12th Mathematics - Supplementary Exam Question Paper
Month/Year: July 2022 | Board: Tamil Nadu State Board
Maximum Marks: 90 | Time: 3.00 Hours
PART - I
Note: (i) All questions are compulsory. (ii) Choose the most appropriate answer from the given four alternatives and write the option code and the corresponding answer.
1. If \(A=\begin{pmatrix}\frac{3}{5}&\frac{4}{5}\\ x&\frac{3}{5}\end{pmatrix}\) and \(A^{T}=A^{-1}\), then the value of x is:
Solution: \(A^T = A^{-1} \implies AA^T = I\).
\(\begin{pmatrix}\frac{3}{5}&\frac{4}{5}\\ x&\frac{3}{5}\end{pmatrix} \begin{pmatrix}\frac{3}{5}&x\\ \frac{4}{5}&\frac{3}{5}\end{pmatrix} = \begin{pmatrix}1&0\\ 0&1\end{pmatrix}\).
Comparing first row, second column term: \(\frac{3}{5}(x) + \frac{4}{5}(\frac{3}{5}) = 0 \implies \frac{3x}{5} + \frac{12}{25} = 0 \implies x = \frac{-4}{5}\).
\(\begin{pmatrix}\frac{3}{5}&\frac{4}{5}\\ x&\frac{3}{5}\end{pmatrix} \begin{pmatrix}\frac{3}{5}&x\\ \frac{4}{5}&\frac{3}{5}\end{pmatrix} = \begin{pmatrix}1&0\\ 0&1\end{pmatrix}\).
Comparing first row, second column term: \(\frac{3}{5}(x) + \frac{4}{5}(\frac{3}{5}) = 0 \implies \frac{3x}{5} + \frac{12}{25} = 0 \implies x = \frac{-4}{5}\).
2. If A is a non-singular matrix such that \(A^{-1}=\begin{pmatrix}5&3\\ -2&-1\end{pmatrix}\) then \((A^{T})^{-1}=\)
Solution: Property: \((A^T)^{-1} = (A^{-1})^T\).
Transpose of \(A^{-1}\) is \(\begin{pmatrix}5&-2\\ 3&-1\end{pmatrix}\).
Transpose of \(A^{-1}\) is \(\begin{pmatrix}5&-2\\ 3&-1\end{pmatrix}\).
3. If \(z=x+iy\) is a complex number such that \(|z+2|=|z-2|\) then the locus of z is:
Solution: This represents the perpendicular bisector of the line segment joining (-2, 0) and (2, 0). The midpoint is origin, and the line is the y-axis (imaginary axis).
4. \(i^{n}+i^{n+1}+i^{n+2}+i^{n+3}\) is:
Solution: The sum of four consecutive powers of \(i\) is always 0.
5. A zero of \(x^{3}+64\) is:
Solution: \(x^3 = -64 \implies x = -4\).
6. The principal value of \(cos^{-1}(cos\frac{\pi}{6})\) is:
Solution: Since \(\frac{\pi}{6} \in [0, \pi]\), the value is \(\frac{\pi}{6}\).
7. The equation of the circle passing through the foci of the ellipse \(\frac{x^{2}}{16}+\frac{y^{2}}{9}=1\) having centre at (0, 3) is:
Solution: \(a^2=16, b^2=9 \implies c^2=7\). Foci are \((\pm\sqrt{7}, 0)\).
Radius \(r = \sqrt{(\sqrt{7}-0)^2 + (0-3)^2} = \sqrt{7+9} = 4\).
Circle: \(x^2 + (y-3)^2 = 16 \implies x^2+y^2-6y+9-16=0 \implies x^2+y^2-6y-7=0\).
Radius \(r = \sqrt{(\sqrt{7}-0)^2 + (0-3)^2} = \sqrt{7+9} = 4\).
Circle: \(x^2 + (y-3)^2 = 16 \implies x^2+y^2-6y+9-16=0 \implies x^2+y^2-6y-7=0\).
8. The eccentricity of the hyperbola \(\frac{x^{2}}{16}-\frac{(y-3)^{2}}{4}=1\) is:
Solution: \(e = \sqrt{1+\frac{b^2}{a^2}} = \sqrt{1+\frac{4}{16}} = \sqrt{\frac{20}{16}} = \frac{\sqrt{5}}{2}\).
9. If a vector \(\vec{\alpha}\) lies in the plane of \(\vec{\beta}\) and \(\vec{\gamma}\) then:
Solution: Coplanar vectors have a scalar triple product of 0.
10. Distance from the origin to the plane \(3x-6y+2z+7=0\) is:
Solution: \(d = \frac{|Ax_1+By_1+Cz_1+D|}{\sqrt{A^2+B^2+C^2}} = \frac{|7|}{\sqrt{9+36+4}} = \frac{7}{7} = 1\).
11. A stone is thrown up vertically. The height it reaches at time t seconds is given by \(x=80t-16t^{2}\). The stone reaches the maximum height in time t seconds is given by:
Solution: \(v = \frac{dx}{dt} = 80 - 32t\). At max height, \(v=0 \implies 32t=80 \implies t=2.5\).
12. The angle between the parabolas \(y^{2}=x\) and \(x^{2}=y\) at the origin is:
Solution: The axes are tangents at the origin. The angle between axes is \(90^\circ\) or \(\frac{\pi}{2}\).
13. The percentage error of fifth root of 31 is approximately how many times the percentage error in 31?
Solution: Let \(y = x^{1/5}\). \(\frac{dy}{y} \times 100 = \frac{1}{5} (\frac{dx}{x} \times 100)\).
14. The value of \(\int_{-1}^{2}|x|dx\) is:
Solution: \(\int_{-1}^0 -x dx + \int_0^2 x dx = [\frac{-x^2}{2}]_{-1}^0 + [\frac{x^2}{2}]_0^2 = \frac{1}{2} + 2 = \frac{5}{2}\).
15. The area between \(y^{2}=4x\) and its latus rectum is:
Solution: Focus at (1,0). Area = \(2\int_0^1 2\sqrt{x} dx = 4[\frac{2}{3}x^{3/2}]_0^1 = \frac{8}{3}\).
16. The order of the differential equation of all circles with centre at (h, k) and radius 'a' (where h, k are arbitrary constants) is:
Solution: There are two arbitrary constants (h and k), so the order is 2.
17. The differential equation representing the family of curves \(y=A~cos(x+B)\), where A and B are parameters, is:
18. If a fair die is thrown once then the probability to get a prime number on the face:
Solution: Primes {2, 3, 5}. 3 outcomes out of 6.
19. A random variable X takes the probability mass function:
The value of \(\lambda\) is:
| X | -2 | 3 | 1 |
| P(X=x) | \(\lambda/6\) | \(\lambda/4\) | \(\lambda/12\) |
Solution: Sum = 1. \(\frac{\lambda}{6} + \frac{\lambda}{4} + \frac{\lambda}{12} = 1 \implies \frac{2\lambda+3\lambda+\lambda}{12}=1 \implies 6\lambda=12 \implies \lambda=2\).
20. Which one of the following is a binary operation on N?
PART - II
Note: Answer any seven questions. Question number 30 is Compulsory.
21. Find df for \(f(x)=x^{2}+3x\) and evaluate it for \(x=3\) and \(dx=0.02\).
Solution:
\(f(x) = x^2 + 3x\)
\(df = f'(x)dx = (2x + 3)dx\)
When \(x=3\) and \(dx=0.02\):
\(df = (2(3) + 3)(0.02) = (9)(0.02) = 0.18\)
\(f(x) = x^2 + 3x\)
\(df = f'(x)dx = (2x + 3)dx\)
When \(x=3\) and \(dx=0.02\):
\(df = (2(3) + 3)(0.02) = (9)(0.02) = 0.18\)
22. If \(\alpha\) and \(\beta\) are the roots of \(x^{2}+5x+6=0\), then show that \(\alpha^{2}+\beta^{2}=13\).
Solution:
From equation: \(\alpha + \beta = -5\) and \(\alpha\beta = 6\).
\(\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta\)
\(= (-5)^2 - 2(6) = 25 - 12 = 13\).
Hence proved.
From equation: \(\alpha + \beta = -5\) and \(\alpha\beta = 6\).
\(\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta\)
\(= (-5)^2 - 2(6) = 25 - 12 = 13\).
Hence proved.
23. Find the value of \(sin^{-1}(1)+cos^{-1}(1).\)
Solution:
\(sin^{-1}(1) = \frac{\pi}{2}\)
\(cos^{-1}(1) = 0\)
Sum = \(\frac{\pi}{2} + 0 = \frac{\pi}{2}\).
\(sin^{-1}(1) = \frac{\pi}{2}\)
\(cos^{-1}(1) = 0\)
Sum = \(\frac{\pi}{2} + 0 = \frac{\pi}{2}\).
24. Find the acute angle between the two straight lines \(\frac{x-4}{2}=\frac{y}{1}=\frac{z+1}{-2}\) and \(\frac{x-1}{4}=\frac{y+1}{-4}=\frac{z-2}{2}\).
Solution:
Vectors are \(\vec{b} = 2\hat{i}+\hat{j}-2\hat{k}\) and \(\vec{d} = 4\hat{i}-4\hat{j}+2\hat{k}\).
\(cos\theta = \frac{|\vec{b} \cdot \vec{d}|}{|\vec{b}||\vec{d}|}\)
\(\vec{b} \cdot \vec{d} = (2)(4) + (1)(-4) + (-2)(2) = 8 - 4 - 4 = 0\).
Since the dot product is 0, the angle \(\theta = 90^\circ\) or \(\frac{\pi}{2}\).
Vectors are \(\vec{b} = 2\hat{i}+\hat{j}-2\hat{k}\) and \(\vec{d} = 4\hat{i}-4\hat{j}+2\hat{k}\).
\(cos\theta = \frac{|\vec{b} \cdot \vec{d}|}{|\vec{b}||\vec{d}|}\)
\(\vec{b} \cdot \vec{d} = (2)(4) + (1)(-4) + (-2)(2) = 8 - 4 - 4 = 0\).
Since the dot product is 0, the angle \(\theta = 90^\circ\) or \(\frac{\pi}{2}\).
25. Find the tangent to the curve \(y=x^{2}-x^{4}\) at (1, 0).
Solution:
\(y' = 2x - 4x^3\).
Slope \(m\) at \(x=1\) is \(2(1) - 4(1)^3 = -2\).
Equation: \(y - y_1 = m(x - x_1) \implies y - 0 = -2(x - 1)\).
\(y = -2x + 2\) or \(2x + y - 2 = 0\).
\(y' = 2x - 4x^3\).
Slope \(m\) at \(x=1\) is \(2(1) - 4(1)^3 = -2\).
Equation: \(y - y_1 = m(x - x_1) \implies y - 0 = -2(x - 1)\).
\(y = -2x + 2\) or \(2x + y - 2 = 0\).
26. If \(z_{1}=3\), \(z_{2}=-7i\) and \(z_{3}=5+4i\), show that \(z_{1}(z_{2}+z_{3})=z_{1}z_{2}+z_{1}z_{3}\).
Solution:
LHS: \(z_2 + z_3 = -7i + 5 + 4i = 5 - 3i\).
\(z_1(z_2 + z_3) = 3(5 - 3i) = 15 - 9i\).
RHS: \(z_1z_2 = 3(-7i) = -21i\).
\(z_1z_3 = 3(5+4i) = 15 + 12i\).
\(z_1z_2 + z_1z_3 = -21i + 15 + 12i = 15 - 9i\).
LHS = RHS.
LHS: \(z_2 + z_3 = -7i + 5 + 4i = 5 - 3i\).
\(z_1(z_2 + z_3) = 3(5 - 3i) = 15 - 9i\).
RHS: \(z_1z_2 = 3(-7i) = -21i\).
\(z_1z_3 = 3(5+4i) = 15 + 12i\).
\(z_1z_2 + z_1z_3 = -21i + 15 + 12i = 15 - 9i\).
LHS = RHS.
27. Show that \(y=ae^{x}+be^{-x}\) is a solution of the differential equation \(y^{\prime\prime}-y=0\).
Solution:
\(y = ae^x + be^{-x}\)
\(y' = ae^x - be^{-x}\)
\(y'' = ae^x + be^{-x}\)
Therefore, \(y'' = y \implies y'' - y = 0\).
\(y = ae^x + be^{-x}\)
\(y' = ae^x - be^{-x}\)
\(y'' = ae^x + be^{-x}\)
Therefore, \(y'' = y \implies y'' - y = 0\).
28. A random variable X has the following probability mass function. Show that the value of k is \(\frac{1}{6}\).
| x | 1 | 2 | 3 | 4 | 5 |
| f(x) | \(k^2\) | \(2k^2\) | \(3k^2\) | 2k | 3k |
Solution:
Sum of probabilities \(\sum f(x) = 1\).
\(k^2 + 2k^2 + 3k^2 + 2k + 3k = 1\)
\(6k^2 + 5k - 1 = 0\)
\((6k - 1)(k + 1) = 0\).
\(k = 1/6\) or \(k = -1\). Since probability cannot be negative, \(k = 1/6\).
Sum of probabilities \(\sum f(x) = 1\).
\(k^2 + 2k^2 + 3k^2 + 2k + 3k = 1\)
\(6k^2 + 5k - 1 = 0\)
\((6k - 1)(k + 1) = 0\).
\(k = 1/6\) or \(k = -1\). Since probability cannot be negative, \(k = 1/6\).
29. Suppose the amount of milk sold daily at a milk booth is distributed with a minimum of 200 litres and a maximum of 600 litres with probability density function \(f(x)=k\) for \(200\le x\le600\) and 0 otherwise. Find k.
Solution:
Total probability \(\int_{-\infty}^{\infty} f(x)dx = 1\).
\(\int_{200}^{600} k dx = 1\)
\(k[x]_{200}^{600} = 1 \implies k(600 - 200) = 1 \implies 400k = 1\).
\(k = \frac{1}{400}\).
Total probability \(\int_{-\infty}^{\infty} f(x)dx = 1\).
\(\int_{200}^{600} k dx = 1\)
\(k[x]_{200}^{600} = 1 \implies k(600 - 200) = 1 \implies 400k = 1\).
\(k = \frac{1}{400}\).
30. Form the differential equation of the curve \(y=ax^{2}+bx+c\) where a, b and c are arbitrary constants.
Solution:
Since there are 3 arbitrary constants, we differentiate 3 times.
\(y = ax^2 + bx + c\)
\(y' = 2ax + b\)
\(y'' = 2a\)
\(y''' = 0\).
The required differential equation is \(\frac{d^3y}{dx^3} = 0\).
Since there are 3 arbitrary constants, we differentiate 3 times.
\(y = ax^2 + bx + c\)
\(y' = 2ax + b\)
\(y'' = 2a\)
\(y''' = 0\).
The required differential equation is \(\frac{d^3y}{dx^3} = 0\).
PART - III
Note: Answer any seven questions. Question number 40 is Compulsory.
31. Verify \((AB)^{-1}=B^{-1}A^{-1}\) with \(A=\begin{pmatrix}0&-3\\ 1&4\end{pmatrix}\) \(B=\begin{pmatrix}-2&-3\\ 0&-1\end{pmatrix}\).
Solution:
1. Find \(AB = \begin{pmatrix}0&-3\\ 1&4\end{pmatrix} \begin{pmatrix}-2&-3\\ 0&-1\end{pmatrix} = \begin{pmatrix}0&3\\ -2&-7\end{pmatrix}\).
2. Find \((AB)^{-1}\). \(|AB| = 0 - (-6) = 6\). \((AB)^{-1} = \frac{1}{6}\begin{pmatrix}-7&-3\\ 2&0\end{pmatrix}\).
3. Find \(A^{-1}\). \(|A|=3\). \(A^{-1} = \frac{1}{3}\begin{pmatrix}4&3\\ -1&0\end{pmatrix}\).
4. Find \(B^{-1}\). \(|B|=2\). \(B^{-1} = \frac{1}{2}\begin{pmatrix}-1&3\\ 0&-2\end{pmatrix}\).
5. Multiply \(B^{-1}A^{-1} = \frac{1}{6} \begin{pmatrix}-1&3\\ 0&-2\end{pmatrix} \begin{pmatrix}4&3\\ -1&0\end{pmatrix} = \frac{1}{6}\begin{pmatrix}-7&-3\\ 2&0\end{pmatrix}\).
Verified.
1. Find \(AB = \begin{pmatrix}0&-3\\ 1&4\end{pmatrix} \begin{pmatrix}-2&-3\\ 0&-1\end{pmatrix} = \begin{pmatrix}0&3\\ -2&-7\end{pmatrix}\).
2. Find \((AB)^{-1}\). \(|AB| = 0 - (-6) = 6\). \((AB)^{-1} = \frac{1}{6}\begin{pmatrix}-7&-3\\ 2&0\end{pmatrix}\).
3. Find \(A^{-1}\). \(|A|=3\). \(A^{-1} = \frac{1}{3}\begin{pmatrix}4&3\\ -1&0\end{pmatrix}\).
4. Find \(B^{-1}\). \(|B|=2\). \(B^{-1} = \frac{1}{2}\begin{pmatrix}-1&3\\ 0&-2\end{pmatrix}\).
5. Multiply \(B^{-1}A^{-1} = \frac{1}{6} \begin{pmatrix}-1&3\\ 0&-2\end{pmatrix} \begin{pmatrix}4&3\\ -1&0\end{pmatrix} = \frac{1}{6}\begin{pmatrix}-7&-3\\ 2&0\end{pmatrix}\).
Verified.
32. Find the rank of the matrix \(\begin{pmatrix}1&-2&3\\ 2&4&-6\\ 5&1&-1\end{pmatrix}.\)
Solution:
\(R_2 \to R_2 - 2R_1\) and \(R_3 \to R_3 - 5R_1\):
\(\begin{pmatrix}1&-2&3\\ 0&8&-12\\ 0&11&-16\end{pmatrix}\).
Determinant of \(3 \times 3\): \(1(-128 - (-132)) \neq 0\). (Actually, let's check det: \(1(-4+6) +2(-2+30) +3(2-20)\) ... calculation easier via determinant).
Det = \(1(-4+6) + 2(-2+30) + 3(2-20) = 2 + 56 - 54 = 4 \neq 0\).
Since determinant is non-zero, Rank = 3.
\(R_2 \to R_2 - 2R_1\) and \(R_3 \to R_3 - 5R_1\):
\(\begin{pmatrix}1&-2&3\\ 0&8&-12\\ 0&11&-16\end{pmatrix}\).
Determinant of \(3 \times 3\): \(1(-128 - (-132)) \neq 0\). (Actually, let's check det: \(1(-4+6) +2(-2+30) +3(2-20)\) ... calculation easier via determinant).
Det = \(1(-4+6) + 2(-2+30) + 3(2-20) = 2 + 56 - 54 = 4 \neq 0\).
Since determinant is non-zero, Rank = 3.
33. Show that the square roots of \(6-8i\) are \(\pm(2\sqrt{2}-i\sqrt{2})\).
Solution:
Let \(z = 6-8i\). \(|z| = \sqrt{36+64} = 10\).
Formula: \(\pm(\sqrt{\frac{|z|+a}{2}} - i\sqrt{\frac{|z|-a}{2}})\) (since b is negative).
\(= \pm(\sqrt{\frac{10+6}{2}} - i\sqrt{\frac{10-6}{2}}) = \pm(\sqrt{8} - i\sqrt{2}) = \pm(2\sqrt{2} - i\sqrt{2})\).
Let \(z = 6-8i\). \(|z| = \sqrt{36+64} = 10\).
Formula: \(\pm(\sqrt{\frac{|z|+a}{2}} - i\sqrt{\frac{|z|-a}{2}})\) (since b is negative).
\(= \pm(\sqrt{\frac{10+6}{2}} - i\sqrt{\frac{10-6}{2}}) = \pm(\sqrt{8} - i\sqrt{2}) = \pm(2\sqrt{2} - i\sqrt{2})\).
34. Prove that the roots of the equation \(x^{4}-3x^{2}-4=0\) are \(\pm2\), \(\pm i.\)
Solution:
Let \(u = x^2\). Equation becomes \(u^2 - 3u - 4 = 0\).
\((u - 4)(u + 1) = 0\).
Case 1: \(u = 4 \implies x^2 = 4 \implies x = \pm 2\).
Case 2: \(u = -1 \implies x^2 = -1 \implies x = \pm i\).
Roots are \(\pm 2, \pm i\).
Let \(u = x^2\). Equation becomes \(u^2 - 3u - 4 = 0\).
\((u - 4)(u + 1) = 0\).
Case 1: \(u = 4 \implies x^2 = 4 \implies x = \pm 2\).
Case 2: \(u = -1 \implies x^2 = -1 \implies x = \pm i\).
Roots are \(\pm 2, \pm i\).
35. Find centre and radius of the circle \(x^{2}+y^{2}+6x-4y+4=0\).
Solution:
Comparing with \(x^2+y^2+2gx+2fy+c=0\):
\(2g = 6 \implies g = 3\).
\(2f = -4 \implies f = -2\).
\(c = 4\).
Centre \((-g, -f) = (-3, 2)\).
Radius \(r = \sqrt{g^2+f^2-c} = \sqrt{9+4-4} = \sqrt{9} = 3\).
Comparing with \(x^2+y^2+2gx+2fy+c=0\):
\(2g = 6 \implies g = 3\).
\(2f = -4 \implies f = -2\).
\(c = 4\).
Centre \((-g, -f) = (-3, 2)\).
Radius \(r = \sqrt{g^2+f^2-c} = \sqrt{9+4-4} = \sqrt{9} = 3\).
36. A particle acted on by constant forces \(8\hat{i}+2\hat{j}-6\hat{k}\) and \(6\hat{i}+2\hat{j}-2\hat{k}\) is displaced from the point (1, 2, 3) to the point (5, 4, 1). Find the total work done by the forces.
Solution:
Resultant Force \(\vec{F} = \vec{F_1} + \vec{F_2} = (8+6)\hat{i} + (2+2)\hat{j} + (-6-2)\hat{k} = 14\hat{i} + 4\hat{j} - 8\hat{k}\).
Displacement \(\vec{d} = (5-1)\hat{i} + (4-2)\hat{j} + (1-3)\hat{k} = 4\hat{i} + 2\hat{j} - 2\hat{k}\).
Work Done \(W = \vec{F} \cdot \vec{d} = (14)(4) + (4)(2) + (-8)(-2) = 56 + 8 + 16 = 80\) units.
Resultant Force \(\vec{F} = \vec{F_1} + \vec{F_2} = (8+6)\hat{i} + (2+2)\hat{j} + (-6-2)\hat{k} = 14\hat{i} + 4\hat{j} - 8\hat{k}\).
Displacement \(\vec{d} = (5-1)\hat{i} + (4-2)\hat{j} + (1-3)\hat{k} = 4\hat{i} + 2\hat{j} - 2\hat{k}\).
Work Done \(W = \vec{F} \cdot \vec{d} = (14)(4) + (4)(2) + (-8)(-2) = 56 + 8 + 16 = 80\) units.
37. Show that \(lim_{x\rightarrow0^{+}}x~log~x\) is 0.
Solution:
Limit is \(0 \times (-\infty)\) form. Rewrite as \(\frac{\log x}{1/x}\) (\(\frac{\infty}{\infty}\) form).
Apply L'Hopital's Rule:
\(= \lim_{x\to 0} \frac{1/x}{-1/x^2} = \lim_{x\to 0} (-x) = 0\).
Limit is \(0 \times (-\infty)\) form. Rewrite as \(\frac{\log x}{1/x}\) (\(\frac{\infty}{\infty}\) form).
Apply L'Hopital's Rule:
\(= \lim_{x\to 0} \frac{1/x}{-1/x^2} = \lim_{x\to 0} (-x) = 0\).
38. Find the approximate increase in the area of a circular plate of radius 10 cm, when its radius increases by 0.02 cm.
Solution:
Let \(A\) be the area of the circle. \(A = \pi r^2\).
Approximate change \(dA \approx \frac{dA}{dr} \Delta r\).
\(\frac{dA}{dr} = 2\pi r\).
Given \(r = 10\) and \(dr = 0.02\).
\(dA = (2\pi r) dr = 2\pi(10)(0.02)\).
\(dA = 20\pi (0.02) = 0.4\pi\) cm\(^2\).
Let \(A\) be the area of the circle. \(A = \pi r^2\).
Approximate change \(dA \approx \frac{dA}{dr} \Delta r\).
\(\frac{dA}{dr} = 2\pi r\).
Given \(r = 10\) and \(dr = 0.02\).
\(dA = (2\pi r) dr = 2\pi(10)(0.02)\).
\(dA = 20\pi (0.02) = 0.4\pi\) cm\(^2\).
39. Let * be defined on \(N\) by \(a*b = a^b\). Check whether * is commutative and associative.
Solution:
Commutative:
\(a*b = a^b\) and \(b*a = b^a\).
Take \(a=2, b=3\). \(2*3 = 2^3 = 8\). \(3*2 = 3^2 = 9\).
\(a*b \neq b*a\). Hence, * is not commutative.
Associative:
Check \((a*b)*c = a*(b*c)\).
LHS: \((a*b)*c = (a^b)*c = (a^b)^c = a^{bc}\).
RHS: \(a*(b*c) = a*(b^c) = a^{(b^c)}\).
Take \(a=2, b=3, c=2\).
LHS \(= 2^{3 \times 2} = 2^6 = 64\).
RHS \(= 2^{(3^2)} = 2^9 = 512\).
LHS \(\neq\) RHS. Hence, * is not associative.
Commutative:
\(a*b = a^b\) and \(b*a = b^a\).
Take \(a=2, b=3\). \(2*3 = 2^3 = 8\). \(3*2 = 3^2 = 9\).
\(a*b \neq b*a\). Hence, * is not commutative.
Associative:
Check \((a*b)*c = a*(b*c)\).
LHS: \((a*b)*c = (a^b)*c = (a^b)^c = a^{bc}\).
RHS: \(a*(b*c) = a*(b^c) = a^{(b^c)}\).
Take \(a=2, b=3, c=2\).
LHS \(= 2^{3 \times 2} = 2^6 = 64\).
RHS \(= 2^{(3^2)} = 2^9 = 512\).
LHS \(\neq\) RHS. Hence, * is not associative.
40. Evaluate \(\int_{0}^{1}x~e^{x}dx\).
Solution:
Integration by parts: \(\int u dv = uv - \int v du\).
Let \(u=x \implies du=dx\).
Let \(dv=e^x dx \implies v=e^x\).
\(\int_0^1 x e^x dx = [xe^x]_0^1 - \int_0^1 e^x dx\)
\(= (1\cdot e^1 - 0) - [e^x]_0^1\)
\(= e - (e^1 - e^0) = e - e + 1 = 1\).
Integration by parts: \(\int u dv = uv - \int v du\).
Let \(u=x \implies du=dx\).
Let \(dv=e^x dx \implies v=e^x\).
\(\int_0^1 x e^x dx = [xe^x]_0^1 - \int_0^1 e^x dx\)
\(= (1\cdot e^1 - 0) - [e^x]_0^1\)
\(= e - (e^1 - e^0) = e - e + 1 = 1\).
PART - IV
Note: Answer all the questions.
41. (a) Solve the system of linear equations by Cramer's Rule \(3x+3y-z=11\), \(2x-y+2z=9\), \(4x+3y+2z=25\).
(i) Compute the maximum height of the particle reached.
(ii) What is the velocity when the particle hits the ground?
Solution:
\(\Delta = \begin{vmatrix}3&3&-1\\ 2&-1&2\\ 4&3&2\end{vmatrix} = 3(-2-6) -3(4-8) -1(6+4) = -24 + 12 - 10 = -22\).
\(\Delta_x = \begin{vmatrix}11&3&-1\\ 9&-1&2\\ 25&3&2\end{vmatrix} = -44\). \(x = \frac{-44}{-22} = 2\).
\(\Delta_y = \begin{vmatrix}3&11&-1\\ 2&9&2\\ 4&25&2\end{vmatrix} = -22\). \(y = \frac{-22}{-22} = 1\).
\(\Delta_z = \begin{vmatrix}3&3&11\\ 2&-1&9\\ 4&3&25\end{vmatrix} = -66\). \(z = \frac{-66}{-22} = 3\).
Solution: \(\{2, 1, 3\}\).
\(\Delta = \begin{vmatrix}3&3&-1\\ 2&-1&2\\ 4&3&2\end{vmatrix} = 3(-2-6) -3(4-8) -1(6+4) = -24 + 12 - 10 = -22\).
\(\Delta_x = \begin{vmatrix}11&3&-1\\ 9&-1&2\\ 25&3&2\end{vmatrix} = -44\). \(x = \frac{-44}{-22} = 2\).
\(\Delta_y = \begin{vmatrix}3&11&-1\\ 2&9&2\\ 4&25&2\end{vmatrix} = -22\). \(y = \frac{-22}{-22} = 1\).
\(\Delta_z = \begin{vmatrix}3&3&11\\ 2&-1&9\\ 4&3&25\end{vmatrix} = -66\). \(z = \frac{-66}{-22} = 3\).
Solution: \(\{2, 1, 3\}\).
OR
41. (b) A particle is fired straight up from the ground to reach a height of s feet in t seconds, where \(s(t)=128t-16t^{2}\).(i) Compute the maximum height of the particle reached.
(ii) What is the velocity when the particle hits the ground?
Solution:
Velocity \(v(t) = s'(t) = 128 - 32t\).
(i) Max height when \(v=0 \implies 128=32t \implies t=4\)s.
Max Height \(s(4) = 128(4) - 16(16) = 512 - 256 = 256\) ft.
(ii) Hits ground when \(s(t)=0 \implies 16t(8-t)=0 \implies t=8\)s.
Velocity at \(t=8\): \(v(8) = 128 - 32(8) = 128 - 256 = -128\) ft/s.
Velocity \(v(t) = s'(t) = 128 - 32t\).
(i) Max height when \(v=0 \implies 128=32t \implies t=4\)s.
Max Height \(s(4) = 128(4) - 16(16) = 512 - 256 = 256\) ft.
(ii) Hits ground when \(s(t)=0 \implies 16t(8-t)=0 \implies t=8\)s.
Velocity at \(t=8\): \(v(8) = 128 - 32(8) = 128 - 256 = -128\) ft/s.
42. (a) Show that \((2+i\sqrt{3})^{10}-(2-i\sqrt{3})^{10}\) is purely imaginary.
Solution:
Let \(z = 2+i\sqrt{3}\). Then \(\bar{z} = 2-i\sqrt{3}\).
Expression is \(z^{10} - \bar{z}^{10}\).
Property: \(w - \bar{w}\) is always purely imaginary (equals \(2i Im(w)\)).
Alternatively, let \(z^{10} = x+iy\), then \(\bar{z}^{10} = x-iy\).
Diff \(= (x+iy)-(x-iy) = 2iy\), which is purely imaginary.
Let \(z = 2+i\sqrt{3}\). Then \(\bar{z} = 2-i\sqrt{3}\).
Expression is \(z^{10} - \bar{z}^{10}\).
Property: \(w - \bar{w}\) is always purely imaginary (equals \(2i Im(w)\)).
Alternatively, let \(z^{10} = x+iy\), then \(\bar{z}^{10} = x-iy\).
Diff \(= (x+iy)-(x-iy) = 2iy\), which is purely imaginary.
OR
42. (b) Find the area of the region bounded by the ellipse \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\) using integration.
Solution:
\(y = \frac{b}{a}\sqrt{a^2-x^2}\).
Area = \(4 \int_0^a y dx = 4\frac{b}{a} \int_0^a \sqrt{a^2-x^2} dx\).
Standard integral: \(\int \sqrt{a^2-x^2}dx = \frac{x}{2}\sqrt{a^2-x^2} + \frac{a^2}{2}sin^{-1}(\frac{x}{a})\).
Limit 0 to a: \((0 + \frac{a^2}{2}(\frac{\pi}{2})) - 0 = \frac{\pi a^2}{4}\).
Total Area = \(4 \times \frac{b}{a} \times \frac{\pi a^2}{4} = \pi ab\).
\(y = \frac{b}{a}\sqrt{a^2-x^2}\).
Area = \(4 \int_0^a y dx = 4\frac{b}{a} \int_0^a \sqrt{a^2-x^2} dx\).
Standard integral: \(\int \sqrt{a^2-x^2}dx = \frac{x}{2}\sqrt{a^2-x^2} + \frac{a^2}{2}sin^{-1}(\frac{x}{a})\).
Limit 0 to a: \((0 + \frac{a^2}{2}(\frac{\pi}{2})) - 0 = \frac{\pi a^2}{4}\).
Total Area = \(4 \times \frac{b}{a} \times \frac{\pi a^2}{4} = \pi ab\).
43. (a) Show that the value of \(sin^{-1}(sin\frac{5\pi}{9}cos\frac{\pi}{9}+cos\frac{5\pi}{9}sin\frac{\pi}{9})\) is \(\frac{\pi}{3}\).
Solution:
Using \(sin(A+B)\): \(sin(\frac{5\pi}{9} + \frac{\pi}{9}) = sin(\frac{6\pi}{9}) = sin(\frac{2\pi}{3})\).
Given expression becomes \(sin^{-1}(sin\frac{2\pi}{3})\).
Range of \(sin^{-1}\) is \([-\pi/2, \pi/2]\). \(\frac{2\pi}{3}\) is outside.
\(sin(\frac{2\pi}{3}) = sin(\pi - \frac{\pi}{3}) = sin(\frac{\pi}{3})\).
Result: \(\frac{\pi}{3}\).
Using \(sin(A+B)\): \(sin(\frac{5\pi}{9} + \frac{\pi}{9}) = sin(\frac{6\pi}{9}) = sin(\frac{2\pi}{3})\).
Given expression becomes \(sin^{-1}(sin\frac{2\pi}{3})\).
Range of \(sin^{-1}\) is \([-\pi/2, \pi/2]\). \(\frac{2\pi}{3}\) is outside.
\(sin(\frac{2\pi}{3}) = sin(\pi - \frac{\pi}{3}) = sin(\frac{\pi}{3})\).
Result: \(\frac{\pi}{3}\).
OR
43. (b) The parabolic communication antenna has a focus at 2 mts distance from the vertex of the antenna. Show that the width of the antenna 3 mts from the vertex is \(4\sqrt{6}\) mts.
Solution:
Equation: \(y^2 = 4ax\). Focus \(a=2\), so \(y^2 = 8x\).
At \(x=3\) (depth), \(y^2 = 8(3) = 24\).
\(y = \sqrt{24} = 2\sqrt{6}\).
Total width = \(2y = 4\sqrt{6}\) mts.
Equation: \(y^2 = 4ax\). Focus \(a=2\), so \(y^2 = 8x\).
At \(x=3\) (depth), \(y^2 = 8(3) = 24\).
\(y = \sqrt{24} = 2\sqrt{6}\).
Total width = \(2y = 4\sqrt{6}\) mts.
44. (a) Solve the differential equation \(\frac{dy}{dx}+\frac{y}{x}=sin~x\).
Solution:
Linear form \(\frac{dy}{dx} + Py = Q\). \(P=1/x, Q=sin~x\).
Integrating Factor (IF) = \(e^{\int (1/x)dx} = e^{ln~x} = x\).
Solution: \(y(IF) = \int Q(IF)dx + c\).
\(xy = \int x~sin~x~dx\).
Integrate by parts: \(-x~cos~x + sin~x + c\).
\(y = \frac{-x~cos~x + sin~x + c}{x}\).
Linear form \(\frac{dy}{dx} + Py = Q\). \(P=1/x, Q=sin~x\).
Integrating Factor (IF) = \(e^{\int (1/x)dx} = e^{ln~x} = x\).
Solution: \(y(IF) = \int Q(IF)dx + c\).
\(xy = \int x~sin~x~dx\).
Integrate by parts: \(-x~cos~x + sin~x + c\).
\(y = \frac{-x~cos~x + sin~x + c}{x}\).
OR
44. (b) Verify whether the following compound proposition is tautology or contradiction or contingency: \((p\rightarrow q)\leftrightarrow(\neg p\rightarrow q)\).
Solution:
Construct Truth Table.
If p=T, q=T: \((T\to T) \leftrightarrow (F \to T) \implies T \leftrightarrow T \implies T\).
If p=T, q=F: \((T\to F) \leftrightarrow (F \to F) \implies F \leftrightarrow T \implies F\).
Since output contains both T and F, it is a Contingency.
Construct Truth Table.
If p=T, q=T: \((T\to T) \leftrightarrow (F \to T) \implies T \leftrightarrow T \implies T\).
If p=T, q=F: \((T\to F) \leftrightarrow (F \to F) \implies F \leftrightarrow T \implies F\).
Since output contains both T and F, it is a Contingency.
45. (a) Prove by using vector method that \(cos(A-B)=cosA~cosB+sinA~sinB.\)
Solution:
Take unit vectors \(\hat{a}\) and \(\hat{b}\) making angles A and B with x-axis.
\(\hat{a} = cosA\hat{i} + sinA\hat{j}\).
\(\hat{b} = cosB\hat{i} + sinB\hat{j}\).
Dot product \(\hat{a} \cdot \hat{b} = |\hat{a}||\hat{b}|cos(A-B) = cos(A-B)\).
Also \(\hat{a} \cdot \hat{b} = cosA~cosB + sinA~sinB\).
Equating both gives the result.
Take unit vectors \(\hat{a}\) and \(\hat{b}\) making angles A and B with x-axis.
\(\hat{a} = cosA\hat{i} + sinA\hat{j}\).
\(\hat{b} = cosB\hat{i} + sinB\hat{j}\).
Dot product \(\hat{a} \cdot \hat{b} = |\hat{a}||\hat{b}|cos(A-B) = cos(A-B)\).
Also \(\hat{a} \cdot \hat{b} = cosA~cosB + sinA~sinB\).
Equating both gives the result.
OR
45. (b) Prove that among all the rectangles of the given perimeter, the square has the maximum area.
Solution:
Perimeter \(2(x+y) = P\) (constant). \(y = P/2 - x\).
Area \(A = xy = x(P/2 - x) = Px/2 - x^2\).
\(dA/dx = P/2 - 2x\). For max, \(P/2 - 2x = 0 \implies x = P/4\).
Then \(y = P/2 - P/4 = P/4\).
Since \(x=y\), the rectangle is a square.
Perimeter \(2(x+y) = P\) (constant). \(y = P/2 - x\).
Area \(A = xy = x(P/2 - x) = Px/2 - x^2\).
\(dA/dx = P/2 - 2x\). For max, \(P/2 - 2x = 0 \implies x = P/4\).
Then \(y = P/2 - P/4 = P/4\).
Since \(x=y\), the rectangle is a square.
46. (a) Find the eccentricity, foci, vertices and centre for the ellipse \(\frac{x^{2}}{25}+\frac{y^{2}}{9}=1\) and draw the rough diagram.
Solution:
\(a^2=25, b^2=9\). Major axis along x-axis.
Centre: (0,0).
Eccentricity \(e = \sqrt{1 - 9/25} = \sqrt{16/25} = 4/5\).
Vertices \((\pm a, 0) = (\pm 5, 0)\).
Foci \((\pm ae, 0) = (\pm 4, 0)\).
\(a^2=25, b^2=9\). Major axis along x-axis.
Centre: (0,0).
Eccentricity \(e = \sqrt{1 - 9/25} = \sqrt{16/25} = 4/5\).
Vertices \((\pm a, 0) = (\pm 5, 0)\).
Foci \((\pm ae, 0) = (\pm 4, 0)\).
OR
46. (b) The cumulative distribution function of a discrete random variable is given. Find (i) The probability mass function (ii) \(P(x<3)\) and (iii) \(P(x\ge2)\).
Solution:
Based on the step values in the standard CDF problem context:
(i) PMF:
\(P(0) = F(0) - F(0^-) = 1/2\)
\(P(1) = F(1) - F(0) = 3/5 - 1/2 = 1/10\)
\(P(2) = F(2) - F(1) = 4/5 - 3/5 = 1/5\)
\(P(3) = F(3) - F(2) = 9/10 - 4/5 = 1/10\)
\(P(4) = 1 - 9/10 = 1/10\)
(ii) \(P(X < 3) = P(0) + P(1) + P(2) = 1/2 + 1/10 + 1/5 = 8/10 = 4/5\).
(iii) \(P(X \ge 2) = 1 - P(X < 2) = 1 - (1/2 + 1/10) = 1 - 0.6 = 0.4\) (or \(2/5\)).
Based on the step values in the standard CDF problem context:
(i) PMF:
\(P(0) = F(0) - F(0^-) = 1/2\)
\(P(1) = F(1) - F(0) = 3/5 - 1/2 = 1/10\)
\(P(2) = F(2) - F(1) = 4/5 - 3/5 = 1/5\)
\(P(3) = F(3) - F(2) = 9/10 - 4/5 = 1/10\)
\(P(4) = 1 - 9/10 = 1/10\)
(ii) \(P(X < 3) = P(0) + P(1) + P(2) = 1/2 + 1/10 + 1/5 = 8/10 = 4/5\).
(iii) \(P(X \ge 2) = 1 - P(X < 2) = 1 - (1/2 + 1/10) = 1 - 0.6 = 0.4\) (or \(2/5\)).
47. (a) Show that the area between the parabola \(y^{2}=16x\) and its latus rectum (using integration) is \(\frac{128}{3}\) sq. units.
Detailed Solution:
Step 1: Identify the curve and limits.
The equation of the parabola is \(y^2 = 16x\). This is a rightward opening parabola of the form \(y^2 = 4ax\).
Comparing \(4a = 16\), we get \(a = 4\).
The latus rectum is the vertical chord passing through the focus \((a, 0)\). Therefore, the equation of the latus rectum is the line \(x = 4\).
Step 2: Set up the area integral.
The parabola is symmetric about the x-axis. The total area is twice the area of the region in the first quadrant (from \(x=0\) to \(x=4\)).
\(Area = 2 \int_{0}^{4} y \, dx\)
From \(y^2 = 16x\), we have \(y = \sqrt{16x} = 4\sqrt{x}\) (taking positive root for the 1st quadrant).
Step 3: Evaluate the integral.
\(Area = 2 \int_{0}^{4} 4\sqrt{x} \, dx\)
\(= 8 \int_{0}^{4} x^{1/2} \, dx\)
\(= 8 \left[ \frac{x^{3/2}}{3/2} \right]_{0}^{4}\)
\(= 8 \times \frac{2}{3} \left[ x^{3/2} \right]_{0}^{4}\)
\(= \frac{16}{3} \left[ 4^{3/2} - 0^{3/2} \right]\)
Step 4: Final Calculation.
\(4^{3/2} = (2^2)^{3/2} = 2^3 = 8\).
\(Area = \frac{16}{3} (8) = \frac{128}{3}\) sq. units.
Hence proved.
Step 1: Identify the curve and limits.
The equation of the parabola is \(y^2 = 16x\). This is a rightward opening parabola of the form \(y^2 = 4ax\).
Comparing \(4a = 16\), we get \(a = 4\).
The latus rectum is the vertical chord passing through the focus \((a, 0)\). Therefore, the equation of the latus rectum is the line \(x = 4\).
Step 2: Set up the area integral.
The parabola is symmetric about the x-axis. The total area is twice the area of the region in the first quadrant (from \(x=0\) to \(x=4\)).
\(Area = 2 \int_{0}^{4} y \, dx\)
From \(y^2 = 16x\), we have \(y = \sqrt{16x} = 4\sqrt{x}\) (taking positive root for the 1st quadrant).
Step 3: Evaluate the integral.
\(Area = 2 \int_{0}^{4} 4\sqrt{x} \, dx\)
\(= 8 \int_{0}^{4} x^{1/2} \, dx\)
\(= 8 \left[ \frac{x^{3/2}}{3/2} \right]_{0}^{4}\)
\(= 8 \times \frac{2}{3} \left[ x^{3/2} \right]_{0}^{4}\)
\(= \frac{16}{3} \left[ 4^{3/2} - 0^{3/2} \right]\)
Step 4: Final Calculation.
\(4^{3/2} = (2^2)^{3/2} = 2^3 = 8\).
\(Area = \frac{16}{3} (8) = \frac{128}{3}\) sq. units.
Hence proved.
OR
47. (b) Show that the Cartesian equation of the plane passing through the points (a, 0, 0), (0, b, 0), and (0, 0, c) is \(\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1\).
Detailed Solution:
Method 1: Using the General Equation of a Plane
Let the equation of the plane be \(Ax + By + Cz + D = 0\).
1. Since it passes through \((a, 0, 0)\):
\(A(a) + B(0) + C(0) + D = 0 \Rightarrow Aa = -D \Rightarrow A = \frac{-D}{a}\).
2. Since it passes through \((0, b, 0)\):
\(A(0) + B(b) + C(0) + D = 0 \Rightarrow Bb = -D \Rightarrow B = \frac{-D}{b}\).
3. Since it passes through \((0, 0, c)\):
\(A(0) + B(0) + C(c) + D = 0 \Rightarrow Cc = -D \Rightarrow C = \frac{-D}{c}\).
Substitute values of A, B, and C back into the general equation:
\(\left(\frac{-D}{a}\right)x + \left(\frac{-D}{b}\right)y + \left(\frac{-D}{c}\right)z + D = 0\)
Divide the entire equation by \(-D\) (where \(D \neq 0\)):
\(\frac{x}{a} + \frac{y}{b} + \frac{z}{c} - 1 = 0\)
\(\Rightarrow \frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1\).
Hence proved.
Method 1: Using the General Equation of a Plane
Let the equation of the plane be \(Ax + By + Cz + D = 0\).
1. Since it passes through \((a, 0, 0)\):
\(A(a) + B(0) + C(0) + D = 0 \Rightarrow Aa = -D \Rightarrow A = \frac{-D}{a}\).
2. Since it passes through \((0, b, 0)\):
\(A(0) + B(b) + C(0) + D = 0 \Rightarrow Bb = -D \Rightarrow B = \frac{-D}{b}\).
3. Since it passes through \((0, 0, c)\):
\(A(0) + B(0) + C(c) + D = 0 \Rightarrow Cc = -D \Rightarrow C = \frac{-D}{c}\).
Substitute values of A, B, and C back into the general equation:
\(\left(\frac{-D}{a}\right)x + \left(\frac{-D}{b}\right)y + \left(\frac{-D}{c}\right)z + D = 0\)
Divide the entire equation by \(-D\) (where \(D \neq 0\)):
\(\frac{x}{a} + \frac{y}{b} + \frac{z}{c} - 1 = 0\)
\(\Rightarrow \frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1\).
Hence proved.
Original Question Paper Pages
