{"id":18514,"date":"2020-01-08T10:55:27","date_gmt":"2020-01-08T05:25:27","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=18514"},"modified":"2021-07-02T15:30:40","modified_gmt":"2021-07-02T10:00:40","slug":"2nd-puc-maths-model-question-paper-2","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/2nd-puc-maths-model-question-paper-2\/","title":{"rendered":"2nd PUC Maths Model Question Paper 2 with Answers"},"content":{"rendered":"

Students can Download 2nd PUC Maths Model Question Paper 2 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 2 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.
    \nA Relation R on A = {1,2,3} defined by R = {(1,1) (1,2) (3,3)} is not symmetric, Why?
    \nAnswer:
    \nThe Relation R is defined by R = {.(1,1) (1,2) (3,3)} for Symmetric Relation if (1,2)\u2208R then (2,1) \u2208R but (2,1) \u2209 R?
    \n\u2234 R is not Symmetric Relation.<\/p>\n

    Question 2.
    \nWrite the domain of f(x) = cos-1<\/sup> x.
    \nAnswer:
    \nThe domain of cos-1<\/sup>x is [-1,1]<\/p>\n

    Question 3.
    \nDefine a scalar matrix.
    \nAnswer:
    \nA diagonal matrix in which all the Principle diagonal Elements are equal is called scalar matrix.<\/p>\n

    \"KSEEB<\/p>\n

    Question 4.
    \nIf A = \\(\\left[ \\begin{matrix} 1 & 2 \\\\ 3 & 4 \\end{matrix} \\right] \\) find |2A|.
    \nAnswer:
    \n\"2nd<\/p>\n

    Question 5.
    \nIf y = log (sin x) find \\(\\frac{\\mathrm{dy}}{\\mathrm{d} \\mathrm{x}}\\)
    \nAnswer:
    \ny = log(sin x)
    \nDifferentiae w.r.t.x, we get
    \n\\(\\frac{\\mathrm{dy}}{\\mathrm{d} \\mathrm{x}}\\) = \\(\\frac{1}{\\sin x}\\) . cosx = cotx
    \n\\(\\frac{\\mathrm{dy}}{\\mathrm{d} \\mathrm{x}}\\) = cot x<\/p>\n

    Question 6.
    \nEvaluate \u222b( sin x + cos x) .dx.
    \nAnswer:
    \n\u222b(sinx + cosx).dx
    \n= \u222b sin x.dx + \u222b cos x.dx.
    \n= -cosx + sin x + c
    \n= sinx – cosx + c<\/p>\n

    \"KSEEB<\/p>\n

    Question 7.
    \nFind the direction cosines of the vector \\(\\hat{\\mathbf{i}}+2 \\hat{\\mathbf{j}}+3 \\hat{\\mathbf{k}}\\)
    \nAnswer:
    \n\\(\\vec{a}=\\hat{i}+2 \\hat{j}+3 \\hat{k}\\)
    \n\"2nd<\/p>\n

    Question 8.
    \nFind the equation of the plane with intercept 4 on z-axis and parallel to xoy plane.
    \nAnswer:
    \n\u2234 The Required Equation of the plane is z =4.<\/p>\n

    Question9.
    \nDefine feasible region.
    \nAnswer:
    \nThe Common Region determined by all the cons traints including the noh-negative constraints (x \u2265 0, y \u2265 0) of a linear programming problem is called the feasible region.<\/p>\n

    Question 10.
    \nIf P(A) = \\(\\frac{4}{5}\\) and P(B\/A) = \\(\\frac{2}{5}\\) find
    \nP(A\u2229B)
    \nAnswer:
    \nP(A \/B) = \\(\\frac{P(A \\cap B)}{P(A)}\\)
    \n\u2234 P(A\u2229B)\u00a0 = P(B\/A).P(A)
    \n= \\(\\frac{2}{5} \\cdot \\frac{4}{5}=\\frac{8}{25}\\)
    \n\u2234 P(A\u2229B) = \\(\\frac{8}{25}\\)<\/p>\n

    \"KSEEB<\/p>\n

    Part – B<\/span><\/p>\n

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

    Question 11.
    \nVerify whether the operation * defined on Q by a*b = \\(\\frac{\\mathrm{ab}}{2}\\) is associative or not.
    \nAnswer:
    \n* is defined on Q by a*b = \\(\\frac{\\mathrm{ab}}{2}\\) for associative we have to prove that
    \na* (b* c) = (a* b)* c
    \n\u2234 a*(b*c) = a * \\(\\frac{\\mathrm{ab}}{2}\\) b*c = \\(\\frac{\\mathrm{bc}}{2}\\)
    \n= \\(\\frac{\\mathrm{abc}}{4}\\) ……………….. (1)
    \n(a*b)*c = \\(\\frac{\\mathrm{ab}}{2}\\) *c
    \n= \\(\\frac{\\mathrm{abc}}{4}\\) ……………….. (2)
    \n\u2234 from (1) and (2)
    \n\u2234 * is Associative
    \n\u2234 * Satisfies the associative property.<\/p>\n

    Question 12.
    \nWrite the simplest form of
    \n\"2nd
    \nAnswer:
    \n\"2nd<\/p>\n

    Question 13.
    \nEvaluate
    \n\"2nd
    \nAnswer:
    \n\"2nd<\/p>\n

    Question 14.
    \nFind the equation of line joining (1,2) and (3,6) using determinants.
    \nAnswer:
    \nLet P (x,y) be any point on AB. Then Area of triangle ABP is is \u201cO\u201d
    \n\"2nd
    \nx(2 – 6) – y(1 – 3) + 1(6 – 6) = 0
    \n-4x + 2y = 0
    \n2x – y = 0<\/p>\n

    \"KSEEB<\/p>\n

    Question 15.
    \nIf y + siny = cosx find \\(\\frac{d y}{d x}\\)
    \nAnswer:
    \ny + siny = cos x
    \nDifferentiate w.r.t. x
    \n\"2nd
    \n\"2nd<\/p>\n

    Question 16.
    \nIf y = xx<\/sup> find \\(\\frac{d y}{d x}\\).
    \nAnswer:
    \ny = xx<\/sup>
    \nTake logarithm on Both side, we get
    \nlogy = logxx<\/sup> \u21d2 logy=xlogx
    \nDifferentiate wr.t. x
    \n\\(\\frac{1}{y} \\cdot \\frac{d y}{d x}\\) = x x \\(\\frac{1}{x}\\) + log x
    \n\\(\\frac{d y}{d x}\\) = y[1 + logx]
    \n\\(\\frac{d y}{d x}\\) = xx<\/sup>[1 + logx]<\/p>\n

    Question 17.
    \nFind the approximate charge in the volume v of a cube of side x metres caused by increasing the sind by 2%.
    \nAnswer:
    \nw.r.t volume of a cube = v = x3<\/sup>.
    \n\u21d2 \\(\\frac{d v}{d x}\\) = 3x2<\/sup> dx
    \n\u0394x<\/sub> = 0.02 x
    \n\u2234 dv = ( \\(\\frac{d v}{d x}\\) ) \u0394x<\/sub> = (3x2<\/sup>) \u0394x<\/sub> = 3x2<\/sup> x 0.02x
    \n= 0.06 x3<\/sup>m3<\/sup><\/p>\n

    \"KSEEB<\/p>\n

    Question 18.
    \nEvaluate \\(\\int \\frac{\\sin ^{2} x}{1+\\cos x} d x\\)
    \nAnswer:
    \n\"2nd<\/p>\n

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

    Question 20.
    \nFind the order and degree of the differential equation.
    \n\"2nd
    \nAnswer:
    \nOrder -2, Degree – 1<\/p>\n

    Question 21.
    \nIf \\(\\overrightarrow{\\mathrm{a}}\\) is a Unit Vector such that
    \n( \\(\\vec{x}-\\vec{a})(\\vec{x}+\\vec{a}\\)) = 8 find\u00a0 | \\(\\overrightarrow{\\mathbf{x}}\\) )
    \nAnswer:
    \n\"2nd<\/p>\n

    Question 22.
    \nFind the area of a parallelogram whose adjacent sides are given by the vectors
    \n\\(\\overrightarrow{\\mathrm{a}}=3 \\hat{\\mathrm{i}}+\\hat{\\mathrm{j}}+4 \\hat{\\mathrm{k}} \\cdot \\overrightarrow{\\mathrm{b}}=\\hat{\\mathrm{i}}-\\hat{\\mathrm{j}}+\\hat{\\mathrm{k}}\\)
    \nAnswer:
    \nThe area of a parallelogram with \\(\\vec{a}\\) and
    \n\\(\\vec{b}\\) as its adjacent sides is given by | \\(\\vec{a} \\times \\vec{b}\\) |
    \n\"2nd<\/p>\n

    Question 23.
    \nFind the angle between the pair of lines.
    \n\\(\\vec{r}\\) = 3\u00ee + 2 \u0135 – 4k\u0302 + \u03bb (\u00ee + 2\u0135 + 2k\u0302) and
    \n\\(\\vec{r}\\) = 5\u00ee + 2 \u0135 + \u03bb (3\u00ee + 2 \u0135 + 6k\u0302)
    \nAnswer:
    \nHere
    \n\\(\\vec{b}_{1}\\) = \u00ee + 2 \u0135 + 2k\u0302 and\u00a0 \\(\\vec{b}_{1}\\) = 3\u00ee + 2 \u0135 + 6k\u0302
    \nThe angle \u03b8 between the two lines is given
    \n\"2nd<\/p>\n

    Question 24.
    \nFind the probability ditribution of number of heads in two tosses of a coin.
    \nAnswer:
    \nWhen a coin is tossed twice the sample space is S = {HH,HT, TH,TT}
    \nLet x denote the number of heads in any out come in s.
    \n\u2234x(HH) = 2
    \nx(HT) = 1
    \nx(TH) = 1
    \nx(TT) = 0
    \nP(x = 0) = P(TT) = \\(\\frac{1}{4}\\)
    \nP(x = 1) P(HT,TH) = \\(\\frac{2}{4}\\) = \\(\\frac{1}{2}\\)
    \nP(x = 2) P(HT,TH) = \\(\\frac{1}{4}\\)
    \nProbability distribution of x is
    \n\"2nd<\/p>\n

    Part – C<\/span><\/p>\n

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

    Question 25.
    \nFind gof and fog if 1:R \u2192 R and g: R \u2192 R are given by f(x) = cosx and g(x) = 3x2<\/sup> show that gof \u2260 fog.
    \nAnswer:
    \nf(x) = cosx
    \ng(x)=3x2<\/sup>
    \ngof(x) = g[f(x)]
    \n= g[cosx]
    \n= 3cos2<\/sup>x
    \nfog(x) = f[g(x)]
    \n= f[3x2<\/sup>]
    \n= cos3x2<\/sup>
    \nfog \u2260 gof<\/p>\n

    Question 26.
    \nP.T. 3cos-1<\/sup>x=cos-1<\/sup>(4x3<\/sup> – 3x) x \u2208 \\(\\frac{1}{2}\\) , 1
    \nAnswer:
    \ncos-1<\/sup> x = A Now \\(\\frac{1}{2}\\) \u2264 x \u2264 1
    \ncosA = x
    \ncos3A = 4cos3<\/sup> A – 3cosA \u21d2\\(\\frac{1}{2}\\) \u2264 A < 1
    \ncos3A=4x3<\/sup> – 3x
    \n3A = cos -1<\/sup>(4x3<\/sup> – 3x) \u21d2 0 \u2264 A \u2264 \\(\\frac{\\pi}{3}\\)
    \n3cos-1<\/sup>x = cos-1<\/sup> [4x3<\/sup> – 3x] \u21d2 0 \u2264 3A \u2264 \u03c0<\/p>\n

    Question 27.
    \nBy using Elementary transformations
    \nfind the iii,e of the matrix
    \nA = missing
    \nIn order to use elementary row operation
    \nA = IA
    \n\"2nd
    \n\"2nd<\/p>\n

    Question 28.
    \nFind \\(\\frac{d y}{d x}\\) if x = a(cos\u03b8 + \u03b8sin\u03b8) and dx
    \ny = a(sin\u03b8 – \u03b8cos\u03b8)
    \nAnswer:
    \nx = a(cos\u03b8 + \u03b8sin\u03b8)
    \nDiff w.r.t \u03b8 we have
    \n\\(\\frac{d x}{d \\theta}\\) = a(-sin\u03b8 + \u03b8cos\u03b8 + sin\u03b8 )
    \n\\(\\frac{d x}{d \\theta}\\) = a[\u03b8cos\u03b8]
    \ny = a(sin\u03b8 – \u03b8cos\u03b8)
    \nw.r.t \u03b8
    \n\\(\\frac{d y}{d \\theta}\\)
    \n= a[cos\u03b8 – \u03b8 x sin \u03b8 – cos\u03b8]
    \n\\(\\frac{d y}{d \\theta}\\) = a[sin\u03b8]
    \n\"2nd
    \n\u2234 \\(\\frac{d y}{d x}\\) = tan \u03b8<\/p>\n

    \"KSEEB<\/p>\n

    Question 29.
    \nVerify Rolle\u2019s theorem for the function y = x2<\/sup> + 2, x \u2208[-2,2]
    \nAnswer:
    \nThe function y = x2<\/sup> +2 is continuous in [-2,2], and differentiable in (-2,2)
    \nAlso f (-2) \u21d2 f (x) = x2<\/sup> + 2
    \nf(- 2) = (-2)2<\/sup> + 2
    \n= 4 + 2 = 6
    \nf(2) = 22<\/sup> + 2 = 6
    \n\u2234 f(-2) = f(2) = 6
    \n\u2234 The value of f(x) at -2 and 2 coincide.
    \n\u2234 Rolle\u2019s Theorem states that there is a point
    \nc \u2208 (-2,2) where f'(c) = 0
    \nSince f'(x) = 2x, we have
    \nf’ (c) = 2c
    \n2c = 0
    \nc = 0
    \n\u2234 we have c = 0 \u2208 (-2,-2)<\/p>\n

    Question 30.
    \nFind the interval in which the function f given by f (x) = x2<\/sup> – 4x + 6 is
    \n(i) Strictly increasing.
    \n(ii) Strictly decreasing.
    \nAnswer:
    \nf(x) = x2<\/sup> – 4x + 6
    \nDifferentiate w.r.t x
    \nf'(x) = 2x – 4
    \n\u2234 f'(x) = 0 gives
    \n0 = 2x – 4
    \n2x = 4
    \nx = 2
    \n\"2nd
    \n\u2234 The point x = 2 divides the real line into two disjoint intervas namely, (-\u221e,2) and (2,\u221e).
    \nThe interval (-\u221e, 2) f'(x) = 2x – 4 < 0.
    \n\u2234 f is strictly decreasing in (-\u221e, 2). Also in the interval (2,\u221e) f'(x)>0 and so the function f is strictly increasing in (2,\u221e).<\/p>\n

    Question 31.
    \nFind \\(\\int \\frac{\\left(x^{2}+1\\right) \\cdot e^{x}}{(x+1)^{2}} \\cdot d x\\)
    \nAnswer:
    \n\"2nd
    \n\"2nd<\/p>\n

    Question 32.
    \nEvaluate \u222btan -1<\/sup>x.dx.
    \nAnswer:
    \n\u222btan -1<\/sup>x.dx.
    \n= \u222bxtan -1<\/sup>x. \u222btan -1<\/sup>dx + c
    \nApply Rule of integration by parts by taking tan-1 <\/sup>x as the first function.
    \n\"2nd<\/p>\n

    Question 33.
    \nFind the area of the region bounded by the curve y = x2<\/sup> and the line y = 4.
    \nAnswer:
    \ny = x2<\/sup> represents the parabola symmetrical about y – axis
    \n\"2nd<\/p>\n

    \u2234 The Area of the region AOBA is given by 2 \\(\\int_{0}^{4} x \\cdot d y=2\\) [Area of the region BONB bounded by curve, y-axis and the lines y=0 and y=4]
    \n\"2nd<\/p>\n

    Question 34.
    \nForm the differential equation representing the family of curves y=mx where m is arbitrary constant.
    \nAnswer:
    \nThe given family of curves represents the equation of straight lines.
    \n\u2234 y = mx ……………………. (1)
    \nDifferentiate w.r.t x, we get dy
    \n\\(\\frac{d y}{d x}\\) = m
    \n\u2234 substitute m = \\(\\frac{d y}{d x}\\), we have
    \ny = mx
    \n\u21d2 y = \\(\\frac{d y}{d x}\\) x
    \n\\(\\frac{d y}{d x}\\) = \\(\\frac{y}{x}\\)
    \n\\(\\frac{d y}{d x} – \\frac{y}{x}\\) = 0
    \nwhich is the required differential equation.<\/p>\n

    Question 35.
    \nProve that
    \n\"2nd
    \nAnswer:
    \n\"2nd<\/p>\n

    Question 36.
    \nIf \\(\\vec{a}=2 \\vec{i}+2 \\hat{j}+3 \\hat{k} \\overrightarrow{\\mathbf{b}}=-\\hat{\\mathbf{i}}+2 \\hat{\\mathbf{j}}+\\hat{\\mathbf{k}} and [latex]\\overrightarrow{\\mathbf{c}}=3\u00a0\\hat{\\mathbf{i}}+\\hat{\\mathbf{j}}\\)\u00a0 are such that \\(\\overrightarrow{\\mathbf{a}}+\\lambda \\overrightarrow{\\mathbf{b}}\\) is perpendicular to \\(\\overrightarrow{\\mathrm{c}}\\). Then find the value of \u03bb.
    \nAnswer:
    \n\"2nd<\/p>\n

    Question 37.
    \nFind the distance of a point (2, 5, -3) from the plane \\(\\vec{r} \\cdot(6 \\hat{i}-3 \\hat{j}+2 \\hat{k})=4\\)
    \nAnswer:
    \n\\(\\vec{a}=2 \\hat{i}+5 \\hat{j}-3 \\hat{k}\\)
    \n\\(\\vec{N}=6 \\hat{i}-3 \\hat{j}+2 \\hat{k}\\)
    \nd = 4
    \ndistance of a point from a plane is
    \n\"2nd
    \n\u2234 The distance of the point (2, 5 – 3) from the
    \ngiven plane is
    \n\"2nd<\/p>\n

    Question 38.
    \nA die is tossed thrice. Find the probability of getting an odd number at least once.
    \nAnswer:
    \nWhen a dice is tossed thrice, the sample space, is S = {(x,y,z):x,y z e (1,2,3,4,5,6}}
    \nwhich contains 6 x 6 x 6 = 216 Equally likely sample points.
    \nLet E : \u201can odd number at least once\u201d
    \nE1<\/sup> : not an odd number any time
    \nE1<\/sup> : “An even number on all throws\u201d
    \nE1<\/sup> = {(x,y,z):x,y,ze{2,4,6}j
    \ne1<\/sup> contains 3x3x3=27 equally likely simple events.
    \nRequired probability = P(E) = 1 – P(E).
    \n= \\(1-\\frac{27}{216}=1-\\frac{1}{8}=\\frac{7}{8}\\)<\/p>\n

    \"KSEEB<\/p>\n

    Part – D<\/span><\/p>\n

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

    Question 39.
    \nProve the function f : N \u2192 Y defined by f(x) = 4x + 3, where Y = {y : Y = 4x + 3, x\u2208N} is invertible. Also write the inverse of f(x).
    \nAnswer:
    \nf:N\u2192y defined by
    \nf(x) = 4x + 3
    \nand Assume a an inverse f-1<\/sup>:y \u2192 N. is defined by f-12<\/sup>y) = x.
    \n\u2234 wk.t
    \nf(x) = 4x + 3
    \ny = 4x + 3
    \ny – 3 = 4x
    \nx = \\(\\frac{y-3}{4}\\)
    \nand wk 7
    \nf-1<\/sup>(y) = x
    \nf-1<\/sup>(y) = \\(\\frac{y-3}{4}\\) f-1<\/sup> = g
    \nor
    \ng(y) = \\(\\frac{y-3}{4}\\)
    \n\u2234 To S.T. it exist inverse.
    \nTo prove it is inversible.
    \nWe have to prove that
    \nfog(x) = x
    \ngof(y) = y
    \n\"2nd<\/p>\n

    To find the inverse of a function :
    \nf-1<\/sup>:y \u2192 N. by f-1<\/sup>(y) = x
    \nf-1<\/sup>(y) = \\(\\frac{y-3}{4}\\)<\/p>\n

    Question 40.
    \nIf A = \\(\\left[ \\begin{matrix} 1 & 2 \\\\ 2 & 1 \\end{matrix} \\right]\\) B = \\(\\left[ \\begin{matrix} 2 & 0 \\\\ 1 & 3 \\end{matrix} \\right]\\) and C = \\(\\left[ \\begin{matrix} 1 & 1 \\\\ 2 & 3 \\end{matrix} \\right]\\) Calculate AC BC and (A+B).C. Also verify that (A+B). C = AC+BC.
    \nAnswer:
    \n\"2nd
    \nFrom (1) and (2)
    \n(A+B). C = AC+BC.<\/p>\n

    Question 41.
    \nSolve the following system of equations by matrix method. x-y + z = 4, 2x + y – 3y = 0 andx + y + z =2.
    \nAnswer:
    \nThe given system can be written as AX = B
    \n\"2nd
    \nThe given system is consistent and has a unique Solution.
    \n\"2nd<\/p>\n

    \"KSEEB<\/p>\n

    Question 42.
    \nIf y = 3e2x<\/sup>+2e3x<\/sup> P.T. \\(\\frac{d^{2} y}{d x^{2}}\\) – \\(\\frac{d y}{d x}\\) + 6y = 0
    \ny = 3e2x<\/sup> + 2e3x<\/sup>
    \nDifferentiate w.r.t x.
    \n\\(\\frac{d y}{d x}\\) = 6e2x<\/sup> + 6e3x<\/sup>
    \nAgain Diff w.r.t x
    \n\\(\\frac{d^{2} y}{d x^{2}}\\) = 12e2x<\/sup> + 18e3x<\/sup>
    \n\"2nd
    \n= 12e2x<\/sup> +183x<\/sup> – 5 ((5e2x<\/sup> + 6e3x<\/sup>) + 6(3e2x <\/sup>+ 2e3x<\/sup>)
    \n= 12e2x<\/sup> + 183x<\/sup> – 30e2x<\/sup> – 30e3x<\/sup>\u00a0+18e2x <\/sup>+ 12e3x<\/sup>
    \n= 30e2x<\/sup> + 30e3x<\/sup>\u00a0– 30e2x<\/sup> – 30e3x<\/sup>
    \n\"2nd<\/p>\n

    Question 43.
    \nA ladder 24 ft long leans against a vertical wall. The lower end is moving away at the rate of 3ft\/sec. Find the rate at which the top of the ladder is moving down wars. If its foot is 8ft from the wall.
    \nAnswer:
    \nThe foot of the ladder is at a distance x from the wall and the top is at a height y at instant of time t. then
    \n(24)2<\/sup> = x2<\/sup> + y2<\/sup>
    \nDiff w.r.t time, we get
    \n\"2nd
    \ngiven \\(\\frac{d x}{d t}\\) = 3ft \/ sec. dt
    \nand x = 8 ft.
    \ny2<\/sup> = (24)2<\/sup> – x2<\/sup>
    \n= (24)2<\/sup> – (8)2<\/sup>
    \n= 576 – 64
    \ny2<\/sup> = 512
    \ny = \u221a512
    \n0 = 2 \u00d7 8 \u00d7 3 + 2 \u00d7 \u221a5I2 \\(\\frac{d y}{d t}\\)
    \n= 48 + 2\u221a512
    \n\"2nd
    \n\"2nd
    \n\u2234 The ladder is moving downwards at rate of \\(\\frac{24}{\\sqrt{512}}\\) ft\/ sec. or 1.06100 ft \/ sec.<\/p>\n

    Question 44.
    \nFind \\(\\int \\frac{d x}{x^{2}-a^{2}}\\) and hence evaluate \\(\\int \\frac{d x}{3 x^{2}+13 x-10}\\)
    \nAnswer:
    \n\"2nd
    \n\"2nd<\/p>\n

    Question 45.
    \nFind the area of the region bounded by the two parabolas y = x2<\/sup> and y2<\/sup> = x.
    \nAnswer:
    \nThe given two curves are parabola y = x2<\/sup> and y2<\/sup> = x
    \nThe point of intersection of these two parabolas are 0 (0,0) and A (1,1) as shown in the fig.
    \nWe can set
    \ny2<\/sup> =x or y = \u221ax – f (x)
    \ny – x2<\/sup> = g(x) where
    \nf(x) \u2265 g(x) in [0,1]
    \n\"2nd
    \n\u2234 The required area of the Shaded Region
    \n\"2nd<\/p>\n

    \u2234The Required area of the Region is = \\(\\frac{1}{3}\\) units.<\/p>\n

    \"KSEEB<\/p>\n

    Question 46.
    \nFind the general solution of the differential equation. e\/tany. dx + (1 -ex<\/sup>). sec2<\/sup>y.dy = 0
    \nAnswer:
    \nex<\/sup>. tany dx + (1-ex<\/sup>). sec2<\/sup> y dy = 0 (1)
    \nSeparating the variables and integrating, we get
    \nThis can be done by divide throughout (l-ex<\/sup>)t any Eqn. (1) becomes and integrate
    \n\"2nd
    \n\"2nd
    \nWhich is the required solution.<\/p>\n

    Question 47.
    \nDerive the equation of the plane in normal form both in the cartesian and vector form.
    \nAnswer:
    \n\"2nd
    \nConsider a plane whose perpendicular distance from the origin is d. where d \u2260 0.
    \nIf \\(\\overrightarrow{\\mathrm{ON}}=d \\hat{\\mathrm{n}}\\)
    \nLet p be any point on the plane Therefore NP.OP=0(l)
    \n\"2nd
    \nlx + my + nz = d
    \nThis is the Required Equation of a plane in normal form.<\/p>\n

    To find P and Q :
    \nx2 <\/sup>= 4y and x = 4y – 2
    \n(4y – 2)2<\/sup> = 4y
    \n\u21d2 16y2 <\/sup>+ 4 – 16y = 4y
    \n\u21d2 16y2<\/sup> – 20y + 4 =0
    \n\u21d2 (y – 1) (16y – 4) =0
    \n\u21d2\u00a0 y =1 (or) y = 1\/4
    \nIfy=1; x = 2 \u21d2 2 (2,1)
    \nand y = 1\/4 x = -1 (-1, 1\/4)
    \nThus , the points of Intersection are
    \n(2,1) 1 (-1.,1\/4)
    \n\u2234 Area of the Region O B Q C = Area of BCQ+ Area of QCO.
    \n\"2nd<\/p>\n

    Question 48.
    \nA person buys a lottery ticket in 50 lotteries in each of which his chance of winning a prize is \\(\\frac{1}{100}\\) what is the probability that he will win a prize exactly once?
    \nAnswer:
    \nGiven it is a lax of Bemoullian trials with n = 50 and P = \\(\\frac{1}{100}[latex]
    \nHence q = 1 – P = 1 – [latex]\\frac{1}{100}\\) = \\(\\frac{99}{100}\\)
    \nThe Probability that he will win a prize exactly once is
    \n\"2nd<\/p>\n

    Part – E<\/span><\/p>\n

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

    Question 49.
    \na) Minimize and maximize Z = x + 2y subject to the constraints,
    \nx + 2y \u2265 100
    \n2x – y \u2264 0
    \n2x + y \u2264 200
    \nx, y \u2265 0 by graphical method.
    \nOur problem is to minimize and maximize
    \nZ = x + 2y …………..(1)
    \nSubject to constraints are
    \nx + 2y \u2265 100 …………….(2)
    \n2 x – y \u2264 0 ……………..(3)
    \n2x + y \u2264 200…………….(4)
    \nx> 0, x>0…………….(5)
    \n\u2234 x + 2y- 100 which passes through the points A( 100,0) and B(0,50)
    \n2x – y = 0 which passes through the points (0,0) and (20,4 0)
    \n2x + y = 200 which passes through the points A(100,0)&C (0,200)
    \n\u2234 The comer points of feasible region are 5(0,50) P(20,40) 0(50,100) and C(0,200).
    \nAt B (0,50) \u21d2 Z = 0 + 2 \u00d7 50 = 100
    \nAt P(20,40) \u21d2 Z = 20 + 2 \u00d7 40 – 100
    \nAt Q(5O,1OO) \u21d2 Z = 50 + 2 \u00d7 100 = 250
    \nAt C(0,200) Z = 0 + 2 \u00d7 200 = 400
    \nHence max Z = 400 at (0, 200) and
    \nMin Z = 100 at B and also at P.
    \n\"2nd
    \n(b)
    \n\"2nd
    \n\"2nd
    \nR1 R1 – (R2 + R3)
    \n\"2nd
    \n= 0 [(c + a) (a + b) – bc) + 2c (b (a + b) – bc) – 2b (bc – (c + a)]
    \n= 0 + 2c [ab + b2<\/sup> -bc] – 2b [bc – c2<\/sup> -ac]
    \n= 2abc + 2cb2<\/sup> – 2bc2<\/sup> – 2b2<\/sup>c + 2bc2<\/sup> + 2abc
    \n= 4 abc = RHS<\/p>\n

    \"KSEEB<\/p>\n

    Question 50.
    \na) Prove that \\(\\int_{0}^{a} f(x) \\cdot d x=\\int_{0}^{a} f(a-x) \\cdot d x\\) and hence evaluate \\(\\int_{0}^{\\pi \/ 4} \\log (1+\\tan x) \\cdot d x\\)
    \nAnswer:
    \n\"2nd
    \nput t = a- x
    \nx = a – t
    \nthan dt = -dx
    \nwhen x = 0 x = a
    \nt = a t = 0
    \nthen
    \n\"2nd
    \n\"2nd<\/p>\n

    (b) Find the value of k, if
    \n\"2nd
    \nAnswer:
    \n\"2nd<\/p>\n","protected":false},"excerpt":{"rendered":"

    Students can Download 2nd PUC Maths Model Question Paper 2 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 2 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\/18514"}],"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=18514"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/18514\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=18514"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=18514"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=18514"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}