(1 - \frac{1}{n_0}\right)t$$ This corresponds to **Option (A)**. --- #### **Case II: Rays become parallel after reflection (Reflection at Focus)** If the virtual source $S'$ is located at the principal focus of one mirror, the reflected light rays emerge parallel to the common principal axis. Setting $u = f$: $$\frac{d}{2} - \frac{t}{2}\left(1 - \frac{1}{n_0}\right) = f$$ $$d = 2f + \left(1 - \frac{1}{n_0}\right)t$$ These parallel rays then travel towards the second mirror. Upon reflection from the second mirror, the rays converge towards its focal point. After re-entering the glass slab, refraction causes these converging rays to focus precisely at the point source $S$. This corresponds to **Option (B)**. --- ### Conclusion The correct options are **A** and **B**.35:T1235,To determine the correct statement(s), we analyze the electric field in each region for both configurations using the principle of superposition. For a thin, infinitely large non-conducting sheet carrying a uniform surface charge density $\sigma$, the electric field produced at any point is: $$E = \frac{\sigma}{2\epsilon_0}$$ directed away from a positively charged sheet and towards a negatively charged sheet. If a region has a total surface charge density $Q_L$ on all sheets to its left and $Q_R$ on all sheets to its right, the net electric field (taking the rightward direction as positive, $+\hat{i}$) in that region is given by: $$E = \frac{Q_L - Q_R}{2\epsilon_0}$$ Given data: - $\sigma_0 = 9\text{ }\mu\text{C/m}^2 = 9 \times 10^{-6}\text{ C/m}^2$ - $\epsilon_0 = 9 \times 10^{-12}\text{ F/m}$ - Distance between consecutive sheets $d = 1\text{ }\mu\text{m} = 10^{-6}\text{ m}$ The fundamental potential factor $\frac{\sigma_0 d}{\epsilon_0}$ is: $$\frac{\sigma_0 d}{\epsilon_0} = \frac{(9 \times 10^{-6}\text{ C/m}^2)(10^{-6}\text{ m})}{9 \times 10^{-12}\text{ F/m}} = 1\text{ V}$$ --- ### **1. Analysis of Configuration I** The surface charge densities on the six sheets from left to right are: $$\sigma_1 = +\sigma_0, \quad \sigma_2 = -\sigma_0, \quad \sigma_3 = +\sigma_0, \quad \sigma_4 = -\sigma_0, \quad \sigma_5 = +\sigma_0, \quad \sigma_6 = -\sigma_0$$ Let us calculate the net electric field in each region: - **Region 1:** $$Q_L = +\sigma_0, \quad Q_R = (-\sigma_0 + \sigma_0 - \sigma_0 + \sigma_0 - \sigma_0) = -\sigma_0$$ $$E_{\text{I}, 1} = \frac{\sigma_0 - (-\sigma_0)}{2\epsilon_0} = \frac{\sigma_0}{\epsilon_0}$$ - **Region 2:** $$Q_L = \sigma_0 - \sigma_0 = 0, \quad Q_R = \sigma_0 - \sigma_0 + \sigma_0 - \sigma_0 = 0$$ $$E_{\text{I}, 2} = 0$$ - **Region 3:** $$Q_L = \sigma_0 - \sigma_0 + \sigma_0 = \sigma_0, \quad Q_R = -\sigma_0 + \sigma_0 - \sigma_0 = -\sigma_0$$ $$E_{\text{I}, 3} = \frac{\sigma_0 - (-\sigma_0)}{2\epsilon_0} = \frac{\sigma_0}{\epsilon_0}$$ - **Region 4:** $$Q_L = \sigma_0 - \sigma_0 + \sigma_0 - \sigma_0 = 0, \quad Q_R = \sigma_0 - \sigma_0 = 0$$ $$E_{\text{I}, 4} = 0$$ - **Region 5:** $$Q_L = \sigma_0, \quad Q_R = -\sigma_0$$ $$E_{\text{I}, 5} = \frac{\sigma_0}{\epsilon_0}$$ #### **Evaluating Options for Configuration I:** - **Statement A:** In region 4 of Configuration I, the electric field magnitude is $E_{\text{I}, 4} = 0$. **(Statement A is CORRECT)** - **Statement C:** Potential difference between the first and the last sheet of Configuration I is: $$\Delta V = V_1 - V_6 = \sum_{k=1}^{5} E_{\text{I}, k} \cdot d = \left(\frac{\sigma_0}{\epsilon_0} + 0 + \frac{\sigma_0}{\epsilon_0} + 0 + \frac{\sigma_0}{\epsilon_0}\right) d = \frac{3\sigma_0 d}{\epsilon_0} = 3 \times 1\text{ V} = 3\text{ V}$$ **(Statement C is INCORRECT)** --- ### **2. Analysis of Configuration II** The surface charge densities on the six sheets from left to right are: $$\sigma_1 = +\frac{\sigma_0}{2}, \quad \sigma_2 = -\sigma_0, \quad \sigma_3 = +\sigma_0, \quad \sigma_4 = -\sigma_0, \quad \sigma_5 = +\sigma_0, \quad \sigma_6 = -\frac{\sigma_0}{2}$$ Let us calculate the net electric field in each region: - **Region 1:** $$Q_L = +\frac{\sigma_0}{2}, \quad Q_R = -\frac{\sigma_0}{2} \implies E_{\text{II}, 1} = \frac{\sigma_0}{2\epsilon_0}$$ - **Region 2:** $$Q_L = -\frac{\sigma_0}{2}, \quad Q_R = +\frac{\sigma_0}{2} \implies E_{\text{II}, 2} = -\frac{\sigma_0}{2\epsilon_0}$$ - **Region 3:** $$Q_L = +\frac{\sigma_0}{2}, \quad Q_R = -\frac{\sigma_0}{2} \implies E_{\text{II}, 3} = \frac{\sigma_0}{2\epsilon_0}$$ - **Region 4:** $$Q_L = -\frac{\sigma_0}{2}, \quad Q_R = +\frac{\sigma_0}{2} \implies E_{\text{II}, 4} = -\frac{\sigma_0}{2\epsilon_0}$$ - **Region 5:** $$Q_L = +\frac{\sigma_0}{2}, \quad Q_R = -\frac{\sigma_0}{2} \implies E_{\text{II}, 5} = \frac{\sigma_0}{2\epsilon_0}$$ #### **Evaluating Options for Configuration II:** - **Statement B:** In region 3 of Configuration II, the electric field magnitude is $|E_{\text{II}, 3}| = \frac{\sigma_0}{2\epsilon_0}$. Statement B claims it is $\frac{\sigma_0}{\epsilon_0}$. **(Statement B is INCORRECT)** - **Statement D:** Potential difference between the first and the last sheet of Configuration II is: $$\Delta V = V_1 - V_6 = \sum_{k=1}^{5} E_{\text{II}, k} \cdot d = \left( \frac{\sigma_0}{2\epsilon_0} - \frac{\sigma_0}{2\epsilon_0} + \frac{\sigma_0}{2\epsilon_0} - \frac{\sigma_0}{2\epsilon_0} + \frac{\sigma_0}{2\epsilon_0} \right) d = \frac{\sigma_0 d}{2\epsilon_0} = 0.5\text{ V} \neq 0$$ **(Statement D is INCORRECT)** --- ### **Conclusion** Only statement **A** is correct.36:T81c,To determine the correct options, we analyze the thermodynamics of the Carnot engine and the heat pump step-by-step. ### 1. Analysis of the Carnot Engine We are given: - Temperature of the hot reservoir, $T_1 = 1000\text{ K}$ - Efficiency of the Carnot engine, $\eta = 0.4$ - Heat extracted per cycle from the hot reservoir, $Q_1 = 150\text{ J}$ #### Work Extracted ($W$): The efficiency of an engine is defined as: $$\eta = \frac{W}{Q_1}$$ Substituting the given values: $$0.4 = \frac{W}{150\text{ J}} \implies W = 0.4 \times 150\text{ J} = 60\text{ J}$$ Thus, **Option (A)** is correct. #### Cold Reservoir Temperature ($T_2$): For a Carnot engine, efficiency is also given by: $$\eta = 1 - \frac{T_2}{T_1}$$ Substituting $\eta = 0.4$ and $T_1 = 1000\text{ K}$: $$0.4 = 1 - \frac{T_2}{1000\text{ K}}$$ $$\frac{T_2}{1000\text{ K}} = 0.6 \implies T_2 = 600\text{ K}$$ Thus, **Option (B)** is correct. --- ### 2. Analysis of the Heat Pump The work input to the heat pump per cycle is $W_{\text{in}} = W = 60\text{ J}$. We are given: - Coefficient of Performance of the heat pump, $(\text{COP})_{\text{hp}} = 10$ - Temperature of the hot reservoir of the heat pump, $T_H = 300\text{ K}$ #### Heat Supplied to Hot Reservoir ($Q_H$): By definition, the coefficient of performance for a heat pump is: $$(\text{COP})_{\text{hp}} = \frac{Q_H}{W_{\text{in}}}$$ Substituting the values: $$10 = \frac{Q_H}{60\text{ J}} \implies Q_H = 600\text{ J}$$ Therefore, the heat supplied to the hot reservoir of the heat pump in one cycle is $600\text{ J}$ (not $540\text{ J}$). Thus, **Option (D)** is incorrect. #### Cold Reservoir Temperature of the Heat Pump ($T_C$): For a Carnot heat pump operating between temperatures $T_H$ and $T_C$: $$(\text{COP})_{\text{hp}} = \frac{T_H}{T_H - T_C}$$ Substituting $(\text{COP})_{\text{hp}} = 10$ and $T_H = 300\text{ K}$: $$10 = \frac{300\text{ K}}{300\text{ K} - T_C}$$ $$300\text{ K} - T_C = 30\text{ K} \implies T_C = 270\text{ K}$$ Thus, **Option (C)** is correct. --- ### Conclusion The correct statements are **A, B, and C**.37:T902,To find the value of $\alpha$, we need to calculate the magnetic dipole moment ($\mu$) and the angular momentum ($L$) of the rotating conducting solid sphere. ### 1. Calculation of Magnetic Dipole Moment ($\mu$) Since the sphere is **conducting**, the charge $Q$ resides entirely on its outer spherical surface with a surface charge density: $$\sigma = \frac{Q}{4\pi R^2}$$ Consider a thin circular ring on the surface of the sphere at an angle $\theta$ relative to the axis of rotation, subtending an angle $d\theta$ at the center. - Radius of the ring: $r = R \sin\theta$ - Area of the ring strip: $dA = 2\pi R \sin\theta \cdot R d\theta = 2\pi R^2 \sin\theta d\theta$ - Charge on the ring strip: $$dq = \sigma dA = \left(\frac{Q}{4\pi R^2}\right) (2\pi R^2 \sin\theta d\theta) = \frac{Q}{2} \sin\theta d\theta$$ As the sphere rotates with uniform angular speed $\omega$, the rotating ring produces a current $dI$: $$dI = \frac{dq}{T} = \frac{\omega}{2\pi} dq = \frac{Q \omega}{4\pi} \sin\theta d\theta$$ The magnetic dipole moment $d\mu$ of this elemental ring is: $$d\mu = dI \cdot (\pi r^2) = \left(\frac{Q \omega}{4\pi} \sin\theta d\theta\right) \pi (R \sin\theta)^2 = \frac{1}{4} Q \omega R^2 \sin^3\theta d\theta$$ Integrating over the entire surface from $\theta = 0$ to $\theta = \pi$: $$\mu = \int_0^{\pi} \frac{1}{4} Q \omega R^2 \sin^3\theta d\theta = \frac{1}{4} Q \omega R^2 \int_0^{\pi} \sin^3\theta d\theta$$ Using the integral identity $\int_0^{\pi} \sin^3\theta d\theta = \frac{4}{3}$: $$\mu = \frac{1}{4} Q \omega R^2 \cdot \frac{4}{3} = \frac{1}{3} Q \omega R^2$$ --- ### 2. Calculation of Angular Momentum ($L$) For a solid sphere of uniform mass $M$ and radius $R$, the moment of inertia about its central axis is: $$I = \frac{2}{5} M R^2$$ The total angular momentum $L$ about the axis of rotation is: $$L = I \omega = \frac{2}{5} M R^2 \omega$$ --- ### 3. Ratio of Magnetic Dipole Moment to Angular Momentum Taking the ratio of $\mu$ to $L$: $$\frac{\mu}{L} = \frac{\frac{1}{3} Q \omega R^2}{\frac{2}{5} M R^2 \omega} = \frac{5}{6} \frac{Q}{M}$$ We can rewrite this in terms of $\frac{Q}{2M}$: $$\frac{\mu}{L} = \frac{5}{3} \left(\frac{Q}{2M}\right)$$ Comparing this with the given expression $\alpha \frac{Q}{2M}$, we get: $$\alpha = \frac{5}{3} \approx 1.67$$38:T858,To find the difference between the energies of the two photons, we analyze each step of the process using energy conservation and the Bohr model of the atom. ### Step 1: Energy of the first photon ($h\nu_1$) A hydrogen atom in its ground state has an ionization energy of $E_I = 13.6\text{ eV}$. When it absorbs a photon of frequency $\nu_1$ and ejects an electron with kinetic energy $K_e = 10\text{ eV}$, the energy of the absorbed photon is given by: $$h\nu_1 = E_I + K_e$$ $$h\nu_1 = 13.6\text{ eV} + 10\text{ eV} = 23.6\text{ eV}$$ --- ### Step 2: Ground state energy of the positronium atom A positronium atom consists of an electron and a positron orbiting their common center of mass. Since both particles have equal mass $m_e$, the reduced mass $\mu$ of the positronium system is: $$\mu = \frac{m_e \cdot m_e}{m_e + m_e} = \frac{m_e}{2}$$ In the Bohr model, the energy levels are directly proportional to the reduced mass of the system: $$E_n \propto \mu$$ Since the ground state energy of hydrogen (where $\mu \approx m_e$) is $-13.6\text{ eV}$, the ground state energy of positronium ($E_{\text{Ps}, 1}$) is: $$E_{\text{Ps}, 1} = \frac{1}{2} \times (-13.6\text{ eV}) = -6.8\text{ eV}$$ --- ### Step 3: Energy of the second photon ($h\nu_2$) The electron (kinetic energy $K_e = 10\text{ eV}$) and the positron at rest ($K_p = 0\text{ eV}$) combine to form a positronium atom moving with center of mass kinetic energy $K_{\text{CM}} = 5\text{ eV}$ and emit a photon of frequency $\nu_2$. By the principle of conservation of energy: $$E_{\text{initial}} = E_{\text{final}}$$ $$K_e + K_p = E_{\text{Ps}, 1} + K_{\text{CM}} + h\nu_2$$ Substituting the known values: $$10\text{ eV} + 0\text{ eV} = -6.8\text{ eV} + 5\text{ eV} + h\nu_2$$ $$10\text{ eV} = -1.8\text{ eV} + h\nu_2$$ $$h\nu_2 = 10\text{ eV} + 1.8\text{ eV} = 11.8\text{ eV}$$ --- ### Step 4: Difference between the two photon energies The difference between the energies of the two photons is: $$\Delta E = h\nu_1 - h\nu_2$$ $$\Delta E = 23.6\text{ eV} - 11.8\text{ eV} = 11.8\text{ eV}$$ Thus, the difference between the two photon energies is **11.8**.39:T8fa,To find the amount of heat absorbed by the ideal monatomic gas along the path $XY$, we analyze the processes in the thermodynamic cycle $WXYZW$. ### 1. Thermodynamic Properties of an Ideal Monatomic Gas For a monatomic ideal gas: * Degrees of freedom: $f = 3$ * Molar heat capacity at constant volume: $C_V = \frac{3}{2}R$ * Molar heat capacity at constant pressure: $C_P = \frac{5}{2}R$ * Adiabatic index (ratio of specific heats): $\gamma = \frac{C_P}{C_V} = \frac{5}{3}$ ### 2. Analysis of the Cyclic Processes The cycle consists of consecutive adiabatic and isobaric processes: * **Path $WX$**: Adiabatic process. * **Path $XY$**: Isobaric process. * **Path $YZ$**: Adiabatic process. * **Path $ZW$**: Isobaric process. ### 3. State Variables at Point $X$ For the adiabatic process $WX$, the relation between volume $V$ and temperature $T$ is given by: $$T_W V_W^{\gamma - 1} = T_X V_X^{\gamma - 1}$$ Since $\gamma - 1 = \frac{5}{3} - 1 = \frac{2}{3}$, we can solve for $T_X$: $$T_X = T_W \left( \frac{V_W}{V_X} \right)^{2/3}$$ Given $V_W = 64\text{ cm}^3 = 4^3\text{ cm}^3$ and $V_X = 125\text{ cm}^3 = 5^3\text{ cm}^3$: $$\frac{V_W}{V_X} = \frac{64}{125} = \left( \frac{4}{5} \right)^3$$ Substituting this ratio into the temperature relation: $$T_X = T_W \left( \left( \frac{4}{5} \right)^3 \right)^{2/3} = T_W \left( \frac{4}{5} \right)^2 = \frac{16}{25} T_W$$ Multiplying both sides by $n R$ and using the given value $n R T_W = 1\text{ J}$: $$n R T_X = \frac{16}{25} n R T_W = \frac{16}{25}\text{ J}$$ ### 4. Heat Absorbed Along Path $XY$ Path $XY$ is an isobaric process. The heat absorbed $Q_{XY}$ along an isobaric path is given by: $$Q_{XY} = n C_P (T_Y - T_X) = n \left( \frac{5}{2} R \right) (T_Y - T_X) = \frac{5}{2} (n R T_Y - n R T_X)$$ By Charles's Law for an isobaric process, $\frac{V_Y}{T_Y} = \frac{V_X}{T_X}$, which gives: $$T_Y = T_X \left( \frac{V_Y}{V_X} \right)$$ Substituting $T_Y$ into the equation for heat absorbed: $$Q_{XY} = \frac{5}{2} n R T_X \left( \frac{V_Y}{V_X} - 1 \right)$$ Given $V_Y = 250\text{ cm}^3$ and $V_X = 125\text{ cm}^3$: $$\frac{V_Y}{V_X} = \frac{250}{125} = 2$$ Now, substituting $n R T_X = \frac{16}{25}\text{ J}$ and $\frac{V_Y}{V_X} = 2$: $$Q_{XY} = \frac{5}{2} \times \frac{16}{25} \times (2 - 1) = \frac{8}{5} = 1.6\text{ J}$$3a:T72d,