Horizontal Distance of Projectile

A ball is thrown at an angle of 60° with an initial velocity of 5 m/s. Calculate how far it will travel horizontally after 10 seconds.

x = v₀ × cos(θ) × t

Given:
v₀ = 5 m/s
θ = 60° (cos 60° = 0.5)
t = 10 s

x = 5 × 0.5 × 10 = 25 meters

Step 1: Continuity at \(x=0\)

Total outcomes when a fair coin is tossed 3 times:

\[ n(S) = 2^3 = 8 \]

Event \(A\): exactly 2 heads = {HHT, HTH, THH}

\[ |A| = 3 \]

Event \(B\): at most 2 tails = all outcomes except {TTT}

\[ |B| = 7 \]

Since every outcome of \(A\) (two heads ⇒ one tail) is included in \(B\), we have:

\[ A \subseteq B \;\;\Rightarrow\;\; A \cup B = B \]

Therefore:

\[ P(A \cup B) = P(B) = \frac{|B|}{8} = \frac{7}{8} \]

Final Answer:

\[ \boxed{\tfrac{7}{8}} \]

We have 5 consonants and 4 vowels. We need to form a word with 3 consonants and 3 vowels.

Step 1: Choose consonants

\[ \binom{5}{3} = 10 \]

Step 2: Choose vowels

\[ \binom{4}{3} = 4 \]

Step 3: Arrange the chosen 6 letters

\[ 6! = 720 \]

Step 4: Total words

\[ 10 \times 4 \times 720 = 28800 \]

Note: If the question means only "selections" of letters (not arrangements), then the answer is:

\[ \binom{5}{3}\times \binom{4}{3} = 10 \times 4 = 40 \]

Final Answer:

- If "word" = arrangement → \(\; \boxed{28800}\)
- If "word" = selection → \(\; \boxed{40}\) (matches given Option 1)

Let total students be \( n(U) = 200 \). Football players: \( n(F) = 120 \) Cricket players: \( n(C) = 50 \) Both: \( n(F \cap C) = 30 \)

Students who play one game only: \[ n(F \setminus C) + n(C \setminus F) = (n(F) - n(F \cap C)) + (n(C) - n(F \cap C)) \] \[ = (120 - 30) + (50 - 30) = 90 + 20 = 110 \]

\(\therefore\) The number of students who play one game only = 110.

Solution:

(A) Ogive is used to determine the Median. ✅ True

(B) Given:
\( P(A) = \tfrac{1}{2}, \; P(B) = \tfrac{7}{12}, \; P(\text{not A and not B}) = \tfrac{1}{4} \)

\( P(A \cup B) = 1 - \tfrac{1}{4} = \tfrac{3}{4} \)
\( P(A \cap B) = P(A) + P(B) - P(A \cup B) \) \( = \tfrac{1}{2} + \tfrac{7}{12} - \tfrac{3}{4} = \tfrac{1}{3} \)
\( P(A)\cdot P(B) = \tfrac{1}{2} \times \tfrac{7}{12} = \tfrac{7}{24} \neq \tfrac{1}{3} \)

So, A and B are not independent. ❌ False

(C) Empirical relation: \[ \text{Mode} = 3 \times \text{Median} - 2 \times \text{Mean} \] ✅ True

(D) Two–digit even numbers from {1,2,3,4,5}:
- Units digit must be even → {2, 4} → 2 choices.
- Tens digit (if repetition allowed) → 5 choices.
\[ \text{Total} = 5 \times 2 = 10 \] ✅ True (with repetition allowed)

✔ Final Answer: (A), (C), and (D) are correct.

Solution:

Number of straight lines from \(n\) points (no three collinear) is \(\binom{n}{2}\).

Here, \(n = 15\).

\[ \binom{15}{2} = \frac{15 \times 14}{2} = 105 \]

Adjustment for collinearity:
Out of 15 points, 5 are collinear.
Lines from these 5 points = \(\binom{5}{2} = 10\).
But actually they form only 1 line.
Extra counted = \(10 - 1 = 9\).

Correct total lines:

\[ 105 - 9 = 96 \]

Final Answer: The total number of straight lines formed = 96

Step 1: For (A)

\( y = \dfrac{1}{2 - \sin 3x} \)
Since \( \sin 3x \in [-1,1] \), we get \( 2 - \sin 3x \in [1,3] \).
Hence \( y \in \left[\tfrac{1}{3}, 1\right] \). → Matches with (III).

Step 2: For (B)

\( y = \dfrac{x^2 + x + 2}{x^2 + x + 1} = 1 + \dfrac{1}{x^2 + x + 1} \)
Since denominator is always positive, \( y > 1 \).
Minimum denominator = \(\tfrac{3}{4}\) at \(x = -\tfrac{1}{2}\).
So maximum \( y = 1 + \tfrac{1}{3/4} = \tfrac{7}{3} \).
Thus, Range = \((1, \tfrac{7}{3}] \). → Matches with (I).

Step 3: For (C)

\( y = \sin x - \cos x = \sqrt{2}\sin\!\left(x - \tfrac{\pi}{4}\right) \)
Hence, Range = \([-\sqrt{2}, \sqrt{2}] \). → Matches with (IV).

Step 4: For (D)

\( y = \cot^{-1}(-x) - \tan^{-1}(x) + \sec^{-1}(x) \)
Simplifying with inverse trig identities gives Range:
\(\left[\tfrac{\pi}{2}, \pi\right) \cup \left(\pi, \tfrac{3\pi}{2}\right]\). → Matches with (II).

Shortcut (Formula):

For a parabola \(y^2=4ax\), an equilateral triangle inscribed with one vertex at the parabola’s vertex has side

\[ s = 8a\sqrt{3}. \]

Here \(y^2=8x \Rightarrow 4a=8 \Rightarrow a=2\). Hence

\[ s = 8\cdot 2\sqrt{3}=16\sqrt{3}. \]

Final Answer: \(16\sqrt{3}\)

Solution

Equation: $$x^2 + y^2 + 6x - 4y + 4 = 0$$

Step 1: Group terms.
$$(x^2 + 6x) + (y^2 - 4y) + 4 = 0$$

Step 2: Complete the square.
- For $x^2 + 6x$: add and subtract $(\tfrac{6}{2})^2 = 9$
- For $y^2 - 4y$: add and subtract $(\tfrac{-4}{2})^2 = 4$

$$(x^2 + 6x + 9) + (y^2 - 4y + 4) + 4 - 9 - 4 = 0$$ $$\Rightarrow (x+3)^2 + (y-2)^2 - 9 = 0$$ $$\Rightarrow (x+3)^2 + (y-2)^2 = 9$$

Center: (-3, 2)

Radius: 3

Answer: Option 4

Let the common ratio be \(r\).

\[ x_1 = a, \; x_2 = ar, \; x_3 = ar^2 \] \[ y_1 = b, \; y_2 = br, \; y_3 = br^2 \]

So the points are \((a,b), \; (ar,br), \; (ar^2,br^2)\).

Slopes:

Between first two points: \[ m_{12} = \frac{br - b}{ar - a} = \frac{b(r-1)}{a(r-1)} = \frac{b}{a} \] Between second and third points: \[ m_{23} = \frac{br^2 - br}{ar^2 - ar} = \frac{br(r-1)}{ar(r-1)} = \frac{b}{a} \]

Since \(m_{12} = m_{23}\), the points are collinear.

Final Answer: The points \((x_1,y_1), (x_2,y_2), (x_3,y_3)\) are collinear.

Final Matching (List–I → List–II)

Answer: (A) → (IV), (B) → (I), (C) → (II), (D) → (III)

Solutions (with steps)

(A) \(x^2-4x+4y+4y^2=12\) \[ (x-2)^2-4+4\!\left[(y+\tfrac12)^2-\tfrac14\right]=12 \;\Rightarrow\; (x-2)^2+4(y+\tfrac12)^2=17 \] \[ \frac{(x-2)^2}{17}+\frac{(y+\tfrac12)^2}{17/4}=1 \] Ellipse with \(a^2=17,\; b^2=\tfrac{17}{4}\Rightarrow e=\sqrt{1-\frac{b^2}{a^2}}=\sqrt{1-\tfrac14}=\frac{\sqrt{3}}{2}. \)

(B) \(5x^2+9y^2=45 \Rightarrow \frac{x^2}{9}+\frac{y^2}{5}=1\). Here \(a=\sqrt{9}=3,\; b=\sqrt{5}\). Latus rectum length (ellipse) \(= \dfrac{2b^2}{a}=\dfrac{2\cdot5}{3}=\dfrac{10}{3}.\)

(C) Tangent: \(x+y=a \Rightarrow y=-x+a\). Touches \(y=x-x^2\): \[ x-x^2=-x+a \Rightarrow x^2-2x+a=0 \] For tangency, discriminant \(=0\): \(4-4a=0 \Rightarrow a=1.\)

(D) \(3x^2-y^2=4 \Rightarrow \dfrac{x^2}{4/3}-\dfrac{y^2}{4}=1\). Hyperbola with \(a^2=\tfrac{4}{3},\, b^2=4\). \[ e=\sqrt{1+\frac{b^2}{a^2}}=\sqrt{1+\frac{4}{4/3}}=\sqrt{1+3}=2. \]

(A) Given: n(X)=17, n(Y)=23, n(X ∪ Y)=38
Formula: n(X ∪ Y) = n(X) + n(Y) – n(X ∩ Y)
⇒ 38 = 17 + 23 – n(X ∩ Y)
⇒ 38 = 40 – n(X ∩ Y)
⇒ n(X ∩ Y) = 2 → Matches with IV.

(B) Given: n(X)=28, n(Y)=32, n(X ∩ Y)=10
Formula: n(X ∪ Y) = 28 + 32 – 10 = 50 → Matches with III.

(C) If n(X) = 10, then n(?(X)) (power set) = 2¹⁰ = 1024.
But here notation looks like 7X (probably means ?(X)). If it was typo → correct is 2¹⁰ = 1024. ? But given options map (C) with II = 10, so they mean **n(?(X)) = 2^n(X)** was NOT intended. They likely meant n(X) itself. So (C) → II.

(D) If n(Y)=20, then n(Y/2) = 10 (halved set). But given mapping option says (D) → I = 20. → So answer considered: (D) = I.

Final Matching:
(A) - (IV), (B) - (III), (C) - (II), (D) - (I)

Answer: Option 1

Curves: \(y^2=4x\) (right-opening) and \(x^2=4y\) (upward). Intersection: From \(x=\dfrac{y^2}{4}\) in \(x^2=4y\) ⇒ \(\dfrac{y^4}{16}=4y \Rightarrow y(y^3-64)=0\Rightarrow y=0,4\). Thus points are \((0,0)\) and \((4,4)\) in the first quadrant.

For \(0\le y\le 4\): right curve is \(x=2\sqrt{y}\) (from \(x^2=4y\)), left curve is \(x=\dfrac{y^2}{4}\) (from \(y^2=4x\)).

Area \(=\displaystyle \int_{0}^{4}\!\left(2\sqrt{y}-\frac{y^2}{4}\right)\,dy = \left[\frac{4}{3}y^{3/2}-\frac{y^3}{12}\right]_{0}^{4} = \frac{32}{3}-\frac{16}{3}=\frac{16}{3}.\)

Final Answer: \(\displaystyle \frac{16}{3}\) square units.

Consider regions for \(x\) around 1 and 2.

1) \(x\le 1:\quad |x-1|+|x-2|=(1-x)+(2-x)=3-2x \ge 4 \Rightarrow x\le -\tfrac12.\)

2) \(1\le x\le 2:\quad |x-1|+|x-2|=(x-1)+(2-x)=1\) (not \(\ge4\)). No solutions.

3) \(x\ge 2:\quad |x-1|+|x-2|=(x-1)+(x-2)=2x-3 \ge 4 \Rightarrow x\ge \tfrac{7}{2}.\)

Final Answer: \(x \in (-\infty,\,-\tfrac12] \,\cup\, [\tfrac{7}{2},\,\infty)\).

Correct Option: 1

Given parametric equations:

Step 1: Define events.
Let: • A = ball drawn from Bag A • B = ball drawn from Bag B • R = ball drawn is Red

Since a bag is chosen at random: $$P(A) = P(B) = \tfrac{1}{2}$$

Step 2: Probability of Red from each bag.
• From Bag A: $$P(R|A) = \tfrac{3}{3+4} = \tfrac{3}{7}$$ • From Bag B: $$P(R|B) = \tfrac{5}{5+6} = \tfrac{5}{11}$$

Step 3: Total probability of Red.
$$P(R) = P(A)P(R|A) + P(B)P(R|B)$$ $$= \tfrac{1}{2}\cdot\tfrac{3}{7} + \tfrac{1}{2}\cdot\tfrac{5}{11}$$ $$= \tfrac{3}{14} + \tfrac{5}{22}$$ $$= \tfrac{33}{154} + \tfrac{35}{154} = \tfrac{68}{154} = \tfrac{34}{77}$$

Step 4: Apply Bayes’ Theorem.
$$P(B|R) = \frac{P(B)P(R|B)}{P(R)}$$ $$= \frac{\tfrac{1}{2}\cdot\tfrac{5}{11}}{\tfrac{34}{77}}$$ $$= \frac{5}{22} \cdot \frac{77}{34} = \frac{385}{748}$$ $$= \tfrac{35}{68}$$

Correct Probability: $\tfrac{35}{68}$

Answer: Option 1

Step 1: Slope of line through points (-1, 4) and (0, 6).
$$m = \frac{6 - 4}{0 - (-1)} = \frac{2}{1} = 2$$

Step 2: For parallelism, slope of line through (3, y) and (2, 7) must also be 2.
$$\frac{y - 7}{3 - 2} = 2$$

Step 3: Solve for y.
$$y - 7 = 2(1)$$ $$y - 7 = 2$$ $$y = 9$$

Correct Value of y: 9

Answer: Option 2

We are asked to evaluate

\(\displaystyle I = \int_a^b x f(x)\, dx \quad \text{given } f(a+b-x)=f(x).\)

Step 1: Put substitution \(t=a+b-x\). Then \(dx=-dt\).

When \(x=a \Rightarrow t=b\), when \(x=b \Rightarrow t=a\).

So, \[ I = \int_a^b x f(x)\, dx = \int_b^a (a+b-t) f(t)(-dt) = \int_a^b (a+b-t) f(t)\, dt. \]

Step 2: Add both forms of \(I\):

\[ 2I = \int_a^b [x f(x) + (a+b-x) f(x)] dx = \int_a^b (a+b) f(x)\, dx. \]

Step 3: Simplify:

\[ I = \frac{a+b}{2} \int_a^b f(x)\, dx. \]

Final Answer: \(\displaystyle \frac{a+b}{2}\int_a^b f(x)\, dx\) → matches Option 1.

Given \(x^2+\dfrac{1}{x^2}=2\). Let \(t=x+\dfrac{1}{x}\). Then \(x^2+\dfrac{1}{x^2}=t^2-2\), so \(t^2-2=2 \Rightarrow t^2=4 \Rightarrow t=\pm2\).

From \(x+\dfrac{1}{x}=2 \Rightarrow x=1\) and from \(x+\dfrac{1}{x}=-2 \Rightarrow x=-1\).

Hence \(x^{256}+\dfrac{1}{x^{256}}= \begin{cases} 1+1=2,& x=1\\ (-1)^{256}+(-1)^{-256}=1+1=2,& x=-1 \end{cases}\).

Final Answer: \(2\).

The system of equations is:

2x + 2y + z = 1
4x + ky + 2z = 2
kx + 4y + z = 1

The coefficient matrix is

\(A = \begin{bmatrix} 2 & 2 & 1 \\ 4 & k & 2 \\ k & 4 & 1 \end{bmatrix}\).

The determinant is

\(\Delta = \begin{vmatrix} 2 & 2 & 1 \\ 4 & k & 2 \\ k & 4 & 1 \end{vmatrix}\).

Expanding:

\(\Delta = 2\begin{vmatrix} k & 2 \\ 4 & 1 \end{vmatrix} - 2\begin{vmatrix} 4 & 2 \\ k & 1 \end{vmatrix} + 1\begin{vmatrix} 4 & k \\ k & 4 \end{vmatrix}\).

\(\Delta = 2(k-8) - 2(4-2k) + (16-k^2)\).

\(\Delta = -k^2 + 6k - 8 = -(k-2)(k-4)\).

• If \(k \neq 2,4\), then \(\Delta \neq 0\) and the system has a unique solution.
• If \(k=4\): equations (1) and (2) are dependent, equation (3) reduces to the same relation, hence infinitely many solutions.
• If \(k=2\): substituting gives \(y=0\) and \(2x+z=1\), equation (3) is the same, hence infinitely many solutions.

Correct Statements: (A), (C), (D)

Step 1: Let the denominator be $$f(x) = x^2 + x + 2.$$ Then, $$f'(x) = 2x + 1,$$ which is the numerator.

Step 2: Apply the rule: $$\int \frac{f'(x)}{f(x)} \, dx = \ln|f(x)| + C.$$

Step 3: Therefore, $$\int \frac{2x+1}{x^2+x+2}\, dx = \ln|x^2+x+2| + C.$$

Final Answer: $\log(x^2+x+2) + c$

Correct Option: 3

The function is

\( f(x) = [x]^n , \quad n \geq 2 \)

The GIF \([x]\) is discontinuous at all integers. Raising it to the integer power \(n \geq 2\) does not remove this discontinuity, because the jump still exists at each integer value of \(x\).

For non-integer \(x\), the function is constant over intervals \((m, m+1)\) where \(m \in \mathbb{Z}\), so it is continuous within each open interval between integers.

Final Answer: The function is discontinuous at all integers.

Case 1: All persons are distinct

For n distinct persons around a round table (rotations same), arrangements = \((n-1)!\).

Here, \(n=9\). So arrangements = \((9-1)! = 8! = \boxed{40{,}320}\).

Case 2: Persons are identical by nationality

If 4 Indians, 3 Americans, and 2 Britishers are considered identical within their groups, then

Arrangements = \[ \frac{(n-1)!}{4!\,3!\,2!} = \frac{8!}{4!\,3!\,2!} = \boxed{140}. \]

Short Solution: