### Related Formula Delta = beginvmatrix a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 endvmatrix$\Delta = \begin{vmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{vmatrix}$ ### Core Logic For the given system, we find the determinant of the coefficient matrix: Delta = beginvmatrix 1 & 1 & 1 \\ 1 & 2 & lambda^2 \\ 1 & 3 & lambda endvmatrix$\Delta = \begin{vmatrix} 1 & 1 & 1 \\ 1 & 2 & \lambda^2 \\ 1 & 3 & \lambda \end{vmatrix}$ Expanding along the first row: Delta = 1(2lambda - 3lambda^2) - 1(lambda - lambda^2) + 1(3 - 2)$\Delta = 1(2\lambda - 3\lambda^2) - 1(\lambda - \lambda^2) + 1(3 - 2)$ Delta = 2lambda - 3lambda^2 - lambda + lambda^2 + 1$\Delta = 2\lambda - 3\lambda^2 - \lambda + \lambda^2 + 1$ Delta = -2lambda^2 + lambda + 1 = 0$\Delta = -2\lambda^2 + \lambda + 1 = 0$ Rightarrow 2lambda^2 - lambda - 1 = 0$\Rightarrow 2\lambda^2 - \lambda - 1 = 0$ Rightarrow (2lambda + 1)(lambda - 1) = 0 Rightarrow lambda = 1, -frac12$\Rightarrow (2\lambda + 1)(\lambda - 1) = 0 \Rightarrow \lambda = 1, -\frac{1}{2}$ ### Step 1: Analyzing Cases For infinite solutions or inconsistency, we check Delta_z$\Delta_z$ (or others): Delta_z = beginvmatrix 1 & 1 & 5 \\ 1 & 2 & 9 \\ 1 & 3 & mu endvmatrix = 0$\Delta_z = \begin{vmatrix} 1 & 1 & 5 \\ 1 & 2 & 9 \\ 1 & 3 & \mu \end{vmatrix} = 0$ Evaluating the determinant: 1(2mu - 27) - 1(mu - 9) + 5(3 - 2) = 0$1(2\mu - 27) - 1(\mu - 9) + 5(3 - 2) = 0$ 2mu - 27 - mu + 9 + 5 = 0 Rightarrow mu - 13 = 0 Rightarrow mu = 13$2\mu - 27 - \mu + 9 + 5 = 0 \Rightarrow \mu - 13 = 0 \Rightarrow \mu = 13$ Thus, if lambda = 1$\lambda = 1$ and mu = 13$\mu = 13$, the system has an infinite number of solutions. If lambda = 1$\lambda = 1$ and mu neq 13$\mu \neq 13$, the system is inconsistent. ### Step 2: Checking Options Option 1 is correct (True statement). Option 2 is correct (True statement). Option 3: If lambda neq 1$\lambda \neq 1$ and lambda neq -frac12$\lambda \neq -\frac{1}{2}$, the system has a unique solution (which is consistent) regardless of mu$\mu$. If lambda = -frac12$\lambda = -\frac{1}{2}$ and mu = 13$\mu = 13$, it may have infinite solutions or be inconsistent, but typically it is consistent for unique solutions if lambda neq 1$\lambda \neq 1$ and lambda neq -1/2$\lambda \neq -1/2$. Wait, the question option 4 says "System has unique solution if lambda neq 1$\lambda \neq 1$ and mu neq 13$\mu \neq 13$". This is NOT correct because if lambda = -frac12$\lambda = -\frac{1}{2}$ and mu neq 13$\mu \neq 13$, the determinant is zero and it does not have a unique solution. ### Pattern Recognition Look for the roots of Delta = 0$\Delta = 0$. Since there are two roots (lambda=1, -1/2$\lambda=1, -1/2$), a blanket statement about "unique solution if lambda neq 1$\lambda \neq 1$" ignores the other root and is inherently false. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Matrices and Determinants

### Related Formula Delta = beginvmatrix a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 endvmatrix$\Delta = \begin{vmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{vmatrix}$ If Delta neq 0$\Delta \neq 0$, unique solution. If Delta = 0$\Delta = 0$ and Delta_x = Delta_y = Delta_z = 0$\Delta_x = \Delta_y = \Delta_z = 0$, infinitely many solutions. If Delta = 0$\Delta = 0$ and at least one of Delta_x, Delta_y, Delta_z neq 0$\Delta_x, \Delta_y, \Delta_z \neq 0$, inconsistent (no solution). ### Core Logic Given system: x + y + z = 4mu$x + y + z = 4\mu$ x + 2y + 2lambda z = 10mu$x + 2y + 2\lambda z = 10\mu$ x + 3y + 4lambda^2 z = mu^2 + 15$x + 3y + 4\lambda^2 z = \mu^2 + 15$ Compute the main determinant Delta$\Delta$: Delta = beginvmatrix 1 & 1 & 1 \\ 1 & 2 & 2lambda \\ 1 & 3 & 4lambda^2 endvmatrix$\Delta = \begin{vmatrix} 1 & 1 & 1 \\ 1 & 2 & 2\lambda \\ 1 & 3 & 4\lambda^2 \end{vmatrix}$ Apply operations: R_2 to R_2 - R_1$R_2 \to R_2 - R_1$, R_3 to R_3 - R_1$R_3 \to R_3 - R_1$ Delta = beginvmatrix 1 & 1 & 1 \\ 0 & 1 & 2lambda-1 \\ 0 & 2 & 4lambda^2-1 endvmatrix = 1 cdot (4lambda^2 - 1 - 2(2lambda - 1))$\Delta = \begin{vmatrix} 1 & 1 & 1 \\ 0 & 1 & 2\lambda-1 \\ 0 & 2 & 4\lambda^2-1 \end{vmatrix} = 1 \cdot (4\lambda^2 - 1 - 2(2\lambda - 1))$ = 4lambda^2 - 1 - 4lambda + 2 = 4lambda^2 - 4lambda + 1 = (2lambda - 1)^2$= 4\lambda^2 - 1 - 4\lambda + 2 = 4\lambda^2 - 4\lambda + 1 = (2\lambda - 1)^2$ ### Step 1: Analyzing Unique Solution For a unique solution, Delta neq 0 Rightarrow 2lambda - 1 neq 0 Rightarrow lambda neq frac12$\Delta \neq 0 \Rightarrow 2\lambda - 1 \neq 0 \Rightarrow \lambda \neq \frac{1}{2}$. Note: For unique solution, mu$\mu$ can be anything. Option (4) states the system is consistent if lambda neq frac12$\lambda \neq \frac{1}{2}$, which is purely correct. Option (1) says "unique solution if lambda neq frac12$\lambda \neq \frac{1}{2}$ and mu neq 1, 15$\mu \neq 1, 15$". While true that it has a unique solution under those conditions, it also has a unique solution for mu = 1, 15$\mu = 1, 15$. Let's check consistency conditions. ### Step 2: Checking Delta components Let Delta = 0$\Delta = 0$, so lambda = frac12$\lambda = \frac{1}{2}$. Compute Delta_x$\Delta_x$ and Delta_z$\Delta_z$ (or Delta_y$\Delta_y$): Wait, substituting lambda = 1/2$\lambda = 1/2$, the equations become: x + y + z = 4mu$x + y + z = 4\mu$ x + 2y + z = 10mu$x + 2y + z = 10\mu$ x + 3y + z = mu^2 + 15$x + 3y + z = \mu^2 + 15$ From the first two, (x+2y+z) - (x+y+z) = 10mu - 4mu Rightarrow y = 6mu$(x+2y+z) - (x+y+z) = 10\mu - 4\mu \Rightarrow y = 6\mu$. From the second and third, (x+3y+z) - (x+2y+z) = mu^2 + 15 - 10mu Rightarrow y = mu^2 - 10mu + 15$(x+3y+z) - (x+2y+z) = \mu^2 + 15 - 10\mu \Rightarrow y = \mu^2 - 10\mu + 15$. For the system to be consistent (infinite solutions since Delta = 0$\Delta = 0$), the two values of y$y$ must match: 6mu = mu^2 - 10mu + 15$6\mu = \mu^2 - 10\mu + 15$ mu^2 - 16mu + 15 = 0$\mu^2 - 16\mu + 15 = 0$ (mu - 1)(mu - 15) = 0$(\mu - 1)(\mu - 15) = 0$ So, if lambda = frac12$\lambda = \frac{1}{2}$, the system is consistent (infinite solutions) ONLY when mu = 1$\mu = 1$ or mu = 15$\mu = 15$. If lambda = frac12$\lambda = \frac{1}{2}$ and mu neq 1, 15$\mu \neq 1, 15$, it is inconsistent. ### Step 3: Checking Options Option (2) states: "The system is inconsistent if lambda = frac12$\lambda = \frac{1}{2}$ and mu neq 1$\mu \neq 1$". If mu = 15$\mu = 15$ (which is neq 1$\neq 1$), the system is actually CONSISTENT (infinite solutions). Therefore, Option (2) is NOT strictly correct because mu=15$\mu=15$ makes it consistent. Thus, statement (2) is the incorrect statement. ### Pattern Recognition Cramer's rule dependencies can be quickly identified using algebraic elimination. When variables align symmetrically (like z$z$ mapping identically), subtracting equations exposes the consistency constraint directly without resolving full 3times3$3\times3$ determinants. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Determinants

### Related Formula Row Operations in Determinants: R_i to R_i + kR_j text does not change the determinant.$R_i \to R_i + kR_j \text{ does not change the determinant.}$ ### Core Logic Apply row operations to simplify the determinant: R_2 to R_2 - R_1$R_2 \to R_2 - R_1$ and R_3 to R_3 - R_1$R_3 \to R_3 - R_1$ f(x) = beginvmatrix 2 cos^4 x & 2 sin^4 x & 3 + sin^2 2x \\ (3 + 2cos^4 x) - 2cos^4 x & 2sin^4 x - 2sin^4 x & sin^2 2x - (3 + sin^2 2x) \\ 2cos^4 x - 2cos^4 x & (3 + 2sin^4 x) - 2sin^4 x & sin^2 2x - (3 + sin^2 2x) endvmatrix$f(x) = \begin{vmatrix} 2 \cos^4 x & 2 \sin^4 x & 3 + \sin^2 2x \\ (3 + 2\cos^4 x) - 2\cos^4 x & 2\sin^4 x - 2\sin^4 x & \sin^2 2x - (3 + \sin^2 2x) \\ 2\cos^4 x - 2\cos^4 x & (3 + 2\sin^4 x) - 2\sin^4 x & \sin^2 2x - (3 + \sin^2 2x) \end{vmatrix}$ ### Step 1: Expanding the simplified determinant f(x) = beginvmatrix 2 cos^4 x & 2 sin^4 x & 3 + sin^2 2x \\ 3 & 0 & -3 \\ 0 & 3 & -3 endvmatrix$f(x) = \begin{vmatrix} 2 \cos^4 x & 2 \sin^4 x & 3 + \sin^2 2x \\ 3 & 0 & -3 \\ 0 & 3 & -3 \end{vmatrix}$ Expand along the 2nd row: f(x) = -3 [ 2sin^4 x(-3) - 3(3 + sin^2 2x) ] + (-3) [ 2cos^4 x(3) - 0 ]$f(x) = -3 [ 2\sin^4 x(-3) - 3(3 + \sin^2 2x) ] + (-3) [ 2\cos^4 x(3) - 0 ]$ Wait, let's expand properly: f(x) = -3 beginvmatrix 2sin^4 x & 3 + sin^2 2x \\ 3 & -3 endvmatrix - (-3) beginvmatrix 2cos^4 x & 2sin^4 x \\ 0 & 3 endvmatrix$f(x) = -3 \begin{vmatrix} 2\sin^4 x & 3 + \sin^2 2x \\ 3 & -3 \end{vmatrix} - (-3) \begin{vmatrix} 2\cos^4 x & 2\sin^4 x \\ 0 & 3 \end{vmatrix}$ f(x) = -3 [ -6sin^4 x - 9 - 3sin^2 2x ] + 3 [ 6cos^4 x ]$f(x) = -3 [ -6\sin^4 x - 9 - 3\sin^2 2x ] + 3 [ 6\cos^4 x ]$ f(x) = 18sin^4 x + 27 + 9sin^2 2x + 18cos^4 x$f(x) = 18\sin^4 x + 27 + 9\sin^2 2x + 18\cos^4 x$ f(x) = 18(sin^4 x + cos^4 x) + 9(4sin^2 x cos^2 x) + 27$f(x) = 18(\sin^4 x + \cos^4 x) + 9(4\sin^2 x \cos^2 x) + 27$ f(x) = 18(sin^2 x + cos^2 x)^2 - 36sin^2 x cos^2 x + 36sin^2 x cos^2 x + 27$f(x) = 18(\sin^2 x + \cos^2 x)^2 - 36\sin^2 x \cos^2 x + 36\sin^2 x \cos^2 x + 27$ f(x) = 18(1) + 27 = 45$f(x) = 18(1) + 27 = 45$ ### Step 2: Differentiating constant Since f(x) = 45$f(x) = 45$ is a constant function, f'(x) = 0$f'(x) = 0$ Therefore, frac15 f'(0) = 0$\frac{1}{5} f'(0) = 0$. ### Pattern Recognition If row subtractions reduce variable rows to pure constants, the entire determinant expansion often resolves into a pure trigonometric identity producing a global constant, immediately rendering its derivative 0. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Determinants Class 12 Maths: Continuity and Differentiability

### Core Logic Let A = beginbmatrix x_1 & y_1 & z_1 \\ x_2 & y_2 & z_2 \\ x_3 & y_3 & z_3 endbmatrix$A = \begin{bmatrix} x_1 & y_1 & z_1 \\ x_2 & y_2 & z_2 \\ x_3 & y_3 & z_3 \end{bmatrix}$. Using the given column transformations: 1) Abeginbmatrix 1 \\ 0 \\ 1 endbmatrix = beginbmatrix 2 \\ 0 \\ 2 endbmatrix implies x_1+z_1=2,\; x_2+z_2=0,\; x_3+z_3=2$A\begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix} = \begin{bmatrix} 2 \\ 0 \\ 2 \end{bmatrix} \implies x_1+z_1=2,\; x_2+z_2=0,\; x_3+z_3=2$. 2) Abeginbmatrix -1 \\ 0 \\ 1 endbmatrix = beginbmatrix -4 \\ 0 \\ 4 endbmatrix implies -x_1+z_1=-4,\; -x_2+z_2=0,\; -x_3+z_3=4$A\begin{bmatrix} -1 \\ 0 \\ 1 \end{bmatrix} = \begin{bmatrix} -4 \\ 0 \\ 4 \end{bmatrix} \implies -x_1+z_1=-4,\; -x_2+z_2=0,\; -x_3+z_3=4$. 3) Abeginbmatrix 0 \\ 1 \\ 0 endbmatrix = beginbmatrix 0 \\ 2 \\ 0 endbmatrix implies y_1=0,\; y_2=2,\; y_3=0$A\begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix} = \begin{bmatrix} 0 \\ 2 \\ 0 \end{bmatrix} \implies y_1=0,\; y_2=2,\; y_3=0$. Solving pairs from 1) and 2): x_1+z_1=2$x_1+z_1=2$ and -x_1+z_1=-4 implies 2z_1=-2 implies z_1=-1, x_1=3$-x_1+z_1=-4 \implies 2z_1=-2 \implies z_1=-1, x_1=3$. x_2+z_2=0$x_2+z_2=0$ and -x_2+z_2=0 implies x_2=0, z_2=0$-x_2+z_2=0 \implies x_2=0, z_2=0$. x_3+z_3=2$x_3+z_3=2$ and -x_3+z_3=4 implies 2z_3=6 implies z_3=3, x_3=-1$-x_3+z_3=4 \implies 2z_3=6 \implies z_3=3, x_3=-1$. So, A = beginbmatrix 3 & 0 & -1 \\ 0 & 2 & 0 \\ -1 & 0 & 3 endbmatrix$A = \begin{bmatrix} 3 & 0 & -1 \\ 0 & 2 & 0 \\ -1 & 0 & 3 \end{bmatrix}$. Now, compute A-3I$A-3I$: A-3I = beginbmatrix 0 & 0 & -1 \\ 0 & -1 & 0 \\ -1 & 0 & 0 endbmatrix$A-3I = \begin{bmatrix} 0 & 0 & -1 \\ 0 & -1 & 0 \\ -1 & 0 & 0 \end{bmatrix}$. Determinant |A-3I| = -1(-1)(-1) = 1 neq 0$|A-3I| = -1(-1)(-1) = 1 \neq 0$. Since the determinant is non-zero, the system (A-3I)X = B$(A-3I)X = B$ has a unique solution. ### Pattern Recognition Whenever independent eigenvectors are given, the matrix can be uniquely constructed. If det != 0 for (A-3I)$(A-3I)$, uniqueness is guaranteed unconditionally. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Matrices