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Page 1 CBSE XII  Mathematics Board Paper 2015 – All India Set – 1 Solution CBSE Board Class XII Mathematics Board Paper – 2015 Solution All India Set – 1 SECTION – A 1. ? ? ? ? ? ? Given that a 2i j 3k and b 3i 5j 2k ? ? ? ? ? ? ? ?? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? 2 2 2 We need to find a b i j k a b 2 1 3 3 5 2 i 2 15 j 4 9 k 10 3 17i 13j 7k Hence, a b 17 13 7 a b 507 2. ? ? ? ? Let a i j; b j k ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? 2 22 2 22 a b i j j k 1 0 1 1 0 1 1 a 1 1 0 2 b 0 1 1 2 We know that a b a b cos a b 1 1 Thus, cos = 2 22 ab cos cos120 120 Page 2 CBSE XII  Mathematics Board Paper 2015 – All India Set – 1 Solution CBSE Board Class XII Mathematics Board Paper – 2015 Solution All India Set – 1 SECTION – A 1. ? ? ? ? ? ? Given that a 2i j 3k and b 3i 5j 2k ? ? ? ? ? ? ? ?? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? 2 2 2 We need to find a b i j k a b 2 1 3 3 5 2 i 2 15 j 4 9 k 10 3 17i 13j 7k Hence, a b 17 13 7 a b 507 2. ? ? ? ? Let a i j; b j k ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? 2 22 2 22 a b i j j k 1 0 1 1 0 1 1 a 1 1 0 2 b 0 1 1 2 We know that a b a b cos a b 1 1 Thus, cos = 2 22 ab cos cos120 120 CBSE XII  Mathematics Board Paper 2015 – All India Set – 1 Solution 3. Consider the vector equation of the plane. ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? 1 1 1 r 6i 3j 2k 4 xi yj zk 6i 3j 2k 4 6x 3y 2z 4 6x 3y 2z 4 0 Thus the Cartesian equation of the plane is 6x 3y 2z 4 0 Let d be the distance between the point 2, 5, 3 to the plane. ax by cz d Thus, d= a ? ? ? ? ?? ? ? ? ? ? ? ? ?? ? ? ? ? ? ? ?? ?? ? ?? ?? 2 2 2 2 22 bc 6 2 3 5 2 3 4 d 6 3 2 12 15 6 4 d 36 9 4 13 d 49 13 d units 7 4. Given that of a ij = e 2ix sin(jx) ? ? ? ? ?? ? ? ? 2 1 x 2x 12 Substitute i = 1 and j = 2 Thus, a e sin 2 x e sin 2x Page 3 CBSE XII  Mathematics Board Paper 2015 – All India Set – 1 Solution CBSE Board Class XII Mathematics Board Paper – 2015 Solution All India Set – 1 SECTION – A 1. ? ? ? ? ? ? Given that a 2i j 3k and b 3i 5j 2k ? ? ? ? ? ? ? ?? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? 2 2 2 We need to find a b i j k a b 2 1 3 3 5 2 i 2 15 j 4 9 k 10 3 17i 13j 7k Hence, a b 17 13 7 a b 507 2. ? ? ? ? Let a i j; b j k ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? 2 22 2 22 a b i j j k 1 0 1 1 0 1 1 a 1 1 0 2 b 0 1 1 2 We know that a b a b cos a b 1 1 Thus, cos = 2 22 ab cos cos120 120 CBSE XII  Mathematics Board Paper 2015 – All India Set – 1 Solution 3. Consider the vector equation of the plane. ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? 1 1 1 r 6i 3j 2k 4 xi yj zk 6i 3j 2k 4 6x 3y 2z 4 6x 3y 2z 4 0 Thus the Cartesian equation of the plane is 6x 3y 2z 4 0 Let d be the distance between the point 2, 5, 3 to the plane. ax by cz d Thus, d= a ? ? ? ? ?? ? ? ? ? ? ? ? ?? ? ? ? ? ? ? ?? ?? ? ?? ?? 2 2 2 2 22 bc 6 2 3 5 2 3 4 d 6 3 2 12 15 6 4 d 36 9 4 13 d 49 13 d units 7 4. Given that of a ij = e 2ix sin(jx) ? ? ? ? ?? ? ? ? 2 1 x 2x 12 Substitute i = 1 and j = 2 Thus, a e sin 2 x e sin 2x CBSE XII  Mathematics Board Paper 2015 – All India Set – 1 Solution 5. Consider the equation, y = mx, where m is the parameter. Thus, the above equation represents the family of lines which pass through the origin. y mx....(1) y m....(2) x ? ?? Differentiating the above equation (1) with respect to x, ? ? ? ?? ?? ?? ? ? ? ?? y mx dy m1 dx dy m dx dy y from equation (2) dx x dy y 0 dx x Thus we have eliminated the constant, m. The required differential equation is dy y 0 dx x 6. Consider the given differential equation: dy xlog x y 2log x dx Dividing the above equation by xlogx, we have, xlog x dy y 2log x xlog x dx xlog x xlog x dy y 2 ....(1) dx xlog x x Consider the general linear differential equation, dy Py Q,where P and Q are funct dx ?? ?? ? ? ? ?? ions of x ? ? ? ? Pdx dx Pdx x log x Comparing equation (1) and the general equation, we have, 12 P x and Q x xlog x x The integrating factor is given by the formula e Thus,I.F. e e ?? ? ? ? ?? Page 4 CBSE XII  Mathematics Board Paper 2015 – All India Set – 1 Solution CBSE Board Class XII Mathematics Board Paper – 2015 Solution All India Set – 1 SECTION – A 1. ? ? ? ? ? ? Given that a 2i j 3k and b 3i 5j 2k ? ? ? ? ? ? ? ?? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? 2 2 2 We need to find a b i j k a b 2 1 3 3 5 2 i 2 15 j 4 9 k 10 3 17i 13j 7k Hence, a b 17 13 7 a b 507 2. ? ? ? ? Let a i j; b j k ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? 2 22 2 22 a b i j j k 1 0 1 1 0 1 1 a 1 1 0 2 b 0 1 1 2 We know that a b a b cos a b 1 1 Thus, cos = 2 22 ab cos cos120 120 CBSE XII  Mathematics Board Paper 2015 – All India Set – 1 Solution 3. Consider the vector equation of the plane. ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? 1 1 1 r 6i 3j 2k 4 xi yj zk 6i 3j 2k 4 6x 3y 2z 4 6x 3y 2z 4 0 Thus the Cartesian equation of the plane is 6x 3y 2z 4 0 Let d be the distance between the point 2, 5, 3 to the plane. ax by cz d Thus, d= a ? ? ? ? ?? ? ? ? ? ? ? ? ?? ? ? ? ? ? ? ?? ?? ? ?? ?? 2 2 2 2 22 bc 6 2 3 5 2 3 4 d 6 3 2 12 15 6 4 d 36 9 4 13 d 49 13 d units 7 4. Given that of a ij = e 2ix sin(jx) ? ? ? ? ?? ? ? ? 2 1 x 2x 12 Substitute i = 1 and j = 2 Thus, a e sin 2 x e sin 2x CBSE XII  Mathematics Board Paper 2015 – All India Set – 1 Solution 5. Consider the equation, y = mx, where m is the parameter. Thus, the above equation represents the family of lines which pass through the origin. y mx....(1) y m....(2) x ? ?? Differentiating the above equation (1) with respect to x, ? ? ? ?? ?? ?? ? ? ? ?? y mx dy m1 dx dy m dx dy y from equation (2) dx x dy y 0 dx x Thus we have eliminated the constant, m. The required differential equation is dy y 0 dx x 6. Consider the given differential equation: dy xlog x y 2log x dx Dividing the above equation by xlogx, we have, xlog x dy y 2log x xlog x dx xlog x xlog x dy y 2 ....(1) dx xlog x x Consider the general linear differential equation, dy Py Q,where P and Q are funct dx ?? ?? ? ? ? ?? ions of x ? ? ? ? Pdx dx Pdx x log x Comparing equation (1) and the general equation, we have, 12 P x and Q x xlog x x The integrating factor is given by the formula e Thus,I.F. e e ?? ? ? ? ?? CBSE XII  Mathematics Board Paper 2015 – All India Set – 1 Solution 4 ? ? ? ? ? ? ? ?? ? ? ? ? ? ? dx log log x x log x dx Consider I= xlog x dx Substituting logx=t; dt x dt Thus I= log t log log x t Hence,I.F. e e log x SECTION – B 7. 2 1 2 2 A 2 1 2 2 2 1 1 2 2 1 2 2 A 2 1 2 2 1 2 2 2 1 2 2 1 1 1 2 2 2 2 1 2 2 1 2 2 1 2 2 2 2 1 2 1 1 2 2 2 2 2 1 1 2 2 2 2 1 2 2 1 2 1 2 2 1 2 2 2 2 1 1 2 2 2 2 2 1 1 1 4 4 2 2 4 2 4 2 2 2 4 4 1 4 4 2 2 2 4 2 4 2 2 4 4 1 9 8 8 8 9 8 8 8 9 2 Consider A 4A 5I 9 8 8 1 2 2 1 0 0 8 9 8 4 2 1 2 5 0 1 0 8 8 9 2 2 1 0 0 1 9 8 8 4 8 8 5 0 0 8 9 8 8 4 8 0 5 0 8 8 9 8 8 4 0 0 5 9 9 8 8 8 8 8 8 9 9 8 8 8 8 8 8 9 9 Page 5 CBSE XII  Mathematics Board Paper 2015 – All India Set – 1 Solution CBSE Board Class XII Mathematics Board Paper – 2015 Solution All India Set – 1 SECTION – A 1. ? ? ? ? ? ? Given that a 2i j 3k and b 3i 5j 2k ? ? ? ? ? ? ? ?? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? 2 2 2 We need to find a b i j k a b 2 1 3 3 5 2 i 2 15 j 4 9 k 10 3 17i 13j 7k Hence, a b 17 13 7 a b 507 2. ? ? ? ? Let a i j; b j k ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? 2 22 2 22 a b i j j k 1 0 1 1 0 1 1 a 1 1 0 2 b 0 1 1 2 We know that a b a b cos a b 1 1 Thus, cos = 2 22 ab cos cos120 120 CBSE XII  Mathematics Board Paper 2015 – All India Set – 1 Solution 3. Consider the vector equation of the plane. ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? 1 1 1 r 6i 3j 2k 4 xi yj zk 6i 3j 2k 4 6x 3y 2z 4 6x 3y 2z 4 0 Thus the Cartesian equation of the plane is 6x 3y 2z 4 0 Let d be the distance between the point 2, 5, 3 to the plane. ax by cz d Thus, d= a ? ? ? ? ?? ? ? ? ? ? ? ? ?? ? ? ? ? ? ? ?? ?? ? ?? ?? 2 2 2 2 22 bc 6 2 3 5 2 3 4 d 6 3 2 12 15 6 4 d 36 9 4 13 d 49 13 d units 7 4. Given that of a ij = e 2ix sin(jx) ? ? ? ? ?? ? ? ? 2 1 x 2x 12 Substitute i = 1 and j = 2 Thus, a e sin 2 x e sin 2x CBSE XII  Mathematics Board Paper 2015 – All India Set – 1 Solution 5. Consider the equation, y = mx, where m is the parameter. Thus, the above equation represents the family of lines which pass through the origin. y mx....(1) y m....(2) x ? ?? Differentiating the above equation (1) with respect to x, ? ? ? ?? ?? ?? ? ? ? ?? y mx dy m1 dx dy m dx dy y from equation (2) dx x dy y 0 dx x Thus we have eliminated the constant, m. The required differential equation is dy y 0 dx x 6. Consider the given differential equation: dy xlog x y 2log x dx Dividing the above equation by xlogx, we have, xlog x dy y 2log x xlog x dx xlog x xlog x dy y 2 ....(1) dx xlog x x Consider the general linear differential equation, dy Py Q,where P and Q are funct dx ?? ?? ? ? ? ?? ions of x ? ? ? ? Pdx dx Pdx x log x Comparing equation (1) and the general equation, we have, 12 P x and Q x xlog x x The integrating factor is given by the formula e Thus,I.F. e e ?? ? ? ? ?? CBSE XII  Mathematics Board Paper 2015 – All India Set – 1 Solution 4 ? ? ? ? ? ? ? ?? ? ? ? ? ? ? dx log log x x log x dx Consider I= xlog x dx Substituting logx=t; dt x dt Thus I= log t log log x t Hence,I.F. e e log x SECTION – B 7. 2 1 2 2 A 2 1 2 2 2 1 1 2 2 1 2 2 A 2 1 2 2 1 2 2 2 1 2 2 1 1 1 2 2 2 2 1 2 2 1 2 2 1 2 2 2 2 1 2 1 1 2 2 2 2 2 1 1 2 2 2 2 1 2 2 1 2 1 2 2 1 2 2 2 2 1 1 2 2 2 2 2 1 1 1 4 4 2 2 4 2 4 2 2 2 4 4 1 4 4 2 2 2 4 2 4 2 2 4 4 1 9 8 8 8 9 8 8 8 9 2 Consider A 4A 5I 9 8 8 1 2 2 1 0 0 8 9 8 4 2 1 2 5 0 1 0 8 8 9 2 2 1 0 0 1 9 8 8 4 8 8 5 0 0 8 9 8 8 4 8 0 5 0 8 8 9 8 8 4 0 0 5 9 9 8 8 8 8 8 8 9 9 8 8 8 8 8 8 9 9 CBSE XII  Mathematics Board Paper 2015 – All India Set – 1 Solution 000 000 000 2 2 2 1 1 1 1 1 1 Now A 4A 5I 0 A 4A 5I A A 4AA 5IA Postmultiply by A A 4I 5A 1 2 2 4 0 0 2 1 2 0 4 0 5A 2 2 1 0 0 4 1 1 3 2 2 2 3 2 5A 2 2 3 3 2 2 555 2 3 2 A 555 2 2 3 555 OR 1 1 1 2 0 1 A 5 1 0 0 1 3 2 3 0 0 15 0 1 5 0 6 0 5 1 0 Hence A exists. A A I 2 0 1 1 0 0 A 5 1 0 0 1 0 0 1 3 0 0 1Read More
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