MATHEMATICS PAPER WITH SOLUTION
SECTION-A
Let f: [–1, 2] → ℝ be given by
f(x) = 2x² + x + [x²] – [x], where [t] denotes the greatest integer less than or equal to t.
The number of points, where f is not continuous, is:
- 6
- 3
- 4
- 5
▶️ Answer/Explanation
Doubtful points: –1, 0, 1, √2, √3, 2
Only x = –1, 0, √2 are points of discontinuity.
The differential equation of the family of circles passing the origin and having center at the line y = x is:
- (x² – y² + 2xy)dx = (x² – y² + 2xy)dy
- (x² + y² + 2xy)dx = (x² + y² – 2xy)dy
- (x² – y² + 2xy)dx = (x² – y² – 2xy)dy
- (x² + y² – 2xy)dx = (x² + y² + 2xy)dy
▶️ Answer/Explanation
The differential equation reduces to
(x² – y² + 2xy)y’ = x² – y² – 2xy
So the answer is option (3).
- \(\frac{125\pi}{6}\)
- \(\frac{125\pi}{24}\)
- \(\frac{125\pi}{4}\)
- \(\frac{125\pi}{12}\)
▶️ Answer/Explanation
\(S_1: x^2 + y^2 \leq 25\) …(1)
\(S_2: \text{Im}\left(\frac{z + (1-\sqrt{3}i)}{1-\sqrt{3}i}\right) \geq 0\)
\(\text{Im}\left(\frac{x + iy}{1-\sqrt{3}i} + 1\right) \geq 0\)
\(\text{Im}\left(\frac{(x+iy)(1+\sqrt{3}i)}{4}\right) \geq 0\)
\(\Rightarrow \sqrt{3}x + y \geq 0\) …(2)
\(S_3: x \geq 0\) …(3)
Area = \(\frac{5}{12}(\pi \times 5^2) = \frac{125\pi}{12}\)
- \(\frac{8}{3}\)
- \(\frac{2}{3}\)
- 1
- \(\frac{4}{3}\)
▶️ Answer/Explanation
For \(x \geq 0\): \(y = x^2\) and \(y = 0\)
For \(x < 0\): \(y = -x^2\) and \(y = 2x\)
Intersection at \((-2, -4)\) and \((0, 0)\)
\[A = \int_{-2}^{0} (-x^2 – 2x) \, dx = \left[-\frac{x^3}{3} – x^2\right]_{-2}^{0}\] \[= 0 – \left(\frac{8}{3} – 4\right) = \frac{4}{3}\]
- OBBHJ
- HBBJO
- OBBJH
- JBBOH
▶️ Answer/Explanation
Letters: B, B, H, J, O (alphabetical order)
Starting with B: \(\frac{4!}{2!} = 24\) words
Starting with H: \(\frac{4!}{2!} = 12\) words
Starting with J: \(\frac{4!}{2!} = 12\) words
Starting with O: 1st word is OBBHJ (37th)
Next sequence: OBBJH (50th rank)
- 10
- 5
- 15
- 12
▶️ Answer/Explanation
Let \(\vec{v} = \vec{a} + \vec{b} + \hat{i} = 5\hat{i} + 3\hat{j} + \hat{k}\)
and \(\vec{c} + \hat{i} = \vec{p}\)
\(\vec{p} \times \vec{v} = \vec{a} \times \vec{p}\)
\(\vec{p} \times (\vec{v} + \vec{a}) = \vec{0}\)
\(\Rightarrow \vec{p} = \lambda(\vec{v} + \vec{a})\)
\(\vec{c} + \hat{i} = \lambda(7\hat{i} + 8\hat{j})\)
\(\vec{a} \cdot \vec{c} + \vec{a} \cdot \hat{i} = \lambda \vec{a} \cdot (7\hat{i} + 8\hat{j})\)
\(-29 + 2 = \lambda(14 + 40)\)
\(\lambda = -\frac{1}{2}\)
\(\vec{c} \cdot (-2\hat{i} + \hat{j} + \hat{k}) = -\frac{1}{2}(-14 + 8) + 2 = 5\)
- 110
- 105
- 124
- 121
▶️ Answer/Explanation
\(|\vec{c} – \vec{a}|^2 = |\vec{c}|^2 + |\vec{a}|^2 – 2\vec{a} \cdot \vec{c}\)
\(= |\vec{c}|^2 + 4 – 0\) (since \(\vec{a} = \vec{b} \times \vec{c}\))
\(|\vec{a}| = |\vec{b} \times \vec{c}|\)
\(2 = 3|\vec{c}| \sin \alpha\)
\(|\vec{c}| = \frac{2}{3 \cos \text{ec} \alpha}\), \(\alpha \in \left[0, \frac{\pi}{3}\right]\)
\(|\vec{c}|_{\min} = \frac{2}{3} \times \frac{2}{\sqrt{3}}\), \(\cos \text{ec} \alpha \in \left[\frac{2}{\sqrt{3}}, \infty\right)\)
\(\Rightarrow 27|\vec{c} – \vec{a}|^2_{\min} = 27\left(\frac{16}{27} + 4\right) = 124\)
Let A(-1, 1) and B(2, 3) be two points and P be a variable point above the line AB such that the area of ΔPAB is 10. If the locus of P is ax + by = 15, then 5a + 2b is:
- \(-\frac{12}{5}\)
- \(-\frac{6}{5}\)
- 4
- 6
▶️ Answer/Explanation
Area of triangle with vertices at (h,k), (-1,1), (2,3):
\[\frac{1}{2}\begin{vmatrix} h & k & 1 \\ -1 & 1 & 1 \\ 2 & 3 & 1 \end{vmatrix} = 10\] \[-2x + 3y = 25\] \(-\frac{6}{5}x + \frac{9}{5}y = 15\)
\(a = -\frac{6}{5}, b = \frac{9}{5}\)
\(5a = -6, 2b = \frac{18}{5}\)
\(5a + 2b = -6 + \frac{18}{5} = -\frac{12}{5}\)
Let \((\alpha, \beta, \gamma)\) be the image of the point \((8, 5, 7)\) in the line \[ \frac{x-1}{2} = \frac{y+1}{3} = \frac{z-2}{5} \] Then the value of \(\alpha + \beta + \gamma\) is:
- 16
- 18
- 14
- 20
▶ Answer & Solution
Let the point on the line be \(M(2\lambda+1,\; 3\lambda-1,\; 5\lambda+2)\).
The vector \( \overrightarrow{AM} = (2\lambda-7, 3\lambda-6, 5\lambda-5) \) must be perpendicular to the direction vector \((2,3,5)\):
\[ (2\lambda-7)\cdot2 + (3\lambda-6)\cdot3 + (5\lambda-5)\cdot5 = 0 \] \[ 4\lambda-14 + 9\lambda-18 + 25\lambda-25 = 0 \implies 38\lambda = 57 \] \( \lambda = \frac{3}{2} \)
Substitute \(\lambda\): \( M(4,\frac{7}{2},\frac{19}{2}) \) is the foot.
So the image point: \(A’ = (0, 2, 12) \implies \alpha+\beta+\gamma = 14\)
- 639
- 724
- 693
- 742
▶️ Answer/Explanation
For \(r=6\), the power terms cancel to give constants; the constant term is
\[ T_7 = \binom{12}{6} 3^{\frac{6}{3}} 2^6 5^{-\frac{6}{3}} = \frac{9 \times 11 \times 7}{25} \times 2^8 \times \sqrt[3]{5}. \] Hence, \[ 25 \alpha = 693. \]
Let \(f, g : \mathbb{R} \to \mathbb{R}\) be defined as: \(f(x) = |x – 1|\) and \[ g(x) = \begin{cases} e^x, & x \geq 0 \\ x + 1, & x \leq 0 \end{cases} \] Then the function \(f(g(x))\) is:
- Neither one-one nor onto
- One-one but not onto
- Both one-one and onto
- Onto but not one-one
▶️ Answer/Explanation
Because of the absolute value and piecewise behavior of \(g\), the composite \(f \circ g\) is neither injective nor surjective.
Answer & Solution
Sol.
Given terms of A.P.:
\[ 2 = a + 6d, \quad p = a + 7d, \quad q = a + 12d, \] with \(p^2 = 2 q\) (since these are also terms of a G.P.).
Solving yields:
\[ p=10, \quad q=50, \quad a=-46, \quad d=8. \] The 5th term of G.P. is: \[ 2 \times 5^4 = 1250, \] which corresponds to the \(n^{th}\) term of A.P.: \[ 1250 = a + (n-1)d = -46 + 8(n-1) \implies n=163. \]
- 1680
- 1520
- 1710
- 1640
▶️ Answer/Explanation
Total partitions = 9!/(3!3!3!) × 3! = 1680x + y + z = 4,
2x + 5y + 5z = 17,
x + 2y + mz = n
has infinitely many solutions, satisfy the equation:
- m² + n² – m – n = 46
- m² + n² + m + n = 64
- m² + n² + mn = 68
- m² + n² – mn = 39
▶️ Answer/Explanation
m² + n² – mn = 4 + 49 – 14 = 39
- 57
- 38
- 19
- 76
▶️ Answer/Explanation
- r = 1
- r² – 8r + 8 = 0
- 2r² – 4r + 1 = 0
- 2r² – 8r + 7 = 0
▶️ Answer/Explanation
- 1021
- 1120
- 2012
- 2120
▶️ Answer/Explanation
Therefore, 100(a + b + c) = 100(1/10 + 1/10 + 21) = 2120.
If \[ B = \begin{bmatrix} 3\alpha & -9 & 3\alpha \\ -\alpha & 7 & -2\alpha \\ -2\alpha & 5 & -2\beta \end{bmatrix} \] is the matrix of cofactors of the elements of \(A\), then \(\det(AB)\) is equal to:
- 343
- 125
- 64
- 216
▶️ Answer/Explanation
Equate the matrix cofactors. By examining, we get:
For \(A_{21}\): \(2\alpha^2 – 3\alpha = \alpha \implies 2\alpha^2 – 4\alpha = 0 \Rightarrow \alpha(\alpha – 2) = 0\). Take \(\alpha = 2\) (since \(\alpha \beta \neq 0\)).
Then \(2\alpha^2 – \alpha \beta = 3\alpha \implies 2 \times 4 – 2 \beta = 6\) ⇒ \(8 – 2\beta = 6\) ⇒ \(\beta = 1\).
Now, \[ |A| = \left| \begin{matrix} 1 & 2 & 3 \\ 2 & 2 & 1 \\ -1 & 2 & 4 \end{matrix} \right| = 6 – 2 \cdot 9 + 3 \cdot 6 = 6 – 18 + 18 = 6 \] Therefore, \[ |AB| = |A|^3 = 6^3 = 216 \]
If \[ y(\theta) = \frac{2 \cos \theta + \cos 2\theta}{\cos 3\theta + 4\cos 2\theta + 5 \cos \theta + 2}, \] then at \(\theta = \frac{\pi}{2}\), \[ y” + y’ + y \] is equal to:
- \(\frac{3}{2}\)
- 1
- \(\frac{1}{2}\)
- 2
▶️ Answer/Explanation
Simplify \(y(\theta)\) to \[ y = \frac{1}{2(1 + \cos\theta)}. \] Evaluate the derivatives at \(\theta = \frac{\pi}{2}\): \[ y\left(\frac{\pi}{2}\right) = \frac{1}{2}, \quad y’\left(\frac{\pi}{2}\right) = \frac{1}{2}, \quad y”\left(\frac{\pi}{2}\right) = 1. \] Thus, \[ y” + y’ + y = 1 + \frac{1}{2} + \frac{1}{2} = 2. \]
For \(x \geq 0\), the least value of \(K\), for which \(4^{1+x}+4^{1-x}, \frac{K}{2}, 16^{x}+16^{-x}\) are three consecutive terms of an A.P., is equal to:
- 10
- 4
- 8
- 16
▶️ Answer/Explanation
Sol.
For three numbers \(a, b, c\) to be in arithmetic progression (A.P.), they must satisfy: \[ 2b = a + c. \] Setting \[ a = 4^{1+x} + 4^{1-x}, \quad b = \frac{K}{2}, \quad c = 16^{x} + 16^{-x}. \] Substituting, \[ 2 \cdot \frac{K}{2} = \left(4^{1+x} + 4^{1-x}\right) + \left(16^{x} + 16^{-x}\right). \] Simplifying, \[ K = 4^{1+x} + 4^{1-x} + 16^{x} + 16^{-x}. \] Note that \(4^{1+x} = 4 \cdot 4^{x}\) and \(16^{x} = (4^{2})^{x} = 4^{2x}\). To find the least value of \(K\), consider \(4^{x} = t \geq 1\): \[ K = 4t + \frac{4}{t} + t^{2} + \frac{1}{t^{2}}. \] Use the AM-GM inequality: \[ 4t + \frac{4}{t} \geq 8, \quad t^{2} + \frac{1}{t^{2}} \geq 2. \] Hence, \[ K \geq 8 + 2 = 10. \] Equality holds when \(t = 1\), i.e., \(x=0\).
| X | \(\alpha\) | 1 | 0 | -3 |
|---|---|---|---|---|
| P(X) | \(\frac{1}{3}\) | k | \(\frac{1}{6}\) | \(\frac{1}{4}\) |
▶️ Answer/Explanation
Solution:
Since total probability is 1: \[ \frac{1}{3} + k + \frac{1}{6} + \frac{1}{4} = 1 \implies k = \frac{1}{4}. \] Calculate mean: \[ \mu = \frac{\alpha}{3} + 1 \cdot k + 0 \cdot \frac{1}{6} + (-3) \cdot \frac{1}{4} = \frac{\alpha}{3} + \frac{1}{4} – \frac{3}{4} = \frac{\alpha}{3} – \frac{1}{2}. \] Given: \[ \sigma – \mu = 2 \implies \sigma = \mu + 2. \] Using definition of variance to solve for \(\alpha\), find \(\alpha = 6\). Then, \[ \mu = \frac{6}{3} – \frac{1}{2} = 2 – \frac{1}{2} = \frac{3}{2}, \quad \sigma = \frac{3}{2} + 2 = \frac{7}{2}. \] Hence, \[ \sigma + \mu = \frac{7}{2} + \frac{3}{2} = 5. \]
Let \(y = y(x)\) be the solution of the differential equation
\[ \frac{dy}{dx} + \frac{2x}{(1+x^2)^2} y = x e^{\frac{1}{1+x^2}}, \quad y(0) = 0. \]Then the area enclosed by the curve \(f(x) = y(x) e^{-\frac{1}{1+x^2}}\) and the line \(y – x = 4\) is _____.
▶️ Answer/Explanation
Solution:
The curve is \(f(x) = y(x) e^{-\frac{1}{1+x^2}}\). Solving the given ODE and evaluating the area between \(f(x)\) and line \(y – x = 4\), we obtain the required area.
▶️ Answer/Explanation
Rewrite: \[ \sin^2 x + (2 + 2x – x^2)\sin x – 3(x-1)^2 = 0 \implies (\sin x – (x-1)^2)\big(\sin x + 3\big) = 0 \] The equation \(\sin x = -3\) has no real solution. Only consider: \[ \sin x = (x-1)^2 \] For \(-\pi \leq x \leq \pi\), \((x-1)^2 \leq 1\) gives two points of intersection.
▶️ Answer/Explanation
The shortest distance line parameters and equation derivation leads to \[ \alpha = -5, \quad \beta = 0, \] so \[ (\alpha – \beta)^2 = (-5 – 0)^2 = 25. \]
▶️ Answer/Explanation
Solution:
Using series expansion and logarithmic identities, it is found that: \[ a = 2, \quad b = 3. \] Hence, \[ 11a + 18b = 11 \times 2 + 18 \times 3 = 22 + 54 = 76. \]
▶️ Answer/Explanation
Solution:
From the quadratic equation, the positive root is \[ a = \frac{-1 + \sqrt{17}}{4}. \] Evaluate the limit using L’Hospital’s Rule or series expansion to get: \[ \alpha = 153, \quad \beta = 17. \] Hence, \[ \alpha + \beta = 153 + 17 = 170. \]
▶️ Answer/Explanation
Solution:
By applying integration techniques including integration by parts and trigonometric identities, the integral evaluates exactly to 1.
▶️ Answer/Explanation
Solution:
From the geometric interpretation, the function represents the sum of squares of distances related to a circle.
The minimum value \(m = 9\) and the maximum value \(M = 41\).
Therefore, \[ M^{2} – m^{2} = 41^{2} – 9^{2} = 1681 – 81 = 1600. \]
▶️ Answer/Explanation
Solution:
Slope of the given line \(2x – y = 10\) is \(2\), so the perpendicular line has slope \(-\frac{1}{2}\).
Equation of the tangent to the parabola \(y^{2} = 4(x – 9)\) with slope \(-\frac{1}{2}\) gives point of contact
\[ P = (13, -4). \] The center of the circle is at \[ C = (7, 4). \] Distance \(CP\) is: \[ \sqrt{(13-7)^{2} + (-4 – 4)^{2}} = \sqrt{36 + 64} = \sqrt{100} = 10. \]
▶️ Answer/Explanation
Solution:
Case I: \(x \geq -5\)
Then \(|x + 5| = x + 5\) and \(|x + 7| = x + 7\), substitute: \[ x (x + 5) + 2 (x + 7) – 2 = 0 \implies x^{2} + 5x + 2x + 14 – 2 = 0, \] \[ x^{2} + 7x + 12 = 0, \] which factors as \[ (x + 3)(x + 4) = 0 \implies x = -3, -4. \] Both satisfy \(x \geq -5\).
Case II: \(-7 < x < -5\)
Then \(|x + 5| = -(x + 5)\) and \(|x + 7| = x + 7\), substitute: \[ x (-(x + 5)) + 2 (x + 7) – 2 = 0 \implies -x^{2} – 5x + 2x + 14 – 2 = 0, \] \[ -x^{2} – 3x + 12 = 0 \implies x^{2} + 3x – 12 = 0, \] with roots \[ x = \frac{-3 \pm \sqrt{9 + 48}}{2} = \frac{-3 \pm \sqrt{57}}{2}. \] Only the root in \((-7, -5)\) is valid, i.e. \(x = \frac{-3 – \sqrt{57}}{2}\).
Case III: \(x \leq -7\)
Then \(|x + 5| = -(x + 5)\), \(|x + 7| = -(x + 7)\), substitute: \[ x (-(x + 5)) + 2 (-(x + 7)) – 2 = 0 \implies -x^{2} – 5x – 2x – 14 – 2 = 0, \] \[ -x^{2} – 7x – 16 = 0 \implies x^{2} + 7x + 16 = 0, \] which has discriminant \[ 49 – 64 = -15 < 0, \] so no real solutions.
Total real solutions = 3.