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Previous year question paper with solutions for Math Mar-2018
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Question paper 1
1. (i) If is a binary operation such that a * b = a2 +b2 then 3 * 5 is
(a) 34 (b) 9 (c) 8 (d) 25Answer:
(ii) If cos-1 x = y then
(a) −๐/2 <= y <= ๐/2 (b) -π <=y <= π(c) 0 <= y <= ๐/2 (d) 0 <= y <= π
Answer:
(iii) If A is a matrix of order 3x3 and |A| =10 then |adj• Aเฅค is
(a) 0 (b) 10 (c) 100 d) 1000Answer:
(iv) If y = sin (sin-1 x + cos-1 x), x € [-1, 1] then dy/dx is
(a) ๐/2 (b) −๐/2 (c) 0 d) 1
Answer:
(v) If
f(x) = { \( {{{sinx\over x}\over k-1} , {x! = 0 \over x=0}}\) } ,๐ฅ = 0 is continuous at x=0 then
(a) 2 (b) 0 (c) -1 (d) 1Answer:
(vi) \(\int e^x ( log x {1\over x}) \) dx is equal to
(a) ex + c (b) ex logx + c (c) ex/x + c (d) log x + c
Answer:
(vii) Integrating factor of differential equation dy/dx + ๐ฆ = 3 is
(a) x (b) e (c) ex (d) logxAnswer:
(vii) This inequality |a . b| <= |a| |b| 5 is called
(a) Cauchy-Schwartz inequality (b) Triangle inequality
(c) Rolle's Theorem (d) Lagrange's Mean Value theoremAnswer:
(ix) Distance between plane 3x +4y-20 = 0 and point (0, 0,-7) is
(a) 4 units (b) 3 units (c) 2 units (d) 1 unitAnswer:
(x) If P(E) denotes probability of occurrence of event E then
(a) P (E) € [-1, 1] (b) P (E) € (1, 2) (c) P (E) € (0, 1) (d) P (E) € [0, 1]Answer:
2. If matrix A = [aij]3 x 2 , and aij = (3i-2j)2 or matrix A find them
Answer:
3. Check whether Lagrange's mean value theorem is applicable on f(x) = sin x + cos x Interval [0, ๐/2 ]
Answer:
4. Evaluate \(\int\limits_0^{x/2} sin^3X / sin^3X + cos^3X { sin^3x \over sin^3x + cos^3x }dx\)
Answer:
5. Evaluate \(\int {7dx \over x(x^7-1)}\)
Answer:
6. Find particular solution of differential equation \({dy \over dx} {1+y^2 \over 1+x^2}\) given that x=0 or y= 1
Answer:
\({dy \over dx} {1+y^2 \over 1+x^2}\) given that y(0) = 1
\({dy \over 1+y^2} = {dx \over 1+ x^2}\)
Integrating both side, we get :
\(\int{dy \over 1+y^2} =\int {dx \over 1+ x^2} + c\)
\(tan^{-1}y = tan^{-1}x + c\)
\(tan^{-1}(1) = tan^{-1}(0) + c\)
\(tan^{-1}tan ({\pi \over 4}) = tan^{-1}tan(0) + c\)
\(c = {\pi \over 4}\)
\(tan^{-1}y = tan^{-1}x + {\pi \over 4}\)
7. Form differential equation representing the family of lines making equal intercepts on the co-ordinate axes.
Answer:
8. Find the angle between the plane 2x+3 y-5z= 10 and the line passing from the points (2, 3,-1) 2 Or (1, 2, 1)
Answer:
9. If P (A) = 7/13, P (B) = 9/13 and P (AUB) = 12/13 then find P(A|B)
Answer:
P (A) = 7/13, P (B) = 9/13
P (AUB) = 12/13 P(A|B) = ?
P(A|B) = \(P(A \bigcap B) \over P(B)\)
\(P (A \bigcup B) = P(A) + P(B) - P (A \bigcap B)\)
\({12 \over 13 } = {7 \over 13 } + {9 \over 13} - P(A \bigcap B)\)
\({12 \over 13 } = {16 \over 13} - P(A \bigcap B)\)
\(P(A \bigcap B) = {16 -12 \over 13} = {4 \over 3}\)
\(P(A | B) = {{4 \over 13} \over {9 \over 13}}\)
\(P(A | B) = {4 \over 9}\)
10 Prove that function f : R --> R, f(x) = \(3โ2๐ฅ \over 7\) in one-one and onto. Also find f-1
Answer:
f(x) = \(3โ2๐ฅ \over 7\)
7 : R --> R x € R
7(x) is 1- 7:
For x1 x2 € R
7(x1) = f(x2)
\(3โ2๐ฅ_1 \over 7\) = \(3โ2๐ฅ_2 \over 7\)
2x1 = 2x2
x1 = x2
.·. f is 1 - 1
f is onto : Let K € R
f(x) = K
\(3โ2๐ฅ \over 7\) = K => 3 - 2x = 7K
x = \(3โ7K \over 2 \) => R
.·. 7(x) is onto
Now, 7-1
For x € R, Let f-1 (x) = K
=> 7(K) = x
=> \(3-2K \over 7 \) = x
3 - 2K = 7x
-2K = 7x - 3
K = \(3 - 7x \over 2 \)
For x € R , 7-1 (x) = \(3 - 7x \over 2 \)
11. Prove that : sin-1 ( \(5\over13\) ) +cos-1 (\(4\over 5\)) = \(1\over 2\) sin-1 \(3696 \over 4225\)
Answer:
12. Express \( \begin{bmatrix} 2 & 5 & -1 \\ 3 & 1 & 5 \\ 7 & 6 & 9 \end{bmatrix}\) as sum of symmetric and skew-symmetric matrices.
Answer:
A = \( \begin{bmatrix} 2 & 5 & -1 \\ 3 & 1 & 5 \\ 7 & 6 & 9 \end{bmatrix}\)
\(A^{t} = \) \( \begin{bmatrix} 2 & 3 & -7 \\ 5 & 1 & 6 \\ -1 & 5 & 9 \end{bmatrix}\)
For Symmetric Matrix : \(A + A^T \over 2\)
\(A + A^T \over 2\) = \(1 \over 2\) = \( \begin{bmatrix} 2 & 5 & -1 \\ 3 & 1 & 5 \\ 7 & 6 & 9 \end{bmatrix}\) + \( \begin{bmatrix} 2 & 3 & -7 \\ 5 & 1 & 6 \\ -1 & 5 & 9 \end{bmatrix}\)
\(A + A^T \over 2\) = \(1 \over 2\) = \( \begin{bmatrix} 4 & 8 & 6 \\ 8 & 2 & 11 \\ 6 & 11 & 18 \end{bmatrix}\)
For Skew-symmetric Matrix = \(A + A^T \over 2\)
\(A + A^T \over 2\) = \(1 \over 2\) = \( \begin{bmatrix} 2 & 5 & -1 \\ 3 & 1 & 5 \\ 7 & 6 & 9 \end{bmatrix}\) + \( \begin{bmatrix} 2 & 3 & -7 \\ 5 & 1 & 6 \\ -1 & 5 & 9 \end{bmatrix}\)
\(A + A^T \over 2\) = \(1 \over 2\) = \( \begin{bmatrix} 0 & 2 & -8 \\ -2 & 0 & -1 \\ 8 & 1 & 0 \end{bmatrix}\)
Or
If x,y,z are different and \( \begin{vmatrix} x & x^2 & 1 + x^3 \\ y & y^2 & 1 + y^3 \\ z & z^2 & 1 + z^3 \end{vmatrix}\) = 0 then prove that xyz =-1
Answer:
\( \begin{vmatrix} x & x^2 & 1 + x^3 \\ y & y^2 & 1 + y^3 \\ z & z^2 & 1 + z^3 \end{vmatrix}\) = 0
\( \begin{vmatrix} x & x^2 & 1 \\ y & y^2 & 1 \\ z & z^2 & 1 \end{vmatrix}\) + \( \begin{vmatrix} x & x^2 & x^3 \\ y & y^2 & y^3 \\ z & z^2 & z^3 \end{vmatrix}\) = 0
=> A1 + A2 = 0
A2 = \( \begin{vmatrix} x & x^2 & x^3 \\ y & y^2 & y^3 \\ z & z^2 & z^3 \end{vmatrix}\)
A2 = xyz \( \begin{vmatrix} 1 & x & x^2 \\ 1 & y & y^2 \\ 1 & z & z^2 \end{vmatrix}\)
A2 = xyz \( \begin{vmatrix} x & x^2 & 1 \\ y & y^2 & 1 \\ z & z^2 & 1 \end{vmatrix}\)
A2 = xyz A1
A1 + A2 = 0
A1 + xyz A1 = 0
A1 ( 1 + xyz ) = 0
A1 = xyz \( \begin{vmatrix} x & x^2 & 1 \\ y & y^2 & 1 \\ z & z^2 & 1 \end{vmatrix}\)
R2 + R2 - R1 & R3 + R3 - R1
A1 = xyz \( \begin{vmatrix} x & x^2 & 1 \\ y-x & y^2-x^2 & 0 \\ z-x & z^2-x^2 & 0 \end{vmatrix}\)
= (y-x) (z-x) = \( \begin{vmatrix} x & x^2 & 1 \\ 1 & y^2-x^2 & 0 \\ 1 & z^2-x^2 & 0 \end{vmatrix}\)
= (y-x) (z-x) (3 + x - y - x)
A1 = (y - x) (z - x) (z - y)
= (y - x) (z - x)(z - y) (1 + xyz) = 0
=> 1 + x y z = 0
xyz = -1
13. If y = (x)tanx + (tanx)x then find ๐๐ฆ/๐๐ฅ
Answer:
y = (x)tanx + (tanx)x
Let u = xtanx , v = tanxx
Now x = u
log u = tanx.logx
\({1 \over u} . {du \over dx} = {tan x \over x }+ logx.sec^2 x\)
\({du \over dx} = u [ {tanx\over x} + sec^2x . logx]\)
\({du \over dx} = x^{tanx} [{tanx \over x} + sec^2 x .logx]\)
\( v = (tanx)^x\)
\(logv = x log (tanx)\)
\({1 \over v} .{dv \over dx} = x.{1 \over tan}.sec^2x + 1 .log(tanx)\)
\({dv \over dx} = (tanx)^x [x .{cosx \over sin x}.{1 \over cos ^2x} + log(tanx)] \)
\({dv \over dx } = (tanx)^x [{1 \over sinx.cosx} + log(tanx)]\)
\({dy \over dx} = {du \over dx} + {dv \over dx}\)
\({dy\over dx} = x^{tanx} [{tanx \over x} + sec^2x .logx] + (tanx)^x [{x\over sinxcosx} + log tanx]\)
14. Using differentials find approximate value of (0.37)1/2
Answer:
\(\sqrt {0.37} = \sqrt {0.36 + 0.01}\)
because \(\sqrt {0.36}\) = 0.6
let y = \(\sqrt x\)
y + Ay = \(\sqrt {x + Ax}\)
Ay = \(\sqrt {x + Ax}\) - \(\sqrt x\)
\(({dy \over dx})\)Ay = \(\sqrt {x + Ax}\) - \(\sqrt x\) ------------ 1
let x = 0.36 Ax = 0.1
from 1 => \({1 \over 2\sqrt x} . Ax = {\sqrt {0.36 + 0.01}} - {\sqrt {0.36}}\)
\({0.1 \over 2 * 0.6} = \sqrt {0.37} - 0.6\)
\({0.1 \over 1.2} + 0.6 = \sqrt {0.37}\)
\((0.37)^{1\over 2} = {0.82 \over 1.2 } = 0.6833\)
15. Evaluate \(\int {x^2 +1 \over x^4 +1}\) dx
Answer:
I = \(\int {x^2 +1 \over x^4 +1}\) dx
I = \(\int {x^2 \over x^4 +1}\). dx + \(\int {dx \over 1 + (x^2)^2}\)
I = \(\int {1 + {1 \over x^2} \over ({x^2 + 1 \over x^2})} dx\)
put \(x - {1 \over x} = b\)
\(({1 + {1 \over x^2}})dx = dt\)
\(\int {dt \over t^2 +2} = \int {dt \over t^2 + (\sqrt2)^2} \)
= \({1 \over \sqrt 2} tan^{-1} (t \sqrt 2) + c\)
=\({1 \over \sqrt 2} tan^{-1} ({x^2 - 1 \over \sqrt 2 x }) + c\)
Or
Evaluate \(\int {dx \over x^2+1}\)Answer:
I = \(\int {dx \over x^2+1}\)
I = \(tan^{-1}x + c\)
16. Find the area of region bounded by the ellipse \({x^2 \over 9 } + {y^2 \over 4} = 1\)
Answer:
The equation and ellipse is \({x^2 \over 9 } + {y^2 \over 4} = 1\)
\( {y^2 \over 4} = 1 -{x^2 \over 9 } \)
\(y^2 = {4 \over 9 } (9 - x^2)\)
\(y = {2\over 3} \sqrt {9-x^2}\)
The ellipse is symmetrical about both the axexs:
Req Area = 4 ( Area AOB)
= 4 \(\int_0^3 y.dx = 4 \int_0^3 {2 \over 3} \sqrt 9 - x^2 dx\)
= \({8 \over 3} \int_0^3 \sqrt{(3)^2- (x)^2} dx\)
= \({8 \over 3} [ {x \sqrt{(3)^2 - (x)^2}\over 2} + {(3)^2 \over 2} sin^{-1} {x \over 3} ]\)
= \(6 \pi \) square units
17. Find the particular solution of differential equation [x sin2 (y/x)-y] dx +xdy = 0;y(1)=๐/4
Answer:
[x sin2 (y/x)-y] dx +xdy = 0;y(1)=๐/4
y(1) = ๐/4
\(sin^2({y \over x}) - {y\over x} + {dy \over dx} = 0 \)
\({dy \over dx} = {y \over x} - sin ^2 ({y \over x})\)-----------1
Put y = vx
\({dy \over dx} = v + x{dv \over dx}\)
1 becomes,
\(v + {x {dv \over dx} } =v - sin^2 v\)
\({x {dv \over dx} } = -sin^2 v\)
\({dv \over sin ^2 v} = -{dx \over x}\)
\(\int cosec^2v dv = - \int {dx \over x}\)
-cotv = -log |x| + c
log |x| - cotv = c
log |x| + cot y/x = c
Now , y(1) = ๐/4
=> log|1| - cot (๐/4) = c
=> c = -1
log |x| - cot (y/x) = -1
Or
Find the particular solution of differential equation " given that tanx \(dy\over dx\) +y = 2x tan x + x2 , x != 0 given that y=0 when x = ๐/2
Answer:
tanx \(dy\over dx\) +y = 2x tan x + x2
\(y ({\pi\over 2 }) = 0\)
\(dy\over dx\) + y.cotx = 2x + x2 cot x
comparing it with \(dy\over dx\) + Py = Q
P = cot x ; Q = x2 cotx + 2x
\(\int P.dx = \int cotx dx = log sinx\)
I.F = \(e^{\int Pdx} e^{logsinx} = sinx\)
y sinx = \(\int ( { x^2 cot x + 2x}) sinx dx +c\)
y sinx = \(\int ( { x^2 cos x + 2xsinx }) dx +c\)
y sinx = \(\int { x^2 cos x dx + 2 \int} xsinx dx +c\)
y sinx = \(x^2 sinx + c\)
18. If à = 2í-3j+4k เจ เจคเฉ 6 = 5i +j-k represents sider parallelogram then find both diagonals and a unit vector perpendicular to both dingonals
Answer:
19. Two cards are drawn (without replacement from a well shulle distribution table and mean of number of kings.
Answer:
20. Solve the following system oflincar equations by matrix mehord:
x - 2y +3z = -5, 3 x +y +c= 8, 2x –y +2z = 1Answer:
x - 2y +3z = -5
3 x +y +c= 8
2x –y +2z = 1
A = \( \begin{vmatrix} 1 & -2 & 3 \\ 3 & 1 & 1 \\ 2 & -1 & 2 \end{vmatrix}\) ; x = \( \begin{vmatrix} x \\ y \\ z & \end{vmatrix}\) ; B = \( \begin{vmatrix} -5 \\ 8 \\ 1 & \end{vmatrix}\)
|A| = \( \begin{vmatrix} 1 & -2 & 3 \\ 3 & 1 & 1 \\ 2 & -1 & 2 \end{vmatrix}\)
= 1(2 + 1) + 2(6 - 2) + 3(-3 - 2)
= 3 + 2(4) + 3 (-5)
= 3 + 8 - 15 = 11 - 15 = -4 != 0
A11 = \( \begin{vmatrix} 1 & 1 & \\ -1 & 2 \end{vmatrix}\) = 2 + 1 = 3
A12 = - \( \begin{vmatrix} 3 & 1 & \\ 2 & 2 \end{vmatrix}\) = - ( 6 - 2 ) = - 4
A13 = \( \begin{vmatrix} 3 & 1 & \\ -2 & -1 \end{vmatrix}\) = -3 - 2 = -5
A21 = \( \begin{vmatrix} -2 & 3 & \\ -1 & 2 \end{vmatrix}\) = -(-4 + 3) = 1
A22 = \( \begin{vmatrix} 1 & 3 & \\ 2 & 2 \end{vmatrix}\) = 2 - 6 = -4
A23 = -\( \begin{vmatrix} 1 & 2 & \\ 2 & -1 \end{vmatrix}\) = -(-1 + 4) = -3
A31 = \( \begin{vmatrix} -2 & 3 & \\ 1 & 1 \end{vmatrix}\) = -2 - 3 = -5
A32 = -\( \begin{vmatrix} 1 & 3 & \\ 3 & 1 \end{vmatrix}\) = -(1 - 9) = 8
A33 = \( \begin{vmatrix} 1 & -2 & \\ 3 & 1 \end{vmatrix}\) = 1 + 6 = 7
adj A = \( \begin{vmatrix} 3 & -4 & 5 \\ 1 & -4 & -3 \\ -5 & 8 & 7 \end{vmatrix}^t\)
= \( \begin{vmatrix} 3 & 1 & -5 \\ -4 &-4 & 8 \\ 5 & -3 & 7 \end{vmatrix}\)
\(A^{-1} = {adj A \over |A|}\)
\(A^{-1} = {1\over 4}\)\( \begin{vmatrix} 3 & 1 & -5 \\ -4 &-4 & 8 \\ 5 & -3 & 7 \end{vmatrix}\)
x = \(A^{-1}B = {1\over 4}\) = \( \begin{vmatrix} 3 & 1 & -5 \\ -4 &-4 & 8 \\ 5 & -3 & 7 \end{vmatrix}\)\( \begin{vmatrix} -5 \\ 8 \\ 1 & \end{vmatrix}\)
X = \(- {1\over 4}\) \( \begin{vmatrix} -15 & 8 & -5 \\ -20 &-32 & 8 \\ 25 & -24 & 7 \end{vmatrix}\)
X = \(- {1\over 4}\)\( \begin{vmatrix} -12 \\ -44 \\ -42 & \end{vmatrix}\) = \( \begin{vmatrix} 3 \\ 11 \\ {42 \over 4} & \end{vmatrix}\)
Or
Using elementary transformations find inverse of \( \begin{vmatrix} 2 & 4 & 1 \\ 1 & 2 & 3 \\ 1 & -3 & 0 \end{vmatrix}\)Answer:
A = \( \begin{vmatrix} 2 & 4 & 1 \\ 1 & 2 & 3 \\ 1 & -3 & 0 \end{vmatrix}\)
21. A window is in the form of rectangle surmounted by a semi-circular opening. The perimeter of window is 30 m. Find the dimensions of window so that it can admit maximum light through the whole opening.
Answer:
Or
Prove that volume of largest cone, which can be inscribed in a sphere, is 8/27 part of sphere.Answer:
22. Find the distance between the point (2, 3,-1) and foot of perpendicular drawn from (3, 1) to the plane X-y+3 z=10
Answer:
23. Find the equation of plane passing from the point A (2.-1, 1), B (4.3, 2) and C (6,5,-? Also prove that point (5. - 1, lies on the plane given by points A, B and C.
Answer:
24. Maximise and minimise : Z=15x + 30y Subject to the constraints : x+y <= 8, 2x +y >= 28, x - 2y >=0, x, y >= 0
Answer:
Or
Maximise and minimize Z = 4x + 3y -7 Subject to the constraints : x+y <= 10, x +y >= 3, x<=8, y <= 9 , x , y >-0Answer:
Question paper 2
1. (i) If y = sin (sin-1 x + cos-1 x), x € [-1, 1] then dy/dx is
(a) ๐/2 (b) −๐/2 (c) 0 d) 1
Answer:
(ii) \(\int e^x ( log x {1\over x}) \) dx is equal to
(a) ex + c (b) ex logx + c (c) ex/x + c (d) log x + c
Answer:
(iv) If P(E) denotes probability of occurrence of event E then
(a) P (E) € [-1, 1] (b) P (E) € (1, 2) (c) P (E) € (0, 1) (d) P (E) € [0, 1]
Answer:
(v) If is a binary operation such that a * b = a2 +b2 then 3 * 5 is
(a) 34 (b) 9 (c) 8 (d) 25Answer:
(vii) If f(x) = { \( {{{sinx\over x}\over k-1} , {x! = 0 \over x=0}}\) } ,๐ฅ = 0 is continuous at x=0 then
(a) 2 (b) 0 (c) -1 (d) 1Answer:
(viii) Distance between plane 3x +4y-20 = 0 and point (0, 0,-7) is
(a) 4 units (b) 3 units (c) 2 units (d) 1 unitAnswer:
(ix) If cos-1 x = y then
(a) −๐/2 <= y <= ๐/2 (b) -π <=y <= π(c) 0 <= y <= ๐/2 (d) 0 <= y <= π
Answer:
(x) Integrating factor of differential equation dy/dx + ๐ฆ = 3 is
(a) x (b) e (c) ex (d) logxAnswer:
2. Evaluate \(\int\limits_0^{x/2} sin^3X / sin^3X + cos^3X { sin^3x \over sin^3x + cos^3x }dx\)
Answer:
3. Find particular solution of differential equation \({dy \over dx} {1+y^2 \over 1+x^2}\) given that x=0 or y= 1
Answer:
4. Find the angle between the plane 2x+3 y-5z= 10 and the line passing from the points (2, 3,-1) Or (1, 2, 1)
Answer:
5. If matrix A = [aij]3 x 2 , and aij = (3i-2j)2 or matrix A find them
Answer:
6. Form differential equation representing the family of lines making equal intercepts on the co-ordinate axes.
Answer:
7. If P (A) = 7/13, P (B) = 9/13 and P (AUB) = 12/13 then find (A|B)
Answer:
8. Check whether Lagrange's mean value theorem is applicable on f(x) = sin x + cos x Interval [0, ๐/2]
Answer:
9. Evaluate \(\int {7dx \over x(x^7-1)}\)
Answer:
10. If y = (x)tanx + (tanx)x then find ๐๐ฆ/๐๐ฅ
Answer:
11. Using differentials find approximate value of (0.37)1/2
Answer:
12. Evaluate \(\int {x^2 +1 \over x^4 +1}\) dx
Or
Evaluate \(\int {dx \over x^2+1}\)Answer:
13 If à = 2í-3j+4k เจ เจคเฉ 6 = 5i +j-k represents sider parallelogram then find both diagonals and a unit vector perpendicular to both dingonals.
Answer:
14. Two cards are drawn (without replacement from a well shulle distribution table and mean of number of kings.
Answer:
15. Find the particular solution of differential equation [x sin2 (y/x)-y] dx +xdy = 0;y(1)=๐/4
Or
Find the particular solution of differential equation " given that tanx ๐๐ฅ/๐๐ฅ +y = 2x tan x + x2 , x != 0 given that y=0 when x = ๐/2
Answer:
16. Prove that function f : R --> R, f(x) = \(3โ2๐ฅ \over 7\) in one-one and onto. Also find f-1
Answer:
17. Express \( \begin{matrix} 2 & 5 & -1 \\ 3 & 1 & 5 \\ 7 & 6 & 9 \end{matrix}\) as sum of symmetric and skew-symmetric matrices.
Or
If x,y,z are different and \( \begin{matrix} x & x^2 & 1 + x^3 \\ y & y^2 & 1 + y^3 \\ z & z^2 & 1 + z^3 \end{matrix}\) = 0 then prove that xyz =-1
Answer:
18. Prove that : sin-1 ( \(5\over13\) ) +cos-1 (\(4\over 5\)) = \(1\over 2\) sin-1 \(3696 \over 4225\)
Answer:
19. Find the area of region bounded by the ellipse \({x^2 \over 9 } + {y^2 \over 4} = 1\)
Answer:
20. A window is in the form of rectangle surmounted by a semi-circular opening. The perimeter of window is 30 m. Find the dimensions of window so that it can admit maximum light through the whole opening.
Or
Prove that volume of largest cone, which can be inscribed in a sphere, is 8/27 part of sphere.Answer:
21. Maximise and minimise : Z=15x + 30y Subject to the constraints : x+y <= 8, 2x +y >= 28, x - 2y >=0, x, y >= 0
Or
Maximise and minimize Z = 4x + 3y -7 Subject to the constraints : x+y <= 10, x +y >= 3, x<=8, y <= 9 , x , y >-0Answer:
22. Solve the following system oflincar equations by matrix mehord:
x - 2y +3z = -5, 3 x +y +c= 8, 2x –y +2z = 1
Or
Using elementary transformations find inverse of \( \begin{matrix} 2 & 4 & 1 \\ 1 & 2 & 3 \\ 1 & -3 & 0 \end{matrix}\)Answer: