{"id":18689,"date":"2020-01-08T16:42:14","date_gmt":"2020-01-08T11:12:14","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=18689"},"modified":"2021-07-02T15:30:35","modified_gmt":"2021-07-02T10:00:35","slug":"2nd-puc-maths-previous-year-question-paper-march-2019","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/2nd-puc-maths-previous-year-question-paper-march-2019\/","title":{"rendered":"2nd PUC Maths Previous Year Question Paper March 2019"},"content":{"rendered":"

Students can Download 2nd PUC Maths Previous Year Question Paper March 2019, Karnataka 2nd PUC Maths Model Question Papers with Answers<\/a> helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.<\/p>\n

Karnataka 2nd PUC Maths Previous Year Question Paper March 2019<\/h2>\n

Time: 3 Hrs 15 Min
\nMax. Marks: 100<\/p>\n

Part-A<\/span><\/p>\n

Answer ALL the following questions: (10 \u00d7 1 = 10)<\/span><\/p>\n

Question 1.
\nDefine Binary Operation.
\nAnswer:
\nA binary operation * on a set ‘A’ is defined as a function.
\n* : A \u00d7 A \u2192 A
\na * b = ab + 1 \u2208 Z \u2200 a, b, \u2208 Z
\n\u2234 * is binary operation on Z.<\/p>\n

Question 2.
\nFind the principal value of cos-1<\/sup>\\(\\left(-\\frac{1}{2}\\right)\\).
\nAnswer:
\n\\(\\frac{2 \\pi}{3}\\)<\/p>\n

Question 3.
\nDefine a Scalar Matrix.
\nAnswer:
\nIn a square matrix the principal diagonal elements are same and all the other elements are zeroes.
\n\"2nd<\/p>\n

\"KSEEB<\/p>\n

Question 4.
\nFind a value of x for which
\n\"2nd
\nAnswer:
\n\"2nd
\n3 – x2<\/sup> = 3 – 8 \u21d2 x2<\/sup> = 8 \u21d2 x = \u00b12\\(\\sqrt{2}\\)<\/p>\n

Question 5.
\nIf y = sin(x2<\/sup> + 5) then find \\(\\frac{d y}{d x}\\)
\nAnswer:
\ny = sin(x2<\/sup> + 5)
\n\\(\\frac{d y}{d x}\\) = cos(x2<\/sup> + 5)(2x)<\/p>\n

Question 6.
\nFind \\(\\int(1-x) \\sqrt{x} d x\\)
\nAnswer:
\n\\(\\int(1-x) \\sqrt{x} d x\\)
\n\"2nd<\/p>\n

Question 7.
\nFind a value of x for which x(\u00ee + \u0135 + k\u0302) is a unit vector.
\nAnswer:
\nx(\u00ee + \u0135 + k\u0302) = 1 \u21d2 x\u00ee+ x\u0135+ xk\u0302 = 1
\n\u21d2\\(\\sqrt{x^{2}+x^{2}+x^{2}}\\) = 1 \u21d2 \\(\\sqrt{3 x^{2}}\\) = 1 \u21d2 x = \\(\\frac{1}{\\sqrt{3}}\\)<\/p>\n

Question 8.
\nIf a line has direction ratios 2, -1, -2 then determine its direction cosines.
\nAnswer:
\n\\(\\left(\\frac{2}{3}, \\frac{-1}{3}, \\frac{-2}{3}\\right)\\)<\/p>\n

Question 9.
\nDefine objective function in Linear Programming Problem.
\nAnswer:
\nThe linear function Z = ax + by, where ‘a’ and ‘b’ are constants which has to be maximised or minimised is called objective function.<\/p>\n

Question 10.
\nIf P(E) = 0.6, P(F) = 0.3 and P(E n f) = 0.2 then find P(F\/E).
\nAnswer:
\n\\(P(F \/ E)=\\frac{P(F \\cap E)}{P(E)}=\\frac{0.2}{0.6}=\\frac{2}{6}=\\frac{1}{3}\\)<\/p>\n

\"KSEEB<\/p>\n

Part-B<\/span><\/p>\n

Answer any TEN questions: (10 \u00d7 2 = 20)<\/span><\/p>\n

Question 11.
\nShow that the function f: N \u2192 N given by f(x) = 2x is one-one but not onto.
\nAnswer:
\nThe function\/is one-one, for f(X1<\/sub>) = f(x2<\/sub>)
\n\u21d2 2x1<\/sub> = 2x2<\/sub> \u21d2 x1<\/sub> = x2<\/sub> Further,\/is not onto, as for 1 \u2208 N, there does not exist any x in N such that f(x) = 2x = 1.<\/p>\n

Question 12.
\nProve that sin-1<\/sup> x + cos-1<\/sup> x = \\(\\frac{\\pi}{2}\\), x \u2208 [-1,1].
\nAnswer:
\nLet sin-1<\/sup> x = \u03b8 then.
\nx = sin \u03b8 \u21d2 x = cos(\\(\\frac{\\pi}{2}\\) – \u03b8)
\n\u21d2 Cos-1<\/sup>x = \\(\\frac{\\pi}{2}\\) – \u03b8 \u21d2 \u03b8 + cos-1<\/sup> x = \\(\\frac{\\pi}{2}\\)
\n\u2234 sin-1<\/sup> x + cos-1<\/sup> x = \\(\\frac{\\pi}{2}\\)<\/p>\n

Question 13.
\nWrite cot-1<\/sup>\\(\\left(\\frac{1}{\\sqrt{x^{2}-1}}\\right)\\), x > 1 simplest form.
\nAnswer:
\nLet x = sec q \u21d2 q = sec-1<\/sup> x.
\n\"2nd<\/p>\n

Question 14.
\nFind the area of the triangle with vertices (2, 7), (1, 1) and (10, 8) using determinant method.
\nAnswer:
\nThe area of triangle is given by
\n\"2nd
\n= \\(\\frac { 1 }{ 2 }\\) |[-2 (1 – 8) -7 (1 – 10) + 1 (8 – 10)]|
\n= \\(\\frac { 1 }{ 2 }\\) |-14 + 63 – 2| = \\(\\frac { 47 }{ 2 }\\) sq.units<\/p>\n

Question 15.
\nFind \\(\\frac{d y}{d x}\\) if y = (logx)cosx<\/sup>.
\nAnswer:
\nLet y = (log x)cosx<\/sup>
\n\u21d2 log y = cos x log (logx)
\n\u21d2 \\(\\frac{d}{d x}\\) log y = \\(\\frac{d}{d x}\\) cos x log(log x)
\n\"2nd<\/p>\n

Question 16.
\nFind ax + by2<\/sup> = cosy then find \\(\\frac{dy}{d x}\\)
\nAnswer:
\nax + by2<\/sup> = cosy
\n\u21d2 \\(\\frac{d}{d x}\\)ax + \\(\\frac{d}{d x}\\)by2<\/sup> = \\(\\frac{d}{d x}\\)cosy
\n\u21d2 a + 2by \\(\\frac{dy}{d x}\\) = (-sin y)\\(\\frac{dy}{d x}\\) dx dx
\n(2by + sin y) \\(\\frac{dy}{d x}\\) = -a \u21d2 \\(\\frac{d}{d x}\\) = \\(-\\frac{a}{2 b y+\\sin y}\\)<\/p>\n

Question 17.
\nFind the approximate change in the volume V of a cube of side x meters caused by increasing side by 2%.
\nAnswer:
\nNote that V = x3<\/sup>
\nor dV = \\(\\left(\\frac{d V}{d x}\\right)\\) \u2206x = (3x2<\/sup>) \u2206x
\n= (3x2<\/sup>) (0.02x) = 0.06x3<\/sup>m3<\/sup>
\nThus, the approximate change in volume is 0.06 x3<\/sup>m3<\/sup><\/p>\n

\"KSEEB<\/p>\n

Question 18.
\nFind \\(\\int \\frac{1}{\\cos ^{2} x(1-\\tan x)^{2}} d x.\\)
\nAnswer:
\nPut (1 – tan x) = t \u21d2 – sec2<\/sup> x dx = dt
\n\"2nd<\/p>\n

Question 19.
\nFind \\(\\int \\sin 2 x \\cos 3 x d x\\)
\nAnswer:
\nW.K.T sin x cosy = \\(\\frac { 1 }{ 2 }\\)[sin(x + y) +sin(x – y)]
\nThen \\(\\int \\sin 2 x \\cos 3 x d x=\\frac{1}{2}\\left[\\int \\sin 5 x d x-\\int \\sin x d x\\right]\\)
\n= \\(\\frac { 1 }{ 2 }\\) [\\(\\frac { 1 }{ 5 }\\)cos 5x + cos x] + C = \\(\\frac { -1 }{ 10 }\\) cos5x + \\(\\frac { 1 }{ 2 }\\) cos x + C.<\/p>\n

Question 20.
\nFind the order and degree (if defined) of the differential equation
\n\\(\\left(\\frac{d^{3} y}{d x^{2}}\\right)^{2}+\\cos \\left(\\frac{d y}{d x}\\right)\\)=0
\nAnswer:
\nOrder = 2, Degree = not defined.<\/p>\n

Question 21.
\nIf \\((\\vec{a}+\\vec{b}) \\cdot(\\vec{a}-\\vec{b})\\) = 8 and \\(|\\vec{a}|\\) = 8\\((\\vec{b})\\) then find \\(|\\vec{b}|\\).
\nAnswer:
\n\"2nd<\/p>\n

Question 22.
\nFind the projection of the vector a\u0304 = 2\u00ee + 3\u0135 + 2k\u0302 and b\u0304 = \u00ee – \u0135 + k\u0302
\nAnswer:
\nProjection of vector a\u0304 on b\u0304 is given by
\n\"2nd<\/p>\n

Question 23.
\nFind the distance of the point (3, -2, 1) from the plane 2x – y + 2Z + 3 = 0.
\nAnswer:
\n\"2nd<\/p>\n

\"KSEEB<\/p>\n

Question 24.
\nProbability of solving specific problem independently by A and B are \\(\\frac { 1 }{ 2 }\\) and \\(\\frac { 1 }{ 3 }\\) respectively. If both try to solve the problem independently, find the probability that the problem is solved.
\nAnswer:
\nGiven P(A) = \\(\\frac { 1 }{ 2 }\\) and P(B) = \\(\\frac { 1 }{ 23}\\)
\n(i) The problem is solved
\n= 1 – P(not solved) = 1 – P(A’ \u2229 B’)
\n= 1 – P(A’) P(B’) = 1 – [(1 – \\(\\frac { 1 }{ 2 }\\) – (1 – \\(\\frac { 1 }{ 3 }\\)]
\n= 1 – (\\(\\frac { 1 }{ 2 }\\) \u00d7 \\(\\frac { 2 }{ 3 }\\)) = 1 – \\(\\frac { 1 }{ 3 }\\) = \\(\\frac { 2 }{ 3 }\\)<\/p>\n

(ii) Exactly one of them solves problem
\n= P(A \u2229 B) + P(A \u2229 B’)
\n= P(A’) + P(B) + P(A) P(B’)
\n\\(=\\left(1-\\frac{1}{2}\\right)\\left(\\frac{1}{3}\\right)+\\frac{1}{2}\\left(1-\\frac{1}{3}\\right)\\)
\n\\(=\\left(\\frac{1}{2}\\right)\\left(\\frac{1}{3}\\right)+\\frac{1}{2}\\left(\\frac{2}{3}\\right)=\\frac{1}{2}\\)<\/p>\n

Part-C<\/span><\/p>\n

Answer any TEN questions: (10 \u00d7 3 = 30)<\/span><\/p>\n

Question 25.
\nCheck whether the relation R in 1R of real numbers defined by R = {(a, b): a < b3<\/sup>} is reflexive, symmetric or transitive.
\nAnswer:
\n(i) For (a, a) \u2208R, a < a3<\/sup> which is false.
\nAs for a = -2, a3<\/sup> = -8
\n\u2234 a \\(\\nleq\\) a3<\/sup> as – 2 \\(\\nleq\\) – 8
\n\u2234 R is not reflexive.<\/p>\n

(ii) For (a, b), a < b3<\/sup> and (b, a), b < a3<\/sup> which is false
\nAs for a = 1 and b = 2
\na < b3<\/sup> as 1 < 8
\nNow, b \\(\\nleq\\) a3<\/sup> as 2 \\(\\nleq\\) 1
\n\u2234 R is not symmetric.<\/p>\n

(iii) For (a, b), (b, c) and (a, c)
\na < b3<\/sup> and b < c3<\/sup>
\nNow a = 25, b = 3, c = 2
\na < b3<\/sup> and 25 < 27
\nb < c3<\/sup> and 3 < 8
\nbut a \\(\\nleq\\) c3<\/sup> as 25 \\(\\nleq\\) 8
\n.\u2019. R is not transitive.
\nHence R is neither reflexive, nor symmetric and nor transitive.<\/p>\n

Question 26.
\nProve that
\ncos-1<\/sup> (\\(\\frac { 4 }{ 5 }\\)) + cos-1<\/sup> (\\(\\frac { 12 }{ 13 }\\)) = cos-1<\/sup> (\\(\\frac { 33 }{ 65 }\\))
\nAnswer:
\nLet
\ncos-1<\/sup> (\\(\\frac { 4 }{ 5 }\\)) \u21d2 cos A = \\(\\frac { 4 }{ 5 }\\) \u21d2 Sin A = \\(\\sqrt{1-\\frac{16}{25}}=\\frac{3}{5}\\)
\ncos-1<\/sup> (\\(\\frac { 12 }{ 13 }\\)) = cos B = \\(\\frac { 12 }{ 13 }\\) \u21d2 sin B = \\(\\sqrt{1-\\frac{144}{169}}=\\frac{5}{13}\\)
\ncos(A + B) = cosA.cosB – sinA.sinB \\(=\\frac{4}{5} \\cdot \\frac{12}{13}-\\frac{3}{5} \\cdot \\frac{5}{13}=\\frac{48}{65}-\\frac{15}{65}=\\frac{33}{65}\\)
\ncos(A + B) = \\(\\frac { 33 }{ 65 }\\) \u21d2 A + B = cos-1<\/sup> (\\(\\frac { 33 }{ 65 }\\))
\ncos-1<\/sup> (\\(\\frac { 4 }{ 5 }\\)) + cos-1<\/sup> (\\(\\frac { 12 }{ 13 }\\)) = cos-1<\/sup> (\\(\\frac { 33 }{ 65 }\\))<\/p>\n

\"KSEEB<\/p>\n

Question 27.
\nBy using elementary transformations, find the inverse of the matrix
\n\"2nd
\nAnswer:
\n\"2nd<\/p>\n

Question 28.
\nIf x = a(\u03b8 + sin \u03b8), y = a(1 – cos\u03b8) then show that \\(\\frac{d y}{d x}=\\tan \\left(\\frac{\\theta}{2}\\right)\\)
\nAnswer:
\nWe have
\n\\(\\frac{d x}{d \\theta}\\) = a(1 + cos \u03b8), \\(\\frac{d y}{d \\theta}\\) = a(sin \u03b8)
\n\"2nd<\/p>\n

\"KSEEB<\/p>\n

Question 29.
\nVerify Rolle’s theorem for the function f(x) = x2<\/sup> + 2x – 8, x \u2208 [-4, 2].
\nAnswer:
\nf(x) is a polynomial in x. Hence it is continuous over [-4, 2] and differentiable over (-4,2).
\nf(-4) = (-4)2<\/sup> + 2 (-4) – 8 = 16 – 8 – 8 = 0
\nf(2) = 4 + 4 – 8 = 0
\n\u2234 f(-4) = f(2)
\n\u2234 All the conditions of the Rolle’s theorem are satisfied.
\n\u2234 there exists a c \u2208[-4,2] such that
\nf'(c) = 0 f'(x) = 2x + 2
\nf(c) = 2c + 2
\nf(c) = 0
\n\u21d2 2c + 2 = 0
\n\u21d2 2c = -2
\n\u21d2 c = -1 \u2208[-4, 2]
\n\u2234 Rolle’s theorem is verified.<\/p>\n

Question 30.
\nFind the intervals in which the function f given by f(x) = 2X3<\/sup> – 3x2<\/sup> – 36x + 7 is strictly increasing.
\nAnswer:
\nf(x) = 2x3<\/sup> – 3x2<\/sup> – 36x + 7
\nf ‘(x) = 6x2<\/sup> – 6x – 36
\nf ‘(x) = 0 \u21d2 6x2<\/sup> – 6x – 36 = 0
\n\"2nd
\n\u21d2 6(x2<\/sup> – x – 6) = 0
\n\u21d2 6(x – 3)(x + 2) = 0
\n6 \u2260 0, x = 3, x = -2
\n\"2nd<\/p>\n

Question 31.
\nFind \\(\\int x \\log x d x\\)
\nAnswer:
\n\"2nd<\/p>\n

\"KSEEB<\/p>\n

Question 32.
\nEvaluate \\(\\int_{0}^{\\frac{\\pi}{2}} \\frac{\\sin x}{1+\\cos ^{2} x} d x\\)
\nAnswer:
\n\"2nd<\/p>\n

Question 33.
\nFind the area of the region bounded by the curve y2<\/sup> = 4x and the line x = 3.
\nAnswer:
\nGiven curve is y2<\/sup> = 4x and line is x = 3
\ny2<\/sup> = 4x
\n\"2nd<\/p>\n

Question 34.
\nForm the differential equation of the family of curves y = ae3x<\/sup> + be-2x<\/sup> by eliminating arbitrary constants a and b.
\nAnswer:
\nGiven, y = a e3x<\/sup> + b e-2x<\/sup> …. (1)
\ny’ = (a e3x<\/sup>) 3 – (be-2x<\/sup>) 2
\ny’ = 3 a e3x<\/sup> – 2 be-2x<\/sup> …. (2)
\nFrom (1) and (2)
\ny = a e3x<\/sup> + b e-2x<\/sup> \u00d7 3
\ny’ = 3a e3x<\/sup> – 2b e-2x<\/sup> \u00d7 1
\n\"2nd
\nDiff
\n\u21d2 3y’ – y” = 5be-2x<\/sup>
\n3y’ – y” = -10be-2x<\/sup> ….(4)
\nFrom (3) & (4)
\n3y’ – y” = (-2) (5be-2x<\/sup>)
\n3y’ – y” = -2[3y – y’]
\n3y’ – y” = -6y + 2y’
\ny” – y’ – 6y = 0.<\/p>\n

\"KSEEB<\/p>\n

Question 35.
\nFind a unit vector perpendicular to each of the vector \\((\\vec{a}+\\vec{b})\\) and \\((\\vec{a}-\\vec{b})\\), where a\u0304 = \u00ee + \u0135 + k\u0302, b\u0304 = \u00ee + 2\u0135 + 3k\u0302.
\nAnswer:
\n\"2nd<\/p>\n

Question 36.
\nShow that the four points with position vectors: 4\u00ee + 8\u0135 + 12k\u0302, 2\u00ee + 4\u0135 + 6k\u0302, 3\u00ee + 5\u0135 + 4k\u0302 and 5\u00ee + 8\u0135 + 5k\u0302 are coplanar.
\nAnswer:
\nLet
\n\"2nd
\n= -2(21 + 0) + 4(7 + 8) – 6(0 + 3)
\n= -42 + 60 – 18 = -60 + 60 = 0
\n\u2234 A, B, C, D are coplanar.<\/p>\n

Question 37.
\nFind the vector equation of the plane passing through the points R(2, 5,-3), S(-2, -3, 5) and T(5, 3, -3).
\nAnswer:
\n\"2nd
\nTheorem: Derive intercept form of the equation of a plane
\nLet the equation of a plane be
\nAx + By + Cz + 0 = 0 (1)
\n\"2nd
\nLet the plane cuts X – axis at p(a, 0, 0), Y-axis at Q(0, b, 0), Z-axis at R(0, 0, c) substituting P(a, 0, 0) in equation (1)
\nAa + D = 0 \u21d2 Aa = -D \u21d2 A = \\(\\frac{-D}{a}\\)
\nsubstituting Q(0, b, 0) in equation (1)
\nBb + D = 0 \u21d2 Bb = D \u21d2 B = \\(\\frac{-D}{b}\\)
\nR(0, 0, c) in equation (1)
\nCc + D = 0 \u21d2 Cc = -D \u21d2 C = \\(\\frac{-D}{c}\\)
\nsubstituting A, B, C in (1)
\n\\(\\left(\\frac{-D}{a}\\right) x+\\left(\\frac{-D}{b}\\right) y+\\left(\\frac{-D}{c}\\right) z=-D\\)
\nDividing through out by -D
\n\\(\\Rightarrow \\frac{x}{a}+\\frac{y}{b}+\\frac{z}{c}=1\\)
\nis the required equation of the plane in intercept form<\/p>\n

\"KSEEB<\/p>\n

Question 38.
\nAn insurance company insured 2000 scooter drivers, 4000 car drivers and, 6000 truck drivers. The probability of an accident is 0.01, 0.03 and 0.15 respectively. One of the insured persons meets with an accident. What is the probability that he is a scooter driver?
\nAnswer:
\nLet S : be Insured person is scooter driver
\nC : be Insured person is car driver
\nT : be Insured person is truck driver
\nLet ‘A’ be the event “meets with an accident”.
\n\"2nd<\/p>\n

Part-D<\/span><\/p>\n

Answer any SIX questions: (6 \u00d7 5 = 30)<\/span><\/p>\n

Question 39.
\nLet f: N \u2192 Y be a function defined as
\nf(x) = 4x + 3, where
\ny = {y \u2208 N: y = 4x + 3 for some x \u2208 N}. Show that f is invertible. Find the inverse of f.
\nAnswer:
\nf:N \u2192 Y
\nf(x) = 4x + 3
\nTo prove that \/ is one – one
\nLet f(x1<\/sub>) = f(x2<\/sub>)
\n4x1<\/sub> + 3 = 4x2<\/sub> + 3
\n4x1<\/sub> = 4x2<\/sub>
\n\u2234 x1 <\/sub>= x2<\/sub>.
\n\u2234 f is one – one
\nTo prove that f is onto
\nLet y \u2208 Y there exits x \u2208 N such that
\nf(x) = y
\n4x + 3 = y
\n4x = y – 3, x = \\(\\frac{y-3}{4}\\)
\n\u2234 f is one-one and onto
\n\u21d2 f is bijective. Hence f is invertible
\n\u2234 f-1<\/sup>(y) = \\(\\frac{y-3}{4}\\)<\/p>\n

Question 40.
\nIf A = \\(\\left[ \\begin{matrix} 1 & 2 & 3 \\\\ 3 & -2 & 1 \\\\ 4 & 2 & 1 \\end{matrix} \\right] \\) then show that A2<\/sup> – 23A – 40I = 0
\nAnswer:
\nWe have
\n\"2nd
\n\"2nd<\/p>\n

\"KSEEB<\/p>\n

Question 41.
\nSolve the following system of linear equations by matrix method:
\n3x – 2y + 3z = 8
\n2x + y – z = 1
\n4x – 3y + 2z = 4
\nAnswer:
\nThe system of equations can be written in the form A X = B where
\n\"2nd
\n|A| = 3(2 – 3) + 2(4 + 4) + 3(-6 -4) = -17
\nHence, A is non singular matrix and so its inverse exists. Now
\nA11<\/sub> = -1 A12<\/sub> = -8 A13<\/sub> = -10
\nA21<\/sub> = -5 A22<\/sub> = -6 A23<\/sub> = +1
\nA31<\/sub> = -1 A32<\/sub> = +9 A33<\/sub> = +7
\n\"2nd<\/p>\n

Question 42.
\nIf y = (sin-1<\/sup> x). show that
\n\\(\\left(1-x^{2}\\right) \\frac{d^{2} y}{d x^{2}}-x\\left(\\frac{d y}{d x}\\right)=0\\)
\nAnswer:
\nGiven that y = sin-1<\/sup> x, we have
\ny1<\/sub> = \\(\\frac{1}{\\sqrt{1-x^{2}}}\\) i.e., (1 – x2<\/sup>)y1<\/sub>2<\/sup> = 1
\nso (1 – x2<\/sup>). 2y1<\/sub>y2<\/sub> + y1<\/sub>2<\/sup>(0 – 2x) = 0
\nHence (1 – x2<\/sub>) y2<\/sub> – xy1<\/sub> = 0<\/p>\n

Question 43.
\nThe length x of a rectangle is decreasing at the rate of 3 cm \/ min and the width y is increasing at the rate of 2 cm \/ min. When x = 10 cm and y = 6 cm, find the rates of change of
\n(i) the perimeter and
\n(ii) the area of the rectangle.
\nAnswer:
\nSince the length x is decreasing and the width y is increasing with respect to time, we have
\n\\(\\frac{d x}{d t}\\) = – 3 cm \/min and \\(\\frac{d y}{d t}\\) = 2 cm\/mm
\n(a) The perimeter P of a rectangel is given by P = 2 (x + y)
\nTherefore \\(\\frac{d P}{d t}\\) = \\(2\\left(\\frac{d x}{d t}+\\frac{d y}{d t}\\right)\\) = 2 (-3 + 2)
\n= – 2 cm \/ min<\/p>\n

(b) The area A of the rectangle is given A = x.y
\nTherefore \\(\\frac{d A}{d t}\\) = \\(\\frac{d x}{d t}\\) . y + x. \\(\\frac{d y}{d t}\\)
\n= -3(6)+ 10(2) (as x= 10 cm and y = 6 cm)
\n= 2 cm2<\/sup>\/min.<\/p>\n

\"KSEEB<\/p>\n

Question 44.
\nFind the integral of \\(\\frac{1}{x^{2}-a^{2}}\\) with respect to x and hence evalauate \\(\\int \\frac{1}{x^{2}-16} d x\\)
\nAnswer:
\n\"2nd
\n\"2nd<\/p>\n

Question 45.
\nUsing the method of integration, find the smaller area enclosed by the circle x2<\/sup> + y2<\/sup> = 4 and the line x + y = 2.
\nAnswer:
\nGiven circle x2<\/sup> + y2<\/sup> = 4 and the line is x + y = 2
\nx2<\/sup> + (2 – x)2<\/sup> = 4
\n[\u2234 y = 2 – x]
\n\"2nd
\n2x2<\/sup> – 4x = 0 \u21d2 2x(x – 2) = 0 \u21d2 x = 0, 2
\nNow, x2<\/sup> + y2<\/sup> = 4 and x + y = 2
\n\u21d2 y2<\/sup> = 4 – x2<\/sup> y = 2 – x,
\n\u21d2 y = \\(y=\\sqrt{2^{2}-x^{2}}\\)
\nRequired Area
\n\"2nd<\/p>\n

Question 46.
\nFind the general solution of the differential equation \\(\\frac{d y}{d x}\\)+ (secx)y = tanx, ( 0 \u2264 x \u2264 \\(\\frac{\\pi}{2}\\)).
\nAnswer:
\nGiven circle x2<\/sup> + y2<\/sup> = 4 and the line is x + y = 2
\nx2<\/sup> + (2 – x)2<\/sup> = 4 [\u2234 y = 2 – x]
\n\"2nd
\n2x2<\/sup> – 4x = 0 \u21d2 2x(x – 2) = 0 \u21d2 x = 0, 2
\nNow, x2<\/sup> + y2<\/sup> = 4 and x + y = 2
\n\u21d2 y2<\/sup> = 4 – x2\u00a0<\/sup> y = 2 – x,
\n\u21d2 y = \\(\\sqrt{2^{2}-x^{2}}\\)
\nRequired Area
\n\"2nd<\/p>\n

\"KSEEB<\/p>\n

Question 47.
\nDerive the equation of a line in space passing through a given point and parallel to a given vector in both vector and Cartesian form.
\nAnswer:
\n\"2nd
\nLet \\(\\overrightarrow{\\mathrm{r}}\\) be any position vector on the line
\n\u2018L\u2019 and \\(\\overrightarrow{\\mathrm{a}}\\) given point on L.
\nclearly, \\(\\overrightarrow{\\mathrm{AP}}\\) || \\(\\overrightarrow{\\mathrm{b}}\\)
\n\\(\\text { ie, } \\overrightarrow{\\mathrm{AP}}=\\lambda \\overrightarrow{\\mathrm{b}}\\)
\nwhere \u03bb – some Real number
\n\"2nd
\n\"2nd
\n\"2nd<\/p>\n

Question 48.
\nFive cards are drawn successively with replacement from a we!! shuffled deck of 52 cards. What is the probability that
\n(i) all the five cards are spades?
\n(ii) only five three cards are spaces?
\n(iii) none of spades?
\nAnswer:
\n\"2nd<\/p>\n

\"KSEEB<\/p>\n

Part-E<\/span><\/p>\n

Answer any ONE question: (1 \u00d7 10 = 10)<\/span><\/p>\n

Question 49.
\n(a) Prove that \\(\\int_{0}^{a} f(x) d x=\\int_{0}^{a} f(a-x) d x\\) and hence evaluate \\(\\int_{0}^{\\frac{\\pi}{2}} \\frac{\\cos ^{5} x}{\\sin ^{5} x+\\cos ^{5} x} d x\\)
\nAnswer:
\nProof: a – x = t
\n– dx = dt
\ndx = -dt
\nL.L: when x = 0, t = a
\nU.L: When x = a, t = 0
\n\"2nd<\/p>\n

(b) Evaluate
\n\"2nd
\nAnswer:
\nTaking a, b, c commn from R1<\/sub>, R2<\/sub>, R3<\/sub>
\n\"2nd
\n\"2nd<\/p>\n

\"KSEEB<\/p>\n

Question 50.
\nMinimise and Maximise
\nz = 5x + 10y
\nSubject to the constraints :
\nx + 2y \u2264 120
\nx + y \u2265 60
\nx – 2y \u2265 0
\nx \u2265 0 and y \u2265 0.
\nAnswer:
\nDraw the graph of the line
\n\"2nd
\nPut (0, 0) in the inequality x + 2y \u2264 120
\n0 + 2(0) \u2264 120
\n0 \u2264 120 (True)
\n\u2234 The half plane lies towards the origin
\nP (0, 0) in the inequality x + y \u2265 0
\n0 + 0 \u2265 0 (False)
\n\u2234 The half plane lies away from the origin
\nPut (0, 1) in the inequality x – 2y \u2265 0
\n0 – 2(1) \u2265 0
\n-2 \u2265 0 (False)
\n\u2234 The half plane does not contain (0, 1)
\nCheck the point (1, 0) in the inequality
\nx – 2y \u2265 0
\n1 – 2 (0) \u2265 0
\n\u21d2 1 \u2265 0 (True)
\n\u2234 The half plane contains (1,0)
\nSo the half plane is towards x – axis
\nCorner points Max & Min z = 5x+ 10y
\nA (40,20) z = 400
\nB (60, 30) z = 600 \u2192 Maximum
\nC (60, 0) z = 300 \u2192 Minimum
\nD (120, 0) z = 600 \u2192 Maximum
\n\u2234 At C (60, 0) z is minimum
\nAt B (60,30), D (120,0) z is maximum<\/p>\n

(b) Find the value of K, if
\nf(x) = \\(\\left\\{\\begin{array}{ll}
\n{K x+1,} & {\\text { if } x \\leq 5} \\\\
\n{3 x-5} & {\\text { if } x>3}
\n\\end{array}\\right.\\)
\nis continuous at x = 5. (4)
\nAnswer:
\nSince given f(x) is continous at x = 5 => lim f(x) = lim f(x)=\/(5)
\n\"2nd
\n\u21d2 K(5) + 1 = 3(5) – 5
\n\u21d2 5K+ 1 = 10 \u21d2 5K = 9
\n\u21d2 K = \\(\\frac { 9 }{ 5 }\\)<\/p>\n","protected":false},"excerpt":{"rendered":"

Students can Download 2nd PUC Maths Previous Year Question Paper March 2019, Karnataka 2nd PUC Maths Model Question Papers with Answers helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations. Karnataka 2nd PUC Maths Previous Year Question Paper March 2019 Time: 3 Hrs 15 Min Max. Marks: …<\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[82],"tags":[],"jetpack_sharing_enabled":true,"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/18689"}],"collection":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/users\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/comments?post=18689"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/18689\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=18689"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=18689"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=18689"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}