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Page 1 CBSE XII  Mathematics Sample Paper – 1 Solution Mathematics Class XII Sample Paper – 1 Solution SECTION – A 1. Let ? ? ? ? and a, b R b, c R ?? ? ? ? ? and R a b b c a c a, c ? ? ? ? ? ?? Hence, R is transitive. 2. Let ? ? 1 x tan 1 ? ?? ? ? ? ?? ? ?? ? ?? ?? ? ? ? ? ? ?? ?? ?? tanx 1 tanx tan 4 tanx tan tan 4 3 p tanx = tan 4 3 p x= 4 3. Matrix of order 3 × 3 has 9 elements. Now the entries have to be either 0 or 1 so that each of the 9 places can be filled with 2 choices 0 or 1. So 2 9 = 512 matrices are possible. 4. ˆ ˆˆ a 5i j 3k; ? ? ? ˆ ˆˆ b i 3j 5k ? ? ? ˆ ˆˆ a b 6i 2j 8k ? ? ? ? ? ab ˆ ˆˆ 3i j 4k 2 ? ? ? ? Page 2 CBSE XII  Mathematics Sample Paper – 1 Solution Mathematics Class XII Sample Paper – 1 Solution SECTION – A 1. Let ? ? ? ? and a, b R b, c R ?? ? ? ? ? and R a b b c a c a, c ? ? ? ? ? ?? Hence, R is transitive. 2. Let ? ? 1 x tan 1 ? ?? ? ? ? ?? ? ?? ? ?? ?? ? ? ? ? ? ?? ?? ?? tanx 1 tanx tan 4 tanx tan tan 4 3 p tanx = tan 4 3 p x= 4 3. Matrix of order 3 × 3 has 9 elements. Now the entries have to be either 0 or 1 so that each of the 9 places can be filled with 2 choices 0 or 1. So 2 9 = 512 matrices are possible. 4. ˆ ˆˆ a 5i j 3k; ? ? ? ˆ ˆˆ b i 3j 5k ? ? ? ˆ ˆˆ a b 6i 2j 8k ? ? ? ? ? ab ˆ ˆˆ 3i j 4k 2 ? ? ? ? CBSE XII  Mathematics Sample Paper – 1 Solution OR Given that the magnitude of each of the two vectors a and b have same magnitude. 2 2 a b ..........(i) the angle between a and b is 60 60 ........(ii) 9 and a b .........(iii) 2 Since, a b a b cos 9 a a cos60 From (i),(ii) and (iii) 2 19 a 22 a9 a3 a b 3 ............... From (i) ?? ? ? ? ? ? ?? ? ? ? ? ? ? ? ? ? ?? ?? ? ? ? SECTION – B 5. Let I = p x dx x p x 0 ? ?? ...(1) According to property, aa 00 f(x)dx f(a x)dx ?? ?? I ? p 0 px dx p x x ? ? ?? ...(2) Adding equations (1) and (2), we get 2I ? p 0 x p x dx x p x ? ?? ?? p 0 1dx ? ? = ? ? p 0 x = p – 0 = p Thus, 2I = p I p 2 ?? Page 3 CBSE XII  Mathematics Sample Paper – 1 Solution Mathematics Class XII Sample Paper – 1 Solution SECTION – A 1. Let ? ? ? ? and a, b R b, c R ?? ? ? ? ? and R a b b c a c a, c ? ? ? ? ? ?? Hence, R is transitive. 2. Let ? ? 1 x tan 1 ? ?? ? ? ? ?? ? ?? ? ?? ?? ? ? ? ? ? ?? ?? ?? tanx 1 tanx tan 4 tanx tan tan 4 3 p tanx = tan 4 3 p x= 4 3. Matrix of order 3 × 3 has 9 elements. Now the entries have to be either 0 or 1 so that each of the 9 places can be filled with 2 choices 0 or 1. So 2 9 = 512 matrices are possible. 4. ˆ ˆˆ a 5i j 3k; ? ? ? ˆ ˆˆ b i 3j 5k ? ? ? ˆ ˆˆ a b 6i 2j 8k ? ? ? ? ? ab ˆ ˆˆ 3i j 4k 2 ? ? ? ? CBSE XII  Mathematics Sample Paper – 1 Solution OR Given that the magnitude of each of the two vectors a and b have same magnitude. 2 2 a b ..........(i) the angle between a and b is 60 60 ........(ii) 9 and a b .........(iii) 2 Since, a b a b cos 9 a a cos60 From (i),(ii) and (iii) 2 19 a 22 a9 a3 a b 3 ............... From (i) ?? ? ? ? ? ? ?? ? ? ? ? ? ? ? ? ? ?? ?? ? ? ? SECTION – B 5. Let I = p x dx x p x 0 ? ?? ...(1) According to property, aa 00 f(x)dx f(a x)dx ?? ?? I ? p 0 px dx p x x ? ? ?? ...(2) Adding equations (1) and (2), we get 2I ? p 0 x p x dx x p x ? ?? ?? p 0 1dx ? ? = ? ? p 0 x = p – 0 = p Thus, 2I = p I p 2 ?? CBSE XII  Mathematics Sample Paper – 1 Solution OR Given ? ? ? 2 2 cos2x 2sin x I dx cos x ? ? ? ? ? ? ?? ? ? ? ? ? ? ? ? ? ?? ? ? ? ? ? 2 2 2 2 2 22 2 22 2 22 2 2 cos2x 2sin x I dx cos x cos x sin x 2sin x I dx ..... cos2x cos x sin x cos x cos x sin x I dx cos x 1 I dx .... cos x sin x 1 cos x I sec x dx I tan x c 6. Area of a parallelogram = Cross product of the vectors representing its adjacent sides. So, required area = (2 i + j + k ) ? (3 i + j +4 k ) Now (2 i + j + k ) ? (3 i + j +4 k ) = i j k 2 1 1 3 1 4 =3 i  5 j  k Area = 3 i 5 j  k  = 9 25 1 35 ? ? ? sq. units. 7. tan (cos –1 4 5 + tan –1 2 3 ) = tan (tan –1 3 4 + tan –1 2 3 ) = 1 32 43 tan tan 1 1 2 ? ?? ?? ? ?? ?? ?? ?? ?? ?? ? ?? ?? ?? ?? = 1 17 tan tan 6 ? ?? ?? ?? ?? ?? ?? = 17 6 Page 4 CBSE XII  Mathematics Sample Paper – 1 Solution Mathematics Class XII Sample Paper – 1 Solution SECTION – A 1. Let ? ? ? ? and a, b R b, c R ?? ? ? ? ? and R a b b c a c a, c ? ? ? ? ? ?? Hence, R is transitive. 2. Let ? ? 1 x tan 1 ? ?? ? ? ? ?? ? ?? ? ?? ?? ? ? ? ? ? ?? ?? ?? tanx 1 tanx tan 4 tanx tan tan 4 3 p tanx = tan 4 3 p x= 4 3. Matrix of order 3 × 3 has 9 elements. Now the entries have to be either 0 or 1 so that each of the 9 places can be filled with 2 choices 0 or 1. So 2 9 = 512 matrices are possible. 4. ˆ ˆˆ a 5i j 3k; ? ? ? ˆ ˆˆ b i 3j 5k ? ? ? ˆ ˆˆ a b 6i 2j 8k ? ? ? ? ? ab ˆ ˆˆ 3i j 4k 2 ? ? ? ? CBSE XII  Mathematics Sample Paper – 1 Solution OR Given that the magnitude of each of the two vectors a and b have same magnitude. 2 2 a b ..........(i) the angle between a and b is 60 60 ........(ii) 9 and a b .........(iii) 2 Since, a b a b cos 9 a a cos60 From (i),(ii) and (iii) 2 19 a 22 a9 a3 a b 3 ............... From (i) ?? ? ? ? ? ? ?? ? ? ? ? ? ? ? ? ? ?? ?? ? ? ? SECTION – B 5. Let I = p x dx x p x 0 ? ?? ...(1) According to property, aa 00 f(x)dx f(a x)dx ?? ?? I ? p 0 px dx p x x ? ? ?? ...(2) Adding equations (1) and (2), we get 2I ? p 0 x p x dx x p x ? ?? ?? p 0 1dx ? ? = ? ? p 0 x = p – 0 = p Thus, 2I = p I p 2 ?? CBSE XII  Mathematics Sample Paper – 1 Solution OR Given ? ? ? 2 2 cos2x 2sin x I dx cos x ? ? ? ? ? ? ?? ? ? ? ? ? ? ? ? ? ?? ? ? ? ? ? 2 2 2 2 2 22 2 22 2 22 2 2 cos2x 2sin x I dx cos x cos x sin x 2sin x I dx ..... cos2x cos x sin x cos x cos x sin x I dx cos x 1 I dx .... cos x sin x 1 cos x I sec x dx I tan x c 6. Area of a parallelogram = Cross product of the vectors representing its adjacent sides. So, required area = (2 i + j + k ) ? (3 i + j +4 k ) Now (2 i + j + k ) ? (3 i + j +4 k ) = i j k 2 1 1 3 1 4 =3 i  5 j  k Area = 3 i 5 j  k  = 9 25 1 35 ? ? ? sq. units. 7. tan (cos –1 4 5 + tan –1 2 3 ) = tan (tan –1 3 4 + tan –1 2 3 ) = 1 32 43 tan tan 1 1 2 ? ?? ?? ? ?? ?? ?? ?? ?? ?? ? ?? ?? ?? ?? = 1 17 tan tan 6 ? ?? ?? ?? ?? ?? ?? = 17 6 CBSE XII  Mathematics Sample Paper – 1 Solution 8. ?? ?? ?? ? ? ? ?? cos sin Let A = ,then sin cos ? ? ?? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? 22 1 1 cos sin A = sin cos cos sin 1 0 Since A 0, therefore A exist. cos sin cos sin AdjA 1 A sin cos sin cos A1 9. ? ? ? 32 Contentment function : C(x)= x 6x 5x 3 (given) ? ? ? ? ? ?? ? ? ? ?? ?? ? ? ? ? x3 2 2 dC(x) Marginal contentment = dx 3x 12x 5 When 3 units of dish are consumed, then dC(x) 3(3) 12(3) 5 dx 27 36 5 68 units. 10. ?? y On differentiating e (x 1) 1 w.r.t x, we get ? ? ? ? ? ? ? ? ? yy y y dy e (x 1)e 0 dx dy e 0 dx dy e dx Page 5 CBSE XII  Mathematics Sample Paper – 1 Solution Mathematics Class XII Sample Paper – 1 Solution SECTION – A 1. Let ? ? ? ? and a, b R b, c R ?? ? ? ? ? and R a b b c a c a, c ? ? ? ? ? ?? Hence, R is transitive. 2. Let ? ? 1 x tan 1 ? ?? ? ? ? ?? ? ?? ? ?? ?? ? ? ? ? ? ?? ?? ?? tanx 1 tanx tan 4 tanx tan tan 4 3 p tanx = tan 4 3 p x= 4 3. Matrix of order 3 × 3 has 9 elements. Now the entries have to be either 0 or 1 so that each of the 9 places can be filled with 2 choices 0 or 1. So 2 9 = 512 matrices are possible. 4. ˆ ˆˆ a 5i j 3k; ? ? ? ˆ ˆˆ b i 3j 5k ? ? ? ˆ ˆˆ a b 6i 2j 8k ? ? ? ? ? ab ˆ ˆˆ 3i j 4k 2 ? ? ? ? CBSE XII  Mathematics Sample Paper – 1 Solution OR Given that the magnitude of each of the two vectors a and b have same magnitude. 2 2 a b ..........(i) the angle between a and b is 60 60 ........(ii) 9 and a b .........(iii) 2 Since, a b a b cos 9 a a cos60 From (i),(ii) and (iii) 2 19 a 22 a9 a3 a b 3 ............... From (i) ?? ? ? ? ? ? ?? ? ? ? ? ? ? ? ? ? ?? ?? ? ? ? SECTION – B 5. Let I = p x dx x p x 0 ? ?? ...(1) According to property, aa 00 f(x)dx f(a x)dx ?? ?? I ? p 0 px dx p x x ? ? ?? ...(2) Adding equations (1) and (2), we get 2I ? p 0 x p x dx x p x ? ?? ?? p 0 1dx ? ? = ? ? p 0 x = p – 0 = p Thus, 2I = p I p 2 ?? CBSE XII  Mathematics Sample Paper – 1 Solution OR Given ? ? ? 2 2 cos2x 2sin x I dx cos x ? ? ? ? ? ? ?? ? ? ? ? ? ? ? ? ? ?? ? ? ? ? ? 2 2 2 2 2 22 2 22 2 22 2 2 cos2x 2sin x I dx cos x cos x sin x 2sin x I dx ..... cos2x cos x sin x cos x cos x sin x I dx cos x 1 I dx .... cos x sin x 1 cos x I sec x dx I tan x c 6. Area of a parallelogram = Cross product of the vectors representing its adjacent sides. So, required area = (2 i + j + k ) ? (3 i + j +4 k ) Now (2 i + j + k ) ? (3 i + j +4 k ) = i j k 2 1 1 3 1 4 =3 i  5 j  k Area = 3 i 5 j  k  = 9 25 1 35 ? ? ? sq. units. 7. tan (cos –1 4 5 + tan –1 2 3 ) = tan (tan –1 3 4 + tan –1 2 3 ) = 1 32 43 tan tan 1 1 2 ? ?? ?? ? ?? ?? ?? ?? ?? ?? ? ?? ?? ?? ?? = 1 17 tan tan 6 ? ?? ?? ?? ?? ?? ?? = 17 6 CBSE XII  Mathematics Sample Paper – 1 Solution 8. ?? ?? ?? ? ? ? ?? cos sin Let A = ,then sin cos ? ? ?? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? 22 1 1 cos sin A = sin cos cos sin 1 0 Since A 0, therefore A exist. cos sin cos sin AdjA 1 A sin cos sin cos A1 9. ? ? ? 32 Contentment function : C(x)= x 6x 5x 3 (given) ? ? ? ? ? ?? ? ? ? ?? ?? ? ? ? ? x3 2 2 dC(x) Marginal contentment = dx 3x 12x 5 When 3 units of dish are consumed, then dC(x) 3(3) 12(3) 5 dx 27 36 5 68 units. 10. ?? y On differentiating e (x 1) 1 w.r.t x, we get ? ? ? ? ? ? ? ? ? yy y y dy e (x 1)e 0 dx dy e 0 dx dy e dx CBSE XII  Mathematics Sample Paper – 1 Solution OR ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? 2 x a 2 sin2 y a 1 cos 2 differentiating w.r.t. dx a 2 cos2 2 d dy a sin2 2 d sin2 2 sin2 dy dx 2 cos2 2 1 cos2 dy 2sin cos dx 2sin dy cos cot dx sin at, 3 dy 1 cot dx 3 3 ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ?? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ?? ? ??Read More
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