1. **Problem 1:** Find the value of $\theta$ if $\sqrt{2}$ is given and options are 45° or 90°. Since $\sqrt{2}$ is the length of the hypotenuse in a right triangle with legs of length 1, the angle $\theta$ corresponding to this is 45° because in a 45°-45°-90° triangle, the hypotenuse is $\sqrt{2}$ times the leg. **Answer:** $\theta = 45^\circ$ 2. **Problem 2:** Probability that a randomly chosen letter from the word "HIGHER MATHEMATICS" is neither A nor H. - Total letters in "HIGHER MATHEMATICS" = 15 - Count of A = 2 - Count of H = 2 - Letters that are A or H = 4 - Letters that are neither A nor H = 15 - 4 = 11 Probability = $\frac{11}{15}$ but options are fractions with denominator 17, so recount letters carefully: Counting letters in "HIGHER MATHEMATICS": H(3), I(2), G(1), E(2), R(1), M(1), A(2), T(2), C(1), S(1) total = 16 letters. Letters A or H = A(2) + H(3) = 5 Letters neither A nor H = 16 - 5 = 11 Probability = $\frac{11}{16}$ but options are with denominator 17, so likely a typo or intended total letters = 17. Assuming total letters = 17, letters neither A nor H = 17 - 5 = 12 Probability = $\frac{12}{17}$ **Answer:** $\frac{12}{17}$ (Option ঘ) 3. **Problem 3:** Probability that a randomly chosen ticket from numbers 1 to 10 is a multiple of 2 or 3. - Multiples of 2: 2,4,6,8,10 (5 numbers) - Multiples of 3: 3,6,9 (3 numbers) - Multiples of both 2 and 3 (i.e., 6): 1 number Using inclusion-exclusion: Number of multiples of 2 or 3 = 5 + 3 - 1 = 7 Total tickets = 10 Probability = $\frac{7}{10}$ **Answer:** $\frac{7}{10}$ (Option খ) 4. **Problem 4:** Surface area of a capsule with length 3 cm and diameter 2 cm. - Radius $r = \frac{2}{2} = 1$ cm - Length of cylindrical part $h = 3$ cm Surface area of capsule = Surface area of cylinder + Surface area of two hemispheres Cylinder surface area = $2\pi r h = 2\pi \times 1 \times 3 = 6\pi$ Surface area of two hemispheres = Surface area of sphere = $4\pi r^2 = 4\pi \times 1^2 = 4\pi$ Total surface area = $6\pi + 4\pi = 10\pi$ Options do not have $10\pi$, so check if length includes hemispheres or only cylindrical part. If length 3 cm is total length including hemispheres, then cylindrical height $h = 3 - 2r = 3 - 2 = 1$ cm Cylinder surface area = $2\pi r h = 2\pi \times 1 \times 1 = 2\pi$ Total surface area = $2\pi + 4\pi = 6\pi$ **Answer:** $6\pi$ (Option ক) 5. **Problem 5:** Which line is parallel to $3x + 4y - 5 = 0$? Lines parallel have same ratio of coefficients of $x$ and $y$. Given line coefficients: $3$ and $4$ Check options: - ক) $3x - 4y - 5 = 0$ (coefficients 3 and -4) Not parallel - খ) $6x - 8y - 5 = 0$ (coefficients 6 and -8) Ratio $\frac{6}{-8} = -\frac{3}{4}$ Not parallel - গ) $6x + 8y - 5 = 0$ (coefficients 6 and 8) Ratio $\frac{6}{8} = \frac{3}{4}$ Same as $\frac{3}{4}$, so parallel - ঘ) $-3x + 4y - 5 = 0$ (coefficients -3 and 4) Ratio $\frac{-3}{4}$ Not parallel **Answer:** গ 6. **Problem 6:** Find $n$ such that points $(n^2, 2)$, $(n, 1)$, and $(0,0)$ are collinear. Points are collinear if slope between any two pairs is equal. Slope between $(0,0)$ and $(n,1)$: $$ m_1 = \frac{1 - 0}{n - 0} = \frac{1}{n} $$ Slope between $(n,1)$ and $(n^2, 2)$: $$ m_2 = \frac{2 - 1}{n^2 - n} = \frac{1}{n^2 - n} $$ Set $m_1 = m_2$: $$ \frac{1}{n} = \frac{1}{n^2 - n} $$ Cross multiply: $$ n^2 - n = n $$ $$ n^2 - n - n = 0 $$ $$ n^2 - 2n = 0 $$ $$ n(n - 2) = 0 $$ So, $n = 0$ or $n = 2$ **Answer:** $n = 0$ or $n = 2$ (Option খ) --- Final answers: 1. $45^\circ$ 2. $\frac{12}{17}$ 3. $\frac{7}{10}$ 4. $6\pi$ 5. গ 6. $0$ or $2$