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SECTION-A, , IÊS-A, , 1., ,  2 1, If A 1  ,  , then A is equal to, 1, 1, , , , [1], ,  2 1, (A) , ,  1 1 , ,  2 1 , (B) , ,  1 1 , ,  1 1, (C) , ,  1 2 , ,  2 1, (D) , ,  1 1, ,  2 1,  h¡, Vmo A ~am~a h¡ :,  1 1, , ¶{X A 1  , , 2., ,  2 1, (A) , ,  1 1 , ,  2 1 , (B) , ,  1 1 , ,  1 1, (C) , ,  1 2 , ,  2 1, (D) , ,  1 1, , cos(tan–1 x) is equal to, , [1], , (A), , 1, 2, x 1, , (B), , (C), , 1, x2  1, , (D), , 60/OSS/1/311-A], , G-607, , 3, , 1, 1  x2, , x2  1, , *60/OSS/1/311-A*, , [ Contd......

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cos(tan–1 x) ~am~a h¡ :, , 3., , (A), , 1, x 1, , (B), , (C), , 1, x 1, , (D), , 2, , 2, , 1, 1  x2, x2  1, , Let f :    be defined by f (x) = 2x + 3, x. Then f is, , [1], , (A) one to one but not onto, (B) onto but not one to one, (C) neither one to one nor onto, (D) one to one and onto, , ‘mZm f :    na f (x) = 2x + 3, x Ûmam n[a^m{fV EH$ ’$bZ h¡& Vmo f EH$, (A) EH¡$H$s naÝVw AmN>mXH$ Zht, ’$bZ h¡, (B) AmN>mXH$ naÝVw EH¡$H$s Zht, ’$bZ h¡, (C) Z Vmo EH¡$H$s Am¡a Z hr AmN>mXH$, ’$bZ h¡, (D) EH¡$H$s Am¡a AmN>mXH$, ’$bZ h¡, , 4., , , , 3, , 2, , (A), , 2 x, dx is equal to, 2 x  x3, , [1], , 1, 2, , (B) 0, (D)  4, , (C) –1, , 60/OSS/1/311-A], , G-607, , 4, , *60/OSS/1/311-A*, , [ Contd......

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, , 3, , 2, , (A), , 2 x, dx ~am~a h¡ :, 2 x  x3, , 1, 2, , (B) 0, (D)  4, , (C) –1, , 5., ,  1, 1 ,   log x   log x 2  dx is equal to, , , , [1], , (A), , x, c, log x, , (B), , 2x, c, log x, , (C), , x, c, 2log x, , (D), , 1, c, log x, ,  1, 1 , , ,   log x  log x 2  dx ~am~a h¡ :, , , , (A), , x, c, log x, , (B), , 2x, c, log x, , (C), , x, c, 2log x, , (D), , 1, c, log x, , 60/OSS/1/311-A], , G-607, , 5, , *60/OSS/1/311-A*, , [ Contd......

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6., , If y = xtan x, then, , dy, is euqal to, dx, , [1], , tan x  x(log x)sec 2 x, (A) y, x, , tan x  x(log x)sec2 x, (B) y, x, ,  tan x  sec x , 2, , (C), , 2, , (D), , x, , ¶{X y = xtan x h¡, Vmo, , x, , dy, ~am~a h¡ :, dx, , tan x  x(log x)sec 2 x, (A) y, x, , tan x  x(log x)sec2 x, (B) y, x, ,  x tan x  sec x , 2, , tan x  sec 2 x, (C), x, , 7., ,  x tan x  sec x , , (D), , General solution of the differential equation is, , x, , dy,  4 tan y equal to, dx, , (A) sin y = cex, , (B) y = sin–1(ce4x), , (C) y = cos–1(ce4x), , (D) y = sin–1(ce–4x), , AdH$b g‘rH$aU, , dy,  4 tan y H$m ì¶mnH$> hb h¡ :, dx, , (A) sin y = cex, , (B) y = sin–1(ce4x), , (C) y = cos–1(ce4x), , (D) y = sin–1(ce–4x), , 60/OSS/1/311-A], , [1], , G-607, , 6, , *60/OSS/1/311-A*, , [ Contd......

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10. If the perpendicular distance from the point (2, K, 0) to the plane 4x – 2y + 3z = 12, is, , 4, unit, then possible values of K are :, 29, , [1], , (A) 1, 4, , (B) –1, 3, , (C) 0, –4, , (D) 0, 4, , {~ÝXþ (2, K, 0) go Vb 4x – 2y + 3z = 12 H$s bå~dV² Xÿar, , 4, BH$mB© h¡& K Ho$ gå^dV… ‘mZ h¢ :, 29, , (A) 1, 4, , (B) –1, 3, , (C) 0, –4, , (D) 0, 4, , SECTION-B, , IÊS - ~,  0 7 43 , , 0 47  , then show that |A| = 0., 11. If A   7,  43 47, 0 , , , [2], ,  0 7 43 , 0 47  h¡, Vmo Xem©BE H$s{OE {H$ |A| = 0, ¶{X A   7,  43 47, 0 , , , OR/AWdm, 2, 1 2, For A  , , verify that A 2   A  ., , 3 2, , 2, 1 2, 2, Ho, $, {bE, g˶m{nV, H$s{OE, {H$, A, , A, A, , , , 3 2, , 60/OSS/1/311-A], , G-607, , 8, , *60/OSS/1/311-A*, , [ Contd......

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16. Write the converse the following statements :, a), , If game is cancelled, then team A is win., , b), , If a is a multiple of b then b is a factor of a., , [2], , {ZåZ H$WZm| Ho$ {dbmo‘ {b{IE :, a), , ¶{X Iob aÔ hmoVm h¡, Vmo Q>r‘ A OrVVr h¡&, , b), , ¶{X a, b H$m JwUO h¡, Vmo b, a H$m JwUZI§S> h¡&, , SECTION - C, , IÊS - g, , a 2  2a 2a  1 1, 2a  1, 17. Show that, 3, , a  2 1   a  1, ., 3, 1, 3, , [4], , a 2  2a 2a  1 1, 3, 2a  1 a  2 1   a  1, Xem©BE {H$, 3, 3, 1, ,  3 1, 18. If A  , , find x and y so that A2 + xI2 = yA., ,  6 5, , [4], ,  3 1, Ho$ {bE A2 + xI2 = yA h¡, Vmo x Am¡a y H$m ‘mZ kmV H$s{OE&, ,  6 5, , ¶{X A  , , 60/OSS/1/311-A], , G-607, , 10, , *60/OSS/1/311-A*, , [ Contd......

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1, 19. Solve the following equation for x(x > 0) : sin, , 1, x(x > 0) Ho$ {bE {ZåZ g‘rH$aU H$mo hb H$s{OE : sin, , 6, 8 ,  sin 1 , x, x 2, , [4], , 6, 8 ,  sin 1 , x, x 2, , 20. Determine the values of a, b for which the function, , [4], , | x  2 |,  x  2  a , x  2,, , f ( x )  a  b, , x  2,, , 2 x  b, , x  2, , , is continuous at x = –2., a Am¡a b Ho$ do ‘mZ kmV H$s{OE, {OZHo$ {bE ’$bZ, | x  2 |,  x  2  a , x  2,, , f ( x )  a  b, , x  2,, , 2 x  b, , x  2, , , x = –2 na gVV h¡&, , 21. Find the interval in which the function f (x) = 2x3 + 9x2 + 12x + 20 are increasing or, decreasing., [4], , do A§Vamb kmV H$s{OE, {OZ na ’$bZ f (x) = 2x3 + 9x2 + 12x + 20 dY©‘mZ ¶m ömg‘mZ h¡&, OR/AWdm, , 60/OSS/1/311-A], , G-607, , 11, , *60/OSS/1/311-A*, , [ Contd......

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25. Solve the following differential equation :, , [4], , , dy,  sin( x  y )  sin( x  y ), given y  when x = 0., dx, 2, , {ZåZ AdH$b g‘rH$aU H$mo hb H$s{OE :, dy,  sin( x  y )  sin( x  y ) ,, dx, , {X¶m h¡ {H$ O~ x = 0 Vmo y , , , 2, OR/AWdm, , Form the differential equation corresponding to y = ex(a cosx + b sinx) by eliminating, 'a' and 'b'., a Am¡a b H$m {dbmonZ H$aVo hþE y = ex(a cosx + b sinx) Ho$ g§JV AdH$b g‘rH$aU ~ZmBE&, , 26. Find the equation of the line passing through the point (–1, –3, –2) and perpendicular, to the lines, , x y z, x  2 y 1 z 1,   and, , , ., 3, 2, 5, 1 2 3, , {~ÝXþ (–1, –3, –2) go JwOaZo dmbr Am¡a aoImAm|, , [4], , x y z, x  2 y 1 z 1,   VWm, , , Ho$ bå~dV², 3, 2, 5, 1 2 3, , aoIm H$m g‘rH$aU kmV H$s{OE&, , 60/OSS/1/311-A], , G-607, , 13, , *60/OSS/1/311-A*, , [ Contd......