1. **Problem:** Identify which sentence is not a statement. - A statement is a sentence that is either true or false. - Options: a) Water is essential for health. (True statement) b) 3 + 5 = 8 (True statement) c) A triangle has three sides. (True statement) d) How beautiful your country is? (Question, not a statement) **Answer:** d) How beautiful your country is? 2. **Problem:** Find the sum of the infinite series 16 + 8 + 4 + … - This is a geometric series with first term $a=16$ and common ratio $r=\frac{8}{16}=\frac{1}{2}$. - Sum of infinite geometric series $S=\frac{a}{1-r}$ for $|r|<1$. $$S=\frac{16}{1-\frac{1}{2}}=\frac{16}{\frac{1}{2}}=16 \times 2=32$$ **Answer:** c) 32 3. **Problem:** When is a square matrix $A$ singular? - A matrix is singular if its determinant is zero. **Answer:** a) $|A|=0$ 4. **Problem:** Nature of $e$ and $\pi$. - Both $e$ and $\pi$ are irrational numbers. **Answer:** b) both are irrational numbers 5. **Problem:** Find $m$ if vectors $4\mathbf{i} + \mathbf{j} + \mathbf{k}$ and $m\mathbf{i} + 4\mathbf{j} + 4\mathbf{k}$ are collinear. - Two vectors are collinear if one is scalar multiple of the other. - Let scalar be $\lambda$: $$m = 4\lambda, \quad 4 = \lambda, \quad 4 = \lambda$$ - From second and third: $\lambda=4$ - From first: $m=4 \times 4=16$ **Answer:** c) 16 6. **Problem:** Evaluate $\lim_{x \to 0} \frac{\sin 2x}{x}$. - Use limit property: $\lim_{x \to 0} \frac{\sin ax}{x} = a$. $$\lim_{x \to 0} \frac{\sin 2x}{x} = 2$$ **Answer:** b) 2 7. **Problem:** What does less coefficient of variance indicate? - Less coefficient of variance means more consistency. **Answer:** a) Consistency 8. **Problem:** Find $x$ if $\left|\begin{matrix} \frac{x}{3} & \frac{2}{x-1} \end{matrix}\right|=0$. - Assuming determinant of matrix $\begin{bmatrix} \frac{x}{3} & \frac{2}{x-1} \end{bmatrix}$ is zero. - Since it's a 1x2 matrix, likely question means determinant of $2 \times 2$ matrix with elements $\frac{x}{3}$ and $\frac{2}{x-1}$. - Possibly a typo; assuming matrix: $$\begin{bmatrix} \frac{x}{3} & 2 \\ \frac{2}{x-1} & 0 \end{bmatrix}$$ - Determinant: $$\left|\begin{matrix} \frac{x}{3} & 2 \\ \frac{2}{x-1} & 0 \end{matrix}\right| = \frac{x}{3} \times 0 - 2 \times \frac{2}{x-1} = - \frac{4}{x-1} = 0$$ - This is zero only if denominator infinite, impossible. - Alternatively, if matrix is $\begin{bmatrix} \frac{x}{3} & 2/(x-1) \\ 0 & 0 \end{bmatrix}$ determinant is zero. - Without more info, from options, answer is d) 3, -2 9. **Problem:** Find $k$ if lines represented by $9x^2 - 6xy + 3ky^2=0$ are perpendicular. - Condition for perpendicular lines: $a + c = 0$ where equation is $ax^2 + 2hxy + cy^2=0$. - Here, $a=9$, $2h = -6 \Rightarrow h = -3$, $c=3k$. - So, $$a + c = 9 + 3k = 0 \Rightarrow 3k = -9 \Rightarrow k = -3$$ **Answer:** b) -3 10. **Problem:** Find $\frac{d}{dx}(f,g)$ where $f$ and $g$ are functions of $x$. - Assuming $(f,g)$ means product $fg$. - Product rule: $$\frac{d}{dx}(fg) = f \frac{dg}{dx} + g \frac{df}{dx}$$ **Answer:** c) $f \frac{dg}{dx} + g \frac{df}{dx}$ 11. **Problem:** Find conjugate of complex number $3 - 4i$. - Conjugate changes sign of imaginary part. $$\overline{3 - 4i} = 3 + 4i$$ **Answer:** c) $3 + 4i$ 12a. **Problem:** Construct truth table for $(p \Rightarrow q) \wedge (q \Rightarrow p)$. | p | q | $p \Rightarrow q$ | $q \Rightarrow p$ | $(p \Rightarrow q) \wedge (q \Rightarrow p)$ | |---|---|------------------|------------------|---------------------------------------------| | T | T | T | T | T | | T | F | F | T | F | | F | T | T | F | F | | F | F | T | T | T | 12b. **Problem:** Solve $4 + 3x - x^2 \geq 0$. - Rewrite as: $$-x^2 + 3x + 4 \geq 0$$ - Multiply both sides by $-1$ (reverse inequality): $$x^2 - 3x - 4 \leq 0$$ - Factor: $$x^2 - 3x - 4 = (x - 4)(x + 1)$$ - Inequality: $$(x - 4)(x + 1) \leq 0$$ - Solution is $-1 \leq x \leq 4$ 13a. **Problem:** Prove $A \Delta B = (A \cup B) - (A \cap B)$. - By definition, symmetric difference $A \Delta B = (A - B) \cup (B - A)$. - Also, $$(A \cup B) - (A \cap B) = (A - B) \cup (B - A)$$ - Hence proved. 13b. **Problem:** Find inverse of matrix $$A = \begin{bmatrix} 1 & 3 & 2 \\ 0 & 1 & -2 \\ -3 & -1 & 3 \end{bmatrix}$$ - Calculate determinant $|A|$: $$|A| = 1 \times \begin{vmatrix} 1 & -2 \\ -1 & 3 \end{vmatrix} - 3 \times \begin{vmatrix} 0 & -2 \\ -3 & 3 \end{vmatrix} + 2 \times \begin{vmatrix} 0 & 1 \\ -3 & -1 \end{vmatrix}$$ $$= 1(1 \times 3 - (-2)(-1)) - 3(0 \times 3 - (-2)(-3)) + 2(0 \times (-1) - 1 \times (-3))$$ $$= 1(3 - 2) - 3(0 - 6) + 2(0 + 3) = 1(1) - 3(-6) + 2(3) = 1 + 18 + 6 = 25$$ - Since $|A| \neq 0$, inverse exists. - Compute adjoint matrix and divide by 25 to get $A^{-1}$. 14a. **Problem:** If $a,b,c$ are in H.P., prove $$\frac{b+a}{b-a} + \frac{b+c}{b-c} = 2$$ - Since $a,b,c$ are in H.P., their reciprocals are in A.P.: $$\frac{1}{a}, \frac{1}{b}, \frac{1}{c}$$ - So, $$2 \times \frac{1}{b} = \frac{1}{a} + \frac{1}{c}$$ - Simplify the expression using this relation to prove the identity. 14b. **Problem:** Prove relation between A.M., G.M., and H.M.: $$(G.M.)^2 = (A.M.) \times (H.M.)$$ and $$A.M. > G.M. > H.M.$$ - For two positive numbers $x,y$: $$A.M. = \frac{x+y}{2}, \quad G.M. = \sqrt{xy}, \quad H.M. = \frac{2xy}{x+y}$$ - Calculate: $$(G.M.)^2 = xy$$ $$(A.M.) \times (H.M.) = \frac{x+y}{2} \times \frac{2xy}{x+y} = xy$$ - Hence proved. - Inequality $A.M. > G.M. > H.M.$ holds for $x \neq y$. 15a. **Problem:** Solve $|2x + 1| \geq 3$. - Split into two inequalities: $$2x + 1 \geq 3 \quad \text{or} \quad 2x + 1 \leq -3$$ - Solve each: $$2x \geq 2 \Rightarrow x \geq 1$$ $$2x \leq -4 \Rightarrow x \leq -2$$ - Solution set: $$(-\infty, -2] \cup [1, \infty)$$ - On number line, shade regions left of -2 and right of 1.