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<ul><li><p>24</p><p>QUESTION PAPER CODE 65/3/RU</p><p>EXPECTED ANSWERS/VALUE POINTS</p><p>SECTION - A</p><p>1. ycotx.y1</p><p>2ydydx</p><p>2 m</p><p>Integrating factor = 2y1log y1ore 2 m</p><p>2.333</p><p>yxzzyxzyxzyx</p><p> m</p><p> = 0 m</p><p>3. order 2, degree 1 (any one correct) m</p><p>sum = 3 m</p><p>4.10</p><p>110</p><p>3z</p><p>151</p><p>157y</p><p>51</p><p>53x</p><p> m</p><p>Direction cosines are 73,</p><p>72,</p><p>76</p><p>or 73,</p><p>72,</p><p>76</p><p> m</p><p>5. a2ccbba22cba2cba2 2222 m</p><p> 6cba2 m</p><p>6. ji011111kji</p><p>b a </p><p> m</p><p>unit vector is 2j</p><p>2i m</p><p>Marks</p></li><li><p>25</p><p>SECTION - B</p><p>7. k5j5ibc;kjiba 1 m</p><p> k4j4551</p><p>111kji</p><p>bcba </p><p>1 m</p><p> Unit vector perpendicular to both of the vectors = 2k</p><p>2j 1 m</p><p>8. let the equation of line passing through (1, 2, 4) be</p><p> kcjbiak4j2ir 1 mSince the line is perpendicular to the two given lines </p><p> 3a 16 b + 7 c = 0 1 m 3a + 8 b 5 c = 0</p><p>Solving we get, 6c</p><p>3b</p><p>2aor</p><p>72c</p><p>36b</p><p>24a</p><p> 1 m</p><p> Equation of line is : k6j3i2k4j2ir mOR</p><p>Equation of plane is : 011112212</p><p>z2y1x</p><p>3 m</p><p>Solving we get, x + 2y + 3z 3 = 0 1 m</p><p>9. Let x = No. of spades in three cards drawn</p><p>x : 0 1 2 3 1 m</p><p>P(x) : </p><p>6427</p><p>433</p><p>3</p><p>C0</p><p> 64</p><p>274</p><p>34</p><p>132</p><p>C1</p><p> 64</p><p>94</p><p>34</p><p>132</p><p>C2</p><p> 64</p><p>14</p><p>34</p><p>1303</p><p>C3</p><p>2 m</p><p>x . P(x) : 0 6427</p><p>6418</p><p>643 m</p></li><li><p>26</p><p>Mean = 43</p><p>6448P(x)x m</p><p>OR</p><p>let p = probability of success ; q = Probability of failure</p><p>then, 9 P(x = 4) = P(x = 2)</p><p> 422624</p><p>46 qpCqpC9 2 m</p><p> 3pqq9p 22 1 m</p><p>Also, p + q = 1 p + 3p = 1 p = 41</p><p>1 m</p><p>10. 2xsinm</p><p>xsinm</p><p>x1em</p><p>dxdyget wex, w.r.t.atedifferenti,ey</p><p>11</p><p> 1 m</p><p>dxdyx1 2 = my, Differentiate again w.r.t. x</p><p>dxdym</p><p>dxdy</p><p>x1x</p><p>dxydx1</p><p>22</p><p>22 1 m</p><p> (my)mdxdyx1m</p><p>dxdyx</p><p>dxydx1 22</p><p>22 </p><p> m</p><p> 0ymdxdyx</p><p>dxydx1 22</p><p>22 m</p><p>11. f (x) = 32x(x)h,1x</p><p>1x(x)g,1x 22 </p><p>Differentiating w.r.t. x, we get</p><p> 2(x)h,</p><p>1xx2x1(x)g,</p><p>1xx(x)f 22</p><p>2</p><p>2</p><p> 1+1+1 m</p><p> 5</p><p>2(x)ghf m</p></li><li><p>27</p><p>12. </p><p> dxxx22x1dx</p><p>21x</p><p>232dxxx22x3 2</p><p>222 2 m</p><p> cxx232</p><p>23</p><p>21x</p><p>sin89xx2</p><p>221x</p><p>2 23212 </p><p> 2 m</p><p>or </p><p> cxx2</p><p>32</p><p>312xsin</p><p>49xx2</p><p>212x 23212</p><p>OR</p><p> dx</p><p>2x1</p><p>53dx</p><p>1x12x</p><p>51dx</p><p>2x1x1xx</p><p>22</p><p>2</p><p>2 m</p><p> dx2x1</p><p>53dx</p><p>1x1</p><p>51dx</p><p>1x2x</p><p>51</p><p>22 m</p><p> c2xlog53xtan</p><p>511xlog</p><p>51 12 1 m</p><p>13. 4</p><p>04</p><p>4</p><p>03</p><p>dxtan x2xcos</p><p>1dx x2sin 2xcos</p><p>11 m</p><p> = </p><p>4</p><p>0</p><p>22</p><p>dxxsecxtan2xtan1</p><p>1 m</p><p> = 1</p><p>0</p><p>2</p><p>dttt1</p><p>21</p><p> Taking, tan x = t; 1 m</p><p> = 1</p><p>0</p><p>25</p><p>t52t2</p><p>21</p><p> m</p><p> = 56</p><p>522</p><p>21</p><p> m</p></li><li><p>28</p><p>14. dx1x1</p><p>x1</p><p>1x1xlogdx</p><p>1x1xlog 2 2 m</p><p>= dx1x1dx1</p><p>1xxlog</p><p>x 1 m</p><p>= c1)(xlogxlog1xxlog</p><p>1 m</p><p>or c1xxlog</p><p>1xxlog</p><p>15.</p><p>390002300030000</p><p>402050</p><p>15040050075250300</p><p>1003004002 m</p><p>cost incurred respectively for three villages is Rs. 30,000, Rs. 23,000, Rs. 39,000 1 m</p><p>One value : Women welfare or Any other relevant value 1 m</p><p>16. </p><p>318tan</p><p>1x1x11 x 1 xtan 11 2 m</p><p>318</p><p>x22x</p><p>2 4x2 + 31x 8 = 0 1 m</p><p>,41x 8 (Rejected) 1 m</p><p>OR</p><p>L.H.S. = </p><p> zx1</p><p>xztanyz1zytan</p><p>xy1yxtan 111 2 m</p><p>RHS0xtanztanztanytanytanxtan 111111</p><p>2 m</p></li><li><p>29</p><p>17.ccbbabba</p><p>cacaabc</p><p>cbcbabacbaba</p><p>cacbca</p><p>22</p><p>22</p><p>22</p><p>Taking a, b &amp; c common from C1 , C</p><p>2 and C</p><p>31 m</p><p>ccbcbabba</p><p>caccaabc2</p><p> 3211 C CCC and taking 2 common from C1 1 m</p><p> bcbcbbbba</p><p>0ccaabc2</p><p> 133 CCC 1 m</p><p>bcbcb0cca0cca</p><p>abc2</p><p> 322 RRR m</p><p>Expand by C3, = 2 abc ( b) ( ac c2 ac + c2) = 4a2 b2 c2 m</p><p>18. Adj A = 27A;366636</p><p>663</p><p>2+1 m</p><p>A. Adj A = 3IA</p><p>100010001</p><p>27366636</p><p>663</p><p>122212221</p><p>1 m</p><p>19. f (x) = 1x1x </p><p>L </p><p> 2</p><p>1x1x2lim</p><p>1x21x1xlim(1)f</p><p> 1)(x1)(x</p><p>1 m</p></li><li><p>30</p><p>R 01x0lim</p><p>1x21x1xlim(1)f</p><p>1)(x1)(x</p><p> 1 m</p><p>(x)f02 is not differentiable at x = 1</p><p>L 01x</p><p>0lim1x</p><p>21x1xlim(1)f 1x1x</p><p>1 m</p><p>R 21x1x2lim</p><p>1x21x1xlim(1)f</p><p>1x1x</p><p> 1 m</p><p>(x)f20 is not differentiable at x = 1</p><p>SECTION - C</p><p>20. correct figure 1 m</p><p>ar (ABDOA) 3</p><p>1612</p><p>dy41</p><p>4</p><p>0</p><p>34</p><p>0</p><p>2 </p><p> yy ......(i) 1 m</p><p> ar (OEBDO) 4</p><p>0</p><p>32</p><p>34</p><p>0</p><p>24</p><p>0 12xx</p><p>34dx</p><p>4xdxx2 </p><p> 316</p><p>316</p><p>332</p><p> .........(ii) 1 m</p><p> ar (OEBCO) 3</p><p>1612xdxx</p><p>41</p><p>4</p><p>0</p><p>34</p><p>0</p><p>2 </p><p> ................(iii) 1 m</p><p>From (i), (ii) and (iii) we get ar (ABDOA) = ar (OEBDO) = ar (OEBCO)</p><p>21.1</p><p>xy</p><p>xy</p><p>dxdy</p><p>xxyy</p><p>dxdy</p><p>2</p><p>2</p><p>2 </p><p> , Hence the differential equation is homogeneous 1 m</p><p>Put y = 1dx</p><p>dxgetwe,dxdx</p><p>dxdyandx</p><p>2</p><p>vvvvvvv 1+1 m</p></li><li><p>31</p><p> 1</p><p>1dxdx</p><p>2</p><p>vvv</p><p>vvv</p><p> 1 m</p><p>cxloglogdxx1d1 vvvv</p><p>v1 m</p><p> cylog</p><p>xyor,cxlog</p><p>xylog</p><p>xy</p><p>1 m</p><p>OR</p><p>Given differential equation can be written as 21</p><p>2 y1ytanx</p><p>y11</p><p>dydx</p><p> 1 m</p><p>Integrating factor = </p><p> dyy1eytanex:issolutionande 2</p><p>ytan1ytanytan</p><p>111</p><p>1+1 m</p><p>x t)ytan(wherec1)y(tanecetedttee 11ytantttytan11</p><p> 1 m</p><p>x = 1, y = 0 c = 2 21)y(taneex 1ytanytan11</p><p> 1 m</p><p>ytan1 1e21ytanxor </p><p>22. E1 : Bolt is manufactured by machine A</p><p>E2 : Bolt is manufactured by machine B</p><p>E3 : Bolt is manufactured by machine C</p><p>A : Bolt is defective</p><p>;10020)P(E;</p><p>10050)P(E;</p><p>10030)P(E 321 </p><p>1001)P(A/E;</p><p>1004)P(A/E;</p><p>1003)P(A/E 321 3 m</p></li><li><p>32</p><p>P(E2/A) = 3120</p><p>2020090200</p><p>1001</p><p>10020</p><p>1004</p><p>10050</p><p>1003</p><p>10030</p><p>1004</p><p>10005</p><p>2 m</p><p>P( E2/A) = 1 P(E2/A) = 3111</p><p>1 m</p><p>23. Equation of line through A and B is (say)6</p><p>5z1</p><p>4y13x</p><p> 2 m</p><p>General point on the line is 564,3, 1 m</p><p>If this is the point of intersection with plane 2x + y + z = 7</p><p>then, 2756432 1 m</p><p> Point of intersection is (1, 2, 7) 1 m</p><p>Required distance = 7742413 222 1 m</p><p>24. Let the two factories I and II be in operation for x and y</p><p>days respectively to produce the order with the minimum cost</p><p>then, the LPP is :</p><p>Minimise cost : z = 12000 x + 15000 y 1 m</p><p>Subject to :</p><p>50x + 40y &gt; 6400 or 5x + 4y &gt; 640</p><p>50x + 20y &gt; 4000 or 5x + 2y &gt; 400 2 m</p><p>30x + 40y &gt; 4800 or 3x + 4y &gt; 480</p><p>x, y &gt; 0</p></li><li><p>33</p><p>correct graph 2 m</p><p> Vertices are A (0, 200) ; B (32, 120)</p><p> C (80, 60) ; D (160, 0) m</p><p>z (A) = Rs. 30,00,000; z (B) = Rs. 21,84,000;</p><p> z (C) = Rs. 18,60,000 (Min.); z (D) = Rs. 19,20,000;</p><p>On plotting z &lt; 1860000</p><p>or 12x + 15y &lt; 1860, we get no</p><p>point common to the feasible region</p><p> Factory I operates for 80 days m</p><p> Factory II operates for 60 days</p><p>25. f : 5</p><p>35y54(y)f;96x5xf(x);)9,R 12</p><p>y95</p><p>35y546</p><p>535y54</p><p>5(y)fof2</p><p>1 </p><p> 3 m</p><p> x</p><p>5396x5x554</p><p>(x)f of2</p><p>1 </p><p> 2 m</p><p>Hence f is invertible with 5</p><p>35y54(y)f 1</p><p> m</p><p>OR</p><p>(i) commutative : let x, y then1R</p><p>x * y = x + y + xy = y + x + yx = y * x * is commutative 1 m</p><p>(ii) Associative : let x, y, z then1R</p><p>x * (y * z) = x * (y + z + yz) = x + (y + z + yz) + x (y + z + yz)</p></li><li><p>34</p><p> = x + y + z + xy + yz + zx + xyz 1 m</p><p>(x * y) * z = (x + y + xy) * z = (x + y + xy) + z + (x + y + xy) . z</p><p> = x + y + z + xy + yz + zx + xyz 1 m</p><p>x * (y * z) = (x * y) * z * is Associative</p><p>(iii) Identity Element : let 1Raaa*ee*athatsuch1Re m</p><p> a + e + ae = a e = 0 m</p><p>(iv) Inverse : let a * b = b * a = e = 0 ; a, b 1R m</p><p> a + b + ab = 0 a1aaor</p><p>a1ab 1</p><p> m</p><p>26. Solving the two curves to get the points of intersection 8,p3 1 m</p><p>m1 = slope of tangent to first curve = 9p2x</p><p>1 m</p><p>m2 = slope of tangent to second curve = p2x</p><p>1 m</p><p>curves cut at right angle iff 1p</p><p>2x9p2x</p><p> m</p><p>)p3x(Put4x9p 22 </p><p>p)(949p2 </p><p> p = 0 ; p = 4 1 m</p></li></ul>

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