{"id":19037,"date":"2020-01-09T14:23:21","date_gmt":"2020-01-09T08:53:21","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=19037"},"modified":"2021-07-02T15:30:33","modified_gmt":"2021-07-02T10:00:33","slug":"2nd-puc-maths-model-question-paper-3","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/2nd-puc-maths-model-question-paper-3\/","title":{"rendered":"2nd PUC Maths Model Question Paper 3 with Answers"},"content":{"rendered":"

Students can Download 2nd PUC Maths Model Question Paper 3 with Answers, 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 Model Question Paper 3 with Answers<\/h2>\n

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

Instructions:<\/p>\n

    \n
  1. The question paper has five parts namely A, B, C, D and E. Answer all the parts.<\/li>\n
  2. Use the graph sheet for the question on Linear programming in PART E.<\/li>\n<\/ol>\n

    Part – A<\/span><\/p>\n

    I. Answer all the questions ( 10 \u00d7 1 =10 )<\/span><\/p>\n

    Question 1.
    \nLet * be a binary operation defined on the set of Rational numbers Q defined by a*b = ab + 1 prove that * is a commutative.
    \nAnswer:
    \nIf * be a commutative binary operation.
    \na*b = b*a
    \n\u21d2 a*b = ab + 1 = ba + 1= b*a
    \n\u2234 * is commutative binary operation on *.<\/p>\n

    Question 2.
    \nFind the principle value of cos-1<\/sup>(-1\/2)
    \nAnswer:
    \n\"2nd<\/p>\n

    Question 3.
    \nDefine a diagonal matrix.
    \nAnswer:
    \nA square matrix in which except the principle diagonal elements all the elements are zero.
    \n\"2nd<\/p>\n

    \"KSEEB<\/p>\n

    Question 4.
    \nIf A = \\(\\left[ \\begin{matrix} 1 & 2 \\\\ 4 & 2 \\end{matrix} \\right]\\) then find the value of |2A|.
    \nAnswer:
    \n\"2nd<\/p>\n

    Question 5.
    \nFind \\(\\frac{\\mathrm{d} y}{\\mathrm{d} x}\\) if y = cos(1 – x).
    \nAnswer:
    \ny = cos(1 – x)
    \nDijf w.rl.x
    \n\\(\\frac{d y}{d x}\\) = \\(\\frac{d}{d x}\\) = cox(1 – x)
    \n= -sin(1 – x) x \\(\\frac{d}{d x}\\) (1 – x) dx
    \n= -sin(1 – x)x – 1 = sin(1 – x)<\/p>\n

    Question 6.
    \nEvaluate \u222b(2x – 3 cos x + ex<\/sup> ).dx
    \n=> \u222b(2x – 3cosx +ex<\/sup> ).dx
    \n= \u222b 2 x.dx – 3 \u222b cos x.dx + \u222b ex<\/sup> .dx
    \n= 2 x \\(\\frac{x^{2}}{2}\\) – 3 x sinx + e +c
    \n= x2<\/sup> – 3sin x + ex<\/sup> + c<\/p>\n

    Question 7.
    \nDefine a Unit vector.
    \nAnswer:
    \nA vector having Unit magnitude along any vector \\(\\vec{a}\\) is Called a unit vector in the direction of \\(\\vec{a}\\) add is denoted \\(\\vec{a}|\\vec{a}|\\) = 1<\/p>\n

    \"KSEEB<\/p>\n

    Question 8.
    \nIf a line makes angle 90\u00b0, 60\u00b0 and 30\u00b0 with positive direction of x, y and z axis respectively. Find its direction cosines.
    \nAnswer:
    \nThe Direction cosines are given by.
    \n=(cos90\u00b0, cos60\u00b0, cos30\u00b0).
    \n= (0, 1\/2, \u221a3\/2 )<\/p>\n

    Question 9.
    \nIn linear programming problems, define linear objectives function.
    \nAnswer:
    \nThe linear function z = ax + by where a, b are constants which has to be maximised or minimized is called linear objective function.<\/p>\n

    Question 10.
    \nIf P(A) = 0.6 P(B) = 0.3 and P(A \u2229B ) = 0.2 find P|A|B).
    \nAnswer:
    \nP(A\/B) = \\(\\frac{P(A \\cap B)}{P(B)}=\\frac{0.2}{0.3}=\\frac{2}{3}\\)<\/p>\n

    Part – B<\/span><\/p>\n

    II. Answer any Ten questions :<\/span><\/p>\n

    Question 11.
    \nS.T. function f : N \u2192 N by f (1) = f(2) = 1 and f(x) = x -1 for every x > 2, is on to but not one-one.
    \nAnswer:
    \nf : N \u2192 N by f(1) = f(2) = 1 and f(x) = x – 1.
    \nf is not one-one because
    \nf (1) = 1 and f(2) = 1
    \n\u2234 f(1) = f(2)
    \nbut
    \n1\u22602
    \n\u2234 f is not one-one
    \nfor every y \u2208 N then
    \nf(x) = y – x – 1 then y = x – 1
    \n\u21d2 x \u2208 N
    \n\u2234 y \u2208 N \u220b x \u2208 N
    \n\u2234 f is onto.<\/p>\n

    \"KSEEB<\/p>\n

    Question 12.
    \nP.T. tan -1<\/sup> – cot -1<\/sup> x = \\(\\frac{\\pi}{2}\\) \u2200 x \u2208R
    \nAnswer:
    \nLet tan -1<\/sup> x = y
    \nx = tan y = cot ( \\(\\pi \/ 2\\) – y)
    \ncot -1<\/sup> x = \\(\\pi \/ 2\\) – y
    \ncot -1<\/sup> x + y = \\(\\pi \/ 2\\)
    \ncot -1<\/sup> x – tan -1<\/sup> x = \\(\\pi \/ 2\\)<\/p>\n

    Question 13.
    \nWrite the simplest form of \\(\\tan ^{-1}(\\sqrt{\\frac{1-\\cos x}{1+\\cos x}})\\) 0 < x < \u03c0
    \nAnswer:
    \n\"2nd<\/p>\n

    Question 14.
    \nFind the equation of a line passing through (3,1) (9,3) using determinants.
    \nLet P(x,y) be the line passing through (3,1) and (9,3)
    \n\u2234 Let A(3,1) B (9,3)
    \n\u2234 If P is a point on the line AB then Area of the \u0394le<\/sup> is zero
    \n\"2nd
    \n\u2234x(1 – 3) – y(3 – 9) + 1(9 – 9) = 0
    \n-2x – y (-6) + (0) = 0
    \n-2x + 6y = 0
    \n6y = 2x
    \nx = 3 y
    \n\u2234 Equation of line is x = 3y<\/p>\n

    Question 15.
    \nif \\(\\sqrt{x}+\\sqrt{y}=\\sqrt{10}\\) S.T \\frac{d y}{d x}+\\sqrt{\\frac{y}{x}}=0[\/latex]
    \nAnswer:
    \n\"2nd
    \n\"2nd<\/p>\n

    \"KSEEB<\/p>\n

    Question 16.
    \nFind \\(\\frac{\\mathrm{d} \\mathbf{y}}{\\mathrm{d} \\mathbf{x}}\\) If y = (log x)cos x<\/sup>
    \nAnswer:
    \ny = (log x)cos x<\/sup>
    \nTake log on both side
    \nlog y – log(log x)cos x<\/sup>
    \nlog y – cos x. log(log x)
    \nDiff w.r.t x, we get
    \n\"2nd<\/p>\n

    Question 17.
    \nApproximate \u221a36.6 by using differential.
    \nAnswer:
    \ny = \u221ax Let x = 36 and Ax = 0.6 then
    \n\u0394y = \\(\\sqrt{x+\\Delta x}-\\sqrt{x}=\\sqrt{36.6}-\\sqrt{36}=\\sqrt{36.6}-6\\)
    \n\u221a36.6 = 6 + \u0394y
    \nNow dy is approximately equal to Ay and is given by.
    \n\"2nd
    \nThe approximate value of \u221a36.6 is 6+0.05 = 6.05.<\/p>\n

    Question 18.
    \nIntegrate sinx. sin(cosx) w.r.t. x
    \n\u222bsinx. sin(cosx).dx
    \nTake cos x = t
    \nDiff w.r.t x, we have
    \n-sinx = \\(\\frac{d t}{d x}\\)
    \n-sin x.dx = dt
    \nsin x.dx = -dt
    \n\u21d2 \u222bsin(cosx).sinx.dx
    \n= \u222bsin(t)\u03bb – dt = – \u222b sin t.dt
    \n= -x – cos t. + c =cost + c = cos(cosx) + c<\/p>\n

    \"KSEEB<\/p>\n

    Question 19.
    \nEvaluate \\(\\int_{0}^{1} \\frac{1}{1+x^{2}} \\cdot d x\\)
    \nAnswer:
    \n\"2nd<\/p>\n

    Question 20.
    \nFind the order and degree of differential equation.
    \n\"2nd
    \nAnswer:
    \norder – 3 degree – 2<\/p>\n

    Question 21.
    \nFind the area of parallelogram whose adjacent sides determine by the vectors.
    \n\\(\\overrightarrow{\\mathrm{a}}=\\hat{\\mathbf{i}}+\\hat{\\mathbf{j}}-\\hat{\\mathbf{k}}\\) and \\(\\overrightarrow{\\mathbf{b}}=\\hat{\\mathbf{i}}-\\hat{\\mathbf{j}}+\\hat{\\mathbf{k}}\\)
    \nAnswer:
    \nArea of the parallelogram whose adjacent sides are given by.
    \n\"2nd<\/p>\n

    Question 22.
    \nObtain the projection of the vector \\(\\vec{a}=2 \\hat{i}+3 \\hat{j}+2 \\hat{k}\\) on the vector \\(\\overrightarrow{\\mathbf{b}}=\\hat{\\mathbf{i}}+2 \\hat{\\mathbf{j}}+\\hat{\\mathbf{k}}\\)
    \nAnswer:
    \n\"2nd<\/p>\n

    Question 23.
    \nFind the equation of the plane through the inter section of planes 3x – y + 2z – 4 = 0 and x + y – z – 2 = 0 and the point (2,2,1).
    \nAnswer:
    \nEquation of any plane through the intersection of the given planes, is of the form.
    \n3x – y + 2z – 4 + \u03bb (x + y – z – 2) = 0
    \n(3 + \u03bb)x + (\u03bb – 1)y + (2 – \u03bb) z – (4 + 2\u03bb) = 0
    \nBy Data it is passes through (2,2,1)
    \n\u21d2 (3 + \u03bb)2 + (\u03bb -1)2 + (2 – \u03bb)1 – (4 + 2\u03bb) = 0
    \n\u21d2 6 + 2\u03bb + 2\u03bb – 2 + 2 – \u03bb – 4 – 2\u03bb = 0
    \n\u21d2 2 + \u03bb = 0
    \n\u03bb = – 2
    \nThus the required equation is
    \n(3 – 2)x + (-2 -1 )y + (+2 – 2)1 – (4 + 2(-2) = 0
    \nx – 3y + (0)(1) – (4 – 4) = 0
    \nx – 3y = 0<\/p>\n

    \"KSEEB<\/p>\n

    Question 24.
    \nA die is thrown. If E is the event the number appearing is a multiple of 3 are F be the event the number appearing is even, then prove that E and F are independent events.
    \nAnswer:
    \nA die is thrown the sample space are {1,2,3,4,5,6}
    \nE is the event \u201cthe number appearing is a multiple of 3\u201d
    \nE = {3,6} P(E) = \\(\\frac{n(E)}{n(S)}=\\frac{2}{6}=\\frac{1}{3}\\)
    \nF be the event the number appearing is even
    \nF = {2, 4, 6} P(F) = \\(\\frac{n(F)}{n(S)}=\\frac{3}{6}=\\frac{1}{2}\\)
    \n\u21d2 (E\u2229F) = {6} =P(E\u2229F) =\u00a0 \\(\\frac{n(E \\cap F)}{n(s)}=\\frac{1}{6}\\)
    \n\u2234P(E)
    \n\u2234 P(E) \u00d7 P(F) = P(E\u2229F)
    \n\u2234 \\(\\frac{1}{3} \\times \\frac{1}{2}=\\frac{1}{6} \\Rightarrow \\frac{1}{6}=\\frac{1}{6}\\)
    \n\u2234 E and F are independent Events.<\/p>\n

    Part – C<\/span><\/p>\n

    III. Answer any Ten questions :<\/span><\/p>\n

    Question 25.
    \nShow that the Relation R in the set Z of integers given by R= {(x y) : 2 divides (x-y)} is an Equivalence Relation.
    \nAnswer:
    \n\u2192 R – {(xy): 2 divides (x – y) }
    \n\u2192 R is Reflexive as 2 divides (a – a) \u2200 a \u2208 z.
    \nIf (a b) \u2208 R then 2 divides a – b.
    \n2 divids b – a Hence (ba)\u2208R
    \n\u21d2R is Symmetric.
    \n(ab) \u2208 R and (be) \u2208 R
    \n\u21d2 a – b and b-c are divisible by 2
    \nNow a – c = (a – b) + (b – c)
    \n= is even
    \n.’. a – c is divisible by 2.
    \n\u21d2 R is Transitive
    \nR is Reflexive, Transitive, Symmetric
    \n.’. R is an Equivalence Relation in Z.<\/p>\n

    \"KSEEB<\/p>\n

    Question 26.
    \nProve that \\(\\tan ^{-1} x \\tan ^{-1}\\left(\\frac{2 x}{1-x^{2}}\\right)\\) = \\(\\tan ^{-1}\\left[\\frac{3 x-x^{3}}{1-3 x^{2}}\\right] \\cdot|x|<\\frac{1}{\\sqrt{3}}\\)
    \nAnswer:
    \n\u2192 Take tan-1<\/sup> x = \u03b8 \u21d2 x = tan\u03b8
    \n\"2nd
    \n\"2nd<\/p>\n

    Question 27.
    \nFor any square matrix A with Real numbers. Prove that A + A1<\/sup> is a symmetric and A – A1<\/sup> is a skew symmetric.
    \nAnswer:
    \nLet B = A + A1<\/sup> then
    \nB1<\/sup> = (A + A1<\/sup>)1<\/sup> = A1<\/sup> + (A1)1<\/sup>
    \n= A1 <\/sup>+ A= A + A1<\/sup> =B
    \n\u2234 B1<\/sup> =B \u2234B is symmetric matrix
    \n\u2234 A + A1<\/sup> is a symmetric matrix.
    \nC = A – A1<\/sup>
    \nC1<\/sup> = (A – A1<\/sup> )1<\/sup> = A1<\/sup> – (A1<\/sup>)1<\/sup>
    \n= A1 <\/sup>– A = -(A – A1<\/sup>) = -C
    \nC1<\/sup> = -C is a skew-symmetric matrix.
    \n\u2234 A – A1<\/sup> is a skew-symmetric matrix.<\/p>\n

    Question 28.
    \nIf x = a(\u03b8 + sin\u03b8) and y = a(1 – cos\u03b8). Prove that \\(\\frac{d y}{d x}=\\tan \\frac{\\theta}{2}\\)
    \nAnswer:
    \n\u21d2 x = a(\u03b8 + sin\u03b8)
    \nDiff.w.r.t \u03b8
    \n\"2nd<\/p>\n

    Question 29.
    \nVerify mean value theorem if f(x) = x2<\/sup> – 4x – 3 in the interval [ab]
    \nwhere a = 1 and b = 4.
    \n\u2192 Given f(x) = x2<\/sup> – 4x – 3
    \n\u21d2 f1<\/sup> (x) = 2x – 4 which exists \u2200 x in R.
    \n\u21d2f is derivable for all x in particular it is derivable in
    \n[a b] = [1 4] and hence continuous in[a b]= [ 1 4] Since f is derivable in [1 4 ] \u21d2 f is derivable in (1 4).
    \n\u2234 Both the condition of L.M.V. Theorem as satisfied.
    \n\u2203 at least one real C in (1 4) \u220b
    \n\"2nd
    \n\u21d2 MVT is verified for f(x) in [14],<\/p>\n

    Question 30.
    \nFind two positive numbers x and y such that x + y = 60 and xy3<\/sup> is maximum.
    \n\u2192 x + y = 60
    \ny = 60 – x
    \nP = xy3<\/sup>
    \nP = x(60 – x)3<\/sup>
    \nP = x(60 – x)3<\/sup>
    \nDiffw.r.t x.
    \n\"2nd
    \n= x.3(60 – x)2<\/sup> x (-1) + (60 – x)2<\/sup>.1
    \n= -3x(60 – x)2<\/sup> + (60 – x)3<\/sup>
    \n= (60 – x)3<\/sup> – 3x(60 – x)2<\/sup>
    \n=(60 – x)2<\/sup> [60-x – 3x)
    \n= (60 – x)2<\/sup> (60 – 4x)
    \nxy3<\/sup> is to be maximum
    \n\\(\\frac{d p}{d x}\\) = 0
    \n0 = (60-x)2<\/sup> (60-4x)
    \n60 – 4x = 0
    \n+4x = -60
    \nx = \\(\\frac{60}{4}\\)
    \nx = 15
    \n\u2234 x = 15 y = 45
    \nx + y = 60
    \ny = 60 – x
    \n= 60 – 15
    \n= 45<\/p>\n

    \"KSEEB<\/p>\n

    Question 31.
    \nEvaluate \u222bsin3x.cos4x.dx.
    \n\u222bcos4x.sin3x.dx
    \n= \u222b \\(\\frac{1}{2}\\) sin(4x + 3x) – sin(4x -3x)]. dx
    \n= \u222b(sin 7x – sin x) dx
    \n= \\(\\frac{1}{2}\\) \u222b sin 7x – \u222b\\(\\frac{1}{2}\\) sin x.dx
    \n\"2nd<\/p>\n

    Question 32.
    \nIntegrate x2<\/sup>.ex<\/sup> w.r.t x.
    \nanswer;
    \n\"2nd<\/p>\n

    Question 33.
    \nDetermine the area of the region bounded by y2<\/sup> = x and the lines x = 1 and x = 4 and the x-axis in the first quadrant.
    \nAnswer:
    \n\u2192 Given that y2<\/sup> = x
    \nIts a Equation of the Parabola and the Lines x = 1 and x = 4<\/p>\n

    \"2nd
    \n\u2234 The Required area of the
    \n\"2nd<\/p>\n

    \"KSEEB<\/p>\n

    Question 34.
    \nForm the differential Equation of the family of circles touching the x-axis at origin.
    \nAnswer:
    \n\u2192 Let C denote the family of circles touching x – axis at origin. Let (0,a) be the co-ordinates of the centre of any member of the family.
    \n\u2234 Equation of family C is. x2<\/sup> + (y – a)2<\/sup> = a2<\/sup>
    \nx2<\/sup> + y2<\/sup> = 2 ay
    \nWhere a is an arbitary constant.
    \nDifferentiating both side w.r.t x, we get
    \n\"2nd
    \nSubstitute the value of a from solution. (2) in eqn. (1), we get.
    \n\"2nd<\/p>\n

    Question 35.
    \nIf the two vector \\(\\overrightarrow{\\mathrm{a}}\\) and \\(\\overrightarrow{\\mathrm{b}}\\) such that \\(|\\overrightarrow{\\mathrm{a}}|\\) = 2
    \n\\(|\\overrightarrow{\\mathrm{b}}|\\) = 3 and \\(\\overrightarrow{\\mathrm{a}} \\cdot \\overrightarrow{\\mathrm{b}}=4\\) final \\(|\\overrightarrow{\\mathbf{a}} \\cdot \\overrightarrow{\\mathbf{b}}|\\)
    \nAnswer:
    \n\"2nd<\/p>\n

    Question 36.
    \nFind a Unit Vector perpendicular to each of the vector \\((\\vec{a}+\\vec{b})\\) and \\((\\vec{a}-\\vec{b})\\).
    \nWhere \\(\\overrightarrow{\\mathbf{a}}=\\hat{\\mathbf{i}}+\\hat{\\mathbf{j}}+\\hat{\\mathbf{k}}\\)\\(\\overrightarrow{\\mathbf{b}}=\\hat{\\mathbf{i}}+2 \\hat{\\mathbf{j}}+3 \\hat{\\mathbf{k}}\\)
    \nAnswer:
    \n\"2nd<\/p>\n

    Question 37.
    \nFind the shortest distance between the lines l1<\/sub>and l2<\/sub> whose vector equations are \\(\\overrightarrow{\\mathbf{r}}=\\hat{\\mathbf{i}}+\\hat{\\mathbf{j}}+\\lambda(2 \\hat{\\mathbf{i}}-\\hat{\\mathbf{j}}+\\hat{\\mathbf{k}}) and \\vec{r}=2 \\hat{i}+\\hat{j}-\\hat{\\mathbf{k}}+\\mu(3 \\hat{\\mathbf{i}}-5 \\hat{\\mathbf{j}}+2 \\hat{\\mathbf{k}})\\)
    \nAnswer:
    \n\"2nd
    \nHence the shortest distance b\/w the given lines is given by.
    \n\"2nd<\/p>\n

    \"KSEEB<\/p>\n

    Question 38.
    \nBag-I contains 3 red and 4 black balls while another Bag-II contains 5 red and 6 black balls. One ball is drawn at random from one of the bags and it is found to be red. Find the probability that it was drawn from Bag-II.
    \nAnswer:
    \n\u2192 Let E1<\/sub> be the event of choosing the bag I. E2 be the event of choosing the bag II and A be the event of drawing a red ball.
    \nThen P(E1<\/sub>) = P(E1<\/sub>) = \\(\\frac{1}{2}\\)
    \nP(A \/ E1<\/sub>) = P (drawing a red ball from Bag I) = \\(\\frac{3}{7}\\)
    \nP(A \/ E2<\/sub>) = P (drawing a red ball from Bag II) = \\(\\frac{5}{11}\\)
    \nNow, the probability of drawing a ball from Bag – III. being given that it is red, is P(E2<\/sub>\/A).
    \nBy using Baye\u2019s theorem, we have
    \n\"2nd
    \n= \\(\\frac{35}{68}\\)<\/p>\n

    Part – D<\/span><\/p>\n

    IV. Answer the following questions<\/span><\/p>\n

    Question 39.
    \nProve that the function f:N \u2192 N defined by ‘ f(x) = x2<\/sup> where y = {y:y = x2<\/sup>, x \u2208 N} is invertible. Also find the inverse of f.
    \nAnswer:
    \nf : N \u2192 N defined by f(x) = x2 <\/sup>then \u2203 another function g : N \u2192N by g(y) = x.
    \nf(x) = x2<\/sup>
    \n\u21d2 y = x2<\/sup>
    \nx = \u221ay
    \n\u2234 g(y) =\u221ay
    \nTo show that f is invertible
    \nfog(x) = x
    \ngof(y) = y
    \nfog(x) = f[g(x)]
    \n\"2nd
    \ngof(y) = g[f\/(y)]
    \n= g[y2<\/sup>] = y \u2208N
    \n\u2234 gof{y) = y \u2208 N
    \n\u2234 f is invertible
    \n\u2234\u00a0 inverse of f exist
    \n\u2234 f-1 <\/sup>: N \u2192 N f-1<\/sup>(y) = x
    \nf-1<\/sup>(y) = \u221ay .<\/p>\n

    Question 40.
    \nIf A = \\(\\left[ \\begin{matrix} 1 & 2 & 3 \\\\ 3 & -2 & 1 \\\\ 4 & 2 & 1 \\end{matrix} \\right]\\) then show that A3<\/sup> – 23 A – 40 I = 0
    \nAnswer:
    \n\"2nd
    \n\"2nd<\/p>\n

    Question 41.
    \nSolve the system of linear equation using matrix method,
    \nx – y + 2z = 7
    \n3x + 4y – 5z = -5
    \n2x – y + 3z = 12
    \nAnswer:
    \nThe given system of equation can be expressed as AX = B, where
    \n\"2nd
    \n\u2234 A is non-singular matrix so A-1<\/sup> exists.
    \nA11<\/sub> = 7, A12<\/sub> = -19, A13<\/sub> =-11, A21<\/sub> =1, A22<\/sub>= -1, A23<\/sub> = -1, A31<\/sub> = -3, A32<\/sub> = 11, A33<\/sub> = 7.
    \n\"2nd
    \non equating, we get x = 2, y = 1, z = 3<\/p>\n

    \"KSEEB<\/p>\n

    Question 42.
    \nGiven y = 3 cos (logx) + 4 sin (logx).
    \nx2<\/sup>y2<\/sub> + xy1 <\/sub>+ y = 0
    \nAnswer:
    \ny = 3 cos (logx) + 4 sin (logx)
    \nDiff w.r.t x
    \n\"2nd
    \n\"2nd<\/p>\n

    Question 43.
    \nThe length x of a rectangle is decreasing at the rate 5 cm\/min and width y is increasing at the rate of 4cm\/ min when x = 8cm and y = 6cm. Find the rate of change of (i) The perimeter (ii) The rate of the rectangle.
    \nAnswer:
    \nGiven \\(\\frac{d x}{d t}\\) = -5 cm \/min and \\(\\frac{d y}{d t}\\) = 4 cm \/min
    \na) perimeter = 2(x+ 5) \u21d2 P = 2 (x + 4)
    \n\"2nd
    \nb) A Area of.rectangle = xy
    \n\u21d2 A = x.y
    \n\"2nd<\/p>\n

    Question 44.
    \nFind the integral of \\(\\frac{1}{\\sqrt{x^{2}+a^{2}}^{2}}\\) with respect to x and hence. Evaluate \\(\\int \\frac{1}{\\sqrt{x^{2}+7}} \\cdot d x\\)
    \nAnswer:
    \n\"2nd
    \n\"2nd
    \nQuestion 45.
    \nFind the area of the region enclosed b\/w the two \u0398les<\/sup>. x2<\/sup> + y2<\/sup> = 4 and (x – 2)2<\/sup> + y2<\/sup> = 4
    \nAnswer:
    \nEquations of the given circles are
    \nx2<\/sup> + y2<\/sup> = 4 …………….. (1)
    \n(x – 2)2<\/sup> + y2<\/sup> = 4 …………………….(2)
    \nEquation (1) is a circle with centre O at the origin and radius 2. Eqn. (2) is a circle with centre C (2,0) and radius 2. Solving Eqn. (1) and (2), we have.
    \n(x – 2 )2 <\/sup>+ y2<\/sup> = x2<\/sup> + y2<\/sup>
    \nx2<\/sup> – 4x + 4 + y2<\/sup> = x2<\/sup> + y2<\/sup>
    \n+ 4x = -4 \u21d2 x = 1
    \nx = 1 which given y = \u00b1 \u221a3
    \nThus the points of intersection of the given circles are A(1,\u221a3) and A1<\/sup>( 1,-\u221a3} as shown in the fig.
    \nRequired area of the enclosed region OACA1<\/sup>O between circle s =2 [area of the region ODCAO]
    \n=2 [area of the region ODAO + area of the region on DCAD]
    \n\"2nd
    \n\"2nd
    \n\"2nd<\/p>\n

    \"KSEEB<\/p>\n

    Question 46.
    \nSolve \\((x \\log x) \\frac{d y}{d x}+y=\\frac{2}{x}(\\log x)\\)
    \nAnswer:
    \n\"2nd<\/p>\n

    Question 47.
    \nDerive equation of a line in a space through a given point and parallel to a vector both in the vector and cartesian form.
    \nAnswer:
    \nLet \\(\\vec{a}\\) be the position vector of the given point A with respect to the origion O of the rectangular co-ordinate system. Let l be the line which passes through the point A and is parallel to a given vector \\(\\vec{b}\\) Let \\(\\vec{r}\\) be the position vector of an arbitary point P on the line.<\/p>\n

    Then \\(\\vec{AP}\\) is parallel to the vector \\(\\vec{b}\\) i.e. \\(\\overrightarrow {A P}=\\lambda \\vec{b}\\) where \u03bb is some real number.
    \nBut \\(\\overrightarrow{A P}=\\overrightarrow{O P}-\\overrightarrow{O A}\\)
    \n\\(\\lambda \\vec{b}=\\vec{r}-\\vec{a}\\)
    \nConversely for each value of the parameter \u03bb, this equation gives the position vector of a point P on the line.
    \nHence the vector equantion of the line is given by
    \n\\(\\vec{r}=\\vec{a}+\\lambda \\vec{b}\\)
    \nDerivation of cartesian form from vector form.
    \nLet the co-ordinates of the given point A be (x, y, z) and the direction ratios of the line a b c consider the co-ordinates of any point P be (x,y,z) then
    \n\"2nd
    \nSubstituting these values in (1) and equating the co-efficients i, \\(\\hat{i}\\) and \\(\\hat{r}\\) we get
    \nx = x1<\/sub> + \u03bba ; y = y1<\/sub> + \u03bbb ; z = z1<\/sub> + \u03bbc ,
    \nThese are parametric equation of the line. Eliminating the Parameter \u03bb from (2) we get
    \n\"2nd
    \nThis is the cartesian equating of the line.<\/p>\n

    Question 48.
    \nIf a fair coins is tossed 10 times. Find the probability of
    \n(i) Exactly six heads and
    \n(ii) at least six heads.
    \nAnswer:
    \nThe repeated toss of a coin are Bernoullis trial s. Let X denote the number of heads in the experiment of 10 trials.
    \nClearly, X has the binomial distribution with
    \nn = 10 and P = \\(\\frac{1}{2}\\)
    \n\u2234 P(X = x) = n<\/sup>Cx<\/sub>qn-x<\/sup> – px. x = 1,2,3 ……………….. n
    \nn = 10 and P = \\(\\frac{1}{2}\\) q = 1 – p = \\(\\frac{1}{2}\\)
    \n\"2nd<\/p>\n

    Part – E<\/span><\/p>\n

    V. Answer any One question :<\/span><\/p>\n

    Question 49.
    \na) Prove that \\(\\int_{a}^{b} f(x) \\cdot d x=\\int_{a}^{b} f(a+b-x) d x\\) and hence. Evaluate \\(\\int_{\\pi \/ 3}^{\\pi \/ 3} \\frac{1}{1+\\sqrt{\\tan x}} d x\\)
    \nAnswer:
    \n\"2nd
    \nLet t = a + b – x then dt – dx
    \nWhen x = a and x = b
    \nt = b t = a
    \n\"2nd
    \n\"2nd
    \n(b) Determine the value of k, if
    \n\"2nd
    \nis continuous x = \\(\\pi \/ 2\\)
    \nAnswer:
    \nIf f is continuous at x = \\(\\pi \/ 2\\)
    \n\u21d2 \\(\\lim _{x \\rightarrow \\pi \/ 2} f(x)=f(\\pi \/ 2)\\)
    \nGiven at x = \\(\\pi \/ 2\\)
    \nf( \\(\\pi \/ 2\\) ) = 3
    \n\"2nd<\/p>\n

    \"KSEEB<\/p>\n

    Question 50.
    \na) Minimize and maximize
    \nZ = 5x + 10y
    \nsubject to the constraints
    \nx + 2y \u2264 120
    \nx + y \u2265 60,
    \nx – 2y \u2265 and x \u2265 0, y \u2265 0 by graphical method.
    \nAnswer:
    \nGiven : Z = 5x + 10y ……………. (1)
    \nsubject to constraints x + 2y \u2264 120 …………….(2)
    \nx + y > 60 …………….(3)
    \nx – 2y \u2265 0 …………….(4)
    \nx \u2265 0, y \u2265 0 …………….(5)
    \nFirst we locate the region represented by (2), (3), (4) & (5) for this consides the lines ;
    \nx + 2y = 120 which passes through A(120, 0) & B(0, 60)
    \nx + y = 60 which passes through C (60, 0) & B (0, 60) and x – 2y = 0 which passes through 0 (0, 0) & P (40, 20)
    \nNow that P(40, 20) lies on BC also. Also x – 2y = 0&x + 2y= 120 meet in Q (60, 30)
    \nThe feasible region is shown shaded in the figure (indicated as feasible region) note that (0,0) does nopt lie in this region as it doesnot satisfy (3) The point C(60,0) lie this region as it saties fies all the constraints (2) (3) (4) and (5).
    \n\"2nd
    \nThe corner points of the feasible region which are to be examined for optimum solution are C(60,0) A( 120,0) Q(60, 30) and P(40,20)
    \nAt C(60, 0), Z = 5 \u00d7 60 + 10 \u00d7 0 = 300
    \nAt A(120, 0), Z = 5 \u00d7 120 + 10 \u00d7 0 = 600
    \nAt Q(60, 30), Z = 5 \u00d7 60 + 10 \u00d7 30 = 600
    \nAt P(40, 20), Z = 5 \u00d7 40 + 10 \u00d7 20 = 400
    \nHence Z is minimum at C(60,0) and min Z = 300 and max Z = 600 at A and also at Q \u21d2 max Z will be attained at every point of segment AQ.
    \nb) Find the value of k, if
    \nf(x) = \\(\\frac{1-\\cos 2 x}{1-\\cos x}\\) , x\u2260 0
    \nf(x) = k ,x= 0
    \nis continuous at x = 0.
    \nAnswer:
    \n\"2nd<\/p>\n","protected":false},"excerpt":{"rendered":"

    Students can Download 2nd PUC Maths Model Question Paper 3 with Answers, 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 Model Question Paper 3 with Answers 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\/19037"}],"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=19037"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/19037\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=19037"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=19037"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=19037"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}