{"id":13097,"date":"2024-01-18T10:23:38","date_gmt":"2024-01-18T04:53:38","guid":{"rendered":"https:\/\/wordpress-505192-1602719.cloudwaysapps.com\/?p=13097"},"modified":"2024-01-18T17:11:27","modified_gmt":"2024-01-18T11:41:27","slug":"tamil-nadu-12th-maths-model-question-paper-1-english-medium","status":"publish","type":"post","link":"https:\/\/samacheerguru.com\/tamil-nadu-12th-maths-model-question-paper-1-english-medium\/","title":{"rendered":"Tamil Nadu 12th Maths Model Question Paper 1 English Medium"},"content":{"rendered":"

Students can Download Tamil Nadu 12th Maths Model Question Paper 5 English Medium Pdf, Tamil Nadu 12th Maths Model Question Papers<\/a>\u00a0helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.<\/p>\n

TN State Board 12th Maths Model Question Paper 5 English Medium<\/h2>\n

Instructions:<\/p>\n

    \n
  1. \u00a0The question paper comprises of four<\/span> parts.<\/li>\n
  2. \u00a0You are to attempt all the parts. An internal choice of questions is provided wherever applicable.<\/li>\n
  3. questions of Part I, II. III and IV are to be attempted separately<\/li>\n
  4. Question numbers 1 to 20 in Part I <\/span>are objective type questions of one -mark<\/span> each. These are to be answered by choosing the most suitable answer from the given four alternatives and writing the option code and the corresponding answer<\/li>\n
  5. Question numbers 21 to 30 in Part II <\/span>are two-marks<\/span> questions. These are to be answered in about one or two sentences.<\/li>\n
  6. Question numbers 31 to 40 in Parr III<\/span> are three-marks<\/span> questions, These are to be answered in about three to five short sentences.<\/li>\n
  7. Question numbers 41 to 47 in Part IV<\/span> are five-marks<\/span> questions. These are to be answered) in detail. Draw diagrams wherever necessary.<\/li>\n<\/ol>\n

    Time: 3 Hours
    \nMaximum Marks: 90<\/p>\n

    Part – I<\/span><\/p>\n

    I. Choose the correct answer. Answer all the questions.\u00a0 [20 \u00d7 1 = 20]<\/span><\/p>\n

    Question 1.
    \nIf A = \\(\\left[\\begin{array}{ll}
    \n2 & 0 \\\\
    \n1 & 5
    \n\\end{array}\\right]\\) and B = \\(\\left[\\begin{array}{ll}
    \n1 & 4 \\\\
    \n2 & 0
    \n\\end{array}\\right]\\)
    \nthen adj (AB)| = ……….
    \n(a) – 40
    \n(b) -80
    \n(c) -60
    \n(b) -20
    \nAnswer:
    \n(b) -80<\/p>\n

    Question 2.
    \nin<\/sup> + 1n+1<\/sup> + in + 2<\/sup> + in +3<\/sup> is ……….
    \n(a) 0
    \n(b) 1
    \n(c)- 1
    \n(d) i
    \nAnswer:
    \n(a) 0<\/p>\n

    \"Tamil<\/p>\n

    Question 3.
    \nIf \u03c9 = \\(\\text { cis } \\frac{2 \\pi}{3}\\) , then the number of distinct roots of
    \n\\(\\left|\\begin{array}{ccc}
    \nz+1 & \\omega & \\omega^{2} \\\\
    \n\\omega & z+\\omega^{2} & 1 \\\\
    \n\\omega^{2} & 1 & z+\\omega
    \n\\end{array}\\right|=0\\)
    \n(a) 1
    \n(b) 2
    \n(c) 3
    \n(d) 4
    \nAnswer:
    \n(a) 1<\/p>\n

    Question 4.
    \nsin-1<\/sup>(cos x) = \\(\\frac{\\pi}{2}-x\\) is valid for ……….
    \n(a) -\u03c0 \u2264 x \u22640
    \n(b) 0 \u2264 x \u2264 \u03c0
    \n(c) \\(-\\frac{\\pi}{2} \\leq x \\leq \\frac{\\pi}{2}\\)
    \n(d) \\(-\\frac{\\pi}{4} \\leq x \\leq \\frac{3 \\pi}{4}\\)
    \nAnswer:
    \n(b) 0 \u2264 x \u2264 \u03c0<\/p>\n

    Question 5.
    \ntan-1<\/sup> x + cot-1<\/sup> = …….
    \n(a) 1
    \n(b) -\u03c0
    \n(c) \u03c0\/2
    \n(d) \u03c0
    \nAnswer:
    \n(c) \u03c0\/2<\/p>\n

    Question 6.
    \nThe equation of the normal to the circle x2<\/sup> + y2<\/sup> – 2x – 2y + 1 = 0 which is parallel to the line 2x + 4y = 3 is …………..
    \n(a) x + 2y = 3.
    \n(b) x + 2y + 3 = 3
    \n(c) 2x + 4y + 3 = 0
    \n(d) x – 2y + 3 = 0
    \nAnswer:
    \n(a) x + 2y = 3.<\/p>\n

    Question 7.
    \nThe axis of the parabola x2<\/sup> = 20y is ……….
    \n(a) y = 5
    \n(b) x = 5
    \n(c) x = 0
    \n(d) y = 0
    \nAnswer:
    \n(c) x = 0<\/p>\n

    \"Tamil<\/p>\n

    Question 8.
    \nIf \\(\\vec{a} \\times(\\vec{b} \\times \\vec{c})=(\\vec{a} \\times \\vec{b}) \\times \\vec{c}\\) for non-coplanar vectors \\(\\vec{a}, \\vec{b}, \\vec{c}\\) then…….
    \n(a) \\(\\vec{a}\\) parallel to \\(\\vec{b}\\)
    \n(b) \\(\\vec{b}\\) parallel to \\(\\vec{c}\\)
    \n(c) \\(\\vec{c}\\) parallel to \\(\\vec{a}\\)
    \n(d) \\(\\vec{a}+\\vec{b}+\\vec{c}=\\overrightarrow{0}\\)<\/p>\n

    Question 9.
    \nThe vector equation of a plane whose distance from the origin is p and perpendicular to a unit vector n\u0302 is ….
    \n(a) \\(\\vec{r} \\cdot \\vec{n}=p\\)
    \n(b) \\(\\begin{aligned}
    \n&-n\\\\
    \n&r \\cdot n=q
    \n\\end{aligned}\\)
    \n(c) \\(\\vec{r} \\times \\vec{n}=p\\)
    \n(d) \\(\\vec{r} \\cdot \\hat{n}=p\\)<\/p>\n

    Question 10.
    \nThe point of inflection of the curve y=(x – 1)3<\/sup> is …………..
    \n(a) (0,0)
    \n(b) (0,1)
    \n(c) (1,0)
    \n(d) (1,1)
    \nAnswer:
    \n(c) (1,0)<\/p>\n

    Question 11.
    \nThe asymptote to the curve y2<\/sup>(1 + x) = x2<\/sup> (1 – x) is …………..
    \n(a) x = 1
    \n(b) y = 1
    \n(c) y = -1
    \n(d) x = -1
    \nAnswer:
    \n(d) x = -1<\/p>\n

    Question 12.
    \nThe solution of a linear differential equation ax + Px = Q where P and Q are function of y, is ……….
    \n(a) y (I.F) = \u222b(I.F) Q dx + c
    \n(b) y (I.F) = \u222b(1.F) Q dy + c
    \n(c) y (I.F) = \u222b(1.F) Q dy + c .
    \n(d) x (I.F) = \u222b(1.F) Q dx + c<\/p>\n

    Question 13.
    \nA circular template has a radius of 10 cm. The measurement of radius has an approximate error of 0.02 cm. Then the percentage error in calculating area of this template is ……..
    \n(a) 0.2%
    \n(b) 0.4%
    \n(c) 0.04%
    \n(d) 0.08%
    \nAnswer:
    \n(b) 0.4%<\/p>\n

    Question 14.
    \nFor any value of n\u2208z, \\(\\int_{0}^{\\pi} e^{\\cos ^{2} x} \\cos ^{3}[(2 n+1) x] d x \\) is……….
    \n(a) \u03c0\/2
    \n(b) \u03c0
    \n(c) 0
    \n(d) 2
    \nAnswer:
    \n(c) 0<\/p>\n

    Question 15.
    \nIf n is odd then \\(\\int_{0}^{\\pi \/ 2} \\sin ^{n} x d x\\) is …. ….
    \n\"Tamil
    \nAnswer:
    \n(d) \\(\\frac{n-1}{n} \\cdot \\frac{n-3}{n-2} \\cdot \\frac{n-5}{n-4} \\dots \\frac{2}{3} \\cdot 1\\)<\/p>\n

    Question 16.
    \nThe solution of \\(\\frac{d y}{d x}=2^{y-x}\\)is ……….
    \n(a) 2x<\/sup> + 2y<\/sup> = c
    \n(b) 2x<\/sup> – 2y<\/sup> = c
    \n(c) \\(\\frac{1}{2^{x}}-\\frac{1}{2^{y}}=c\\)
    \n(d) x + y = c
    \nAnswer:
    \n(c) \\(\\frac{1}{2^{x}}-\\frac{1}{2^{y}}=c\\)<\/p>\n

    Question 17.
    \nIf p and q are the order and degree of the differential equation \\(y \\frac{d y}{d x}+x^{3}\\left(\\frac{d^{2} y}{d x^{2}}\\right)+x y=\\cos x\\) when…..
    \n(a) p < q
    \n(b) p = q
    \n(c) p > q
    \n(d) p exists and q does not exist
    \nAnswer:
    \n(c) p > q<\/p>\n

    Question 18.
    \nA pair of dice numbered 1, 2, 3, 4, 5, 6 of a six-sided die and 1, 2, 3, 4 of a four-sided die is rolled and the sum is determined. Let the random variable X denote this sum. Then the number of elements in the inverse image of 7 is ……….
    \n(a) 1
    \n(6) 2
    \n(c) 3
    \n(d) 4
    \nAnswer:
    \n(d) 4<\/p>\n

    \"Tamil<\/p>\n

    Question 19.
    \nIf in 6 trials, X is a binomial variate which follows the relation 9P(X = 4) = P(X = 2),
    \nthen the probability of success is …….
    \n(a) 0.125 .
    \n(b) 0.25
    \n(c) 0.375
    \n(d) 0.75
    \nAnswer:
    \n(b) 0.25<\/p>\n

    Question 20.
    \nWhich one of the following statements has truth value F ?
    \n(a) Chennai is in India or \u221a2 is an integer
    \n(b) Chennai is in India or \u221a2 is an irrational number
    \n(c) Chennai is in China or \u221a2 is an integer
    \n(d) Chennai is in China or \u221a2 is an irrational number
    \nAnswer:
    \n(c) Chennai is in China or \u221a2 is an integer<\/p>\n

    Part – II<\/span><\/p>\n

    II. Answer any seven questions. Question No. 30 is compulsory. [7 x 2 = 14)<\/span><\/p>\n

    Question 21.
    \nIf A = \\(\\left[\\begin{array}{rr}
    \n8 & -4 \\\\
    \n-5 & 3
    \n\\end{array}\\right]\\), verify that A(adj A) = (adj A)A = |A|I2<\/sub>,
    \n\"Tamil<\/p>\n

    Question 22.
    \nFind the value of \\(\\sum_{k=1}^{8}\\left(\\cos \\frac{2 k \\pi}{9}+i \\sin \\frac{2 k \\pi}{9}\\right)\\)
    \nAnswer:
    \nWe know that the 9th roots of unity are 1, 6, 62, …, 08 Sum of the roots:
    \n1 + \u03c9 + \u03c92<\/sup> + … + \u03c98<\/sup> = 0
    \n\u21d2 \u03c9 + \u03c92<\/sup> + … + \u03c98<\/sup> = -1
    \n\"Tamil<\/p>\n

    Question 23.
    \nSolve the equation \ud83d\ude214<\/sup> – 14x2<\/sup> + 45 = 0.
    \nAnswer:
    \nThe given equation is x4<\/sup> – 14x2<\/sup> + 45 = 0
    \nLet x2<\/sup> = y
    \ny2<\/sup> – 14y + 45 = 0 \u21d2 (y – 9)(y – 5) = 0
    \ny = 9,5
    \n\u2234 x2<\/sup> = 9, x2<\/sup>= 5
    \nx = \u00b13, x = \u00b1\u221a5
    \n\u2234 The roots are \u00b13, \u00b1\u221a5<\/p>\n

    Question 24.
    \nFind the principal value of sin \\(\\sin ^{-1}\\left(-\\frac{1}{2}\\right)\\) (in radians and degrees).
    \nAnswer:
    \n\"Tamil<\/p>\n

    Question 25.
    \nShow that the points (2,3,4),(-1,4,5) and (8,1,2) are collinear.
    \nAnswer:
    \nGiven points are (2, 3, 4), (-1,4,5) and (8,1,2). Equation of the line joining of the first and
    \nsecond point is
    \n\\(\\frac{x-2}{-3}=\\frac{y-3}{1}=\\frac{z-4}{1}=m(\\text { say })\\)
    \n(-3m + 2, m + 3, m + 4)
    \nOn putting m = -2, we get the third point is (8, 1, 2)
    \n\u2234 Given points are collinear.<\/p>\n

    \"Tamil<\/p>\n

    Question 26.
    \nEvaluate: \\(\\int_{0}^{2 \\pi} \\frac{\\cos x}{\\sqrt{4+3 \\sin x}} d x\\)
    \nAnswer:
    \nI = \\(\\int_{0}^{2 \\pi} \\frac{\\cos x}{\\sqrt{4+3 \\sin x}} d x\\)
    \nPut 4 + 3 sin x = t, so that 3 cos x dx = dt = cosx dx = \\(\\frac{1}{3} dt\\) When x = 0, t = 4 + 3(0) = 4 and when x = 2\u03c0, t = 4+ 3(0) = 4
    \n\"Tamil<\/p>\n

    Question 27.
    \nFind the order and degre of the diferential equation center equation \\(\\frac{d^{2} y}{d x^{2}}-y+\\left(\\frac{d y}{d x}+\\frac{d^{3} y}{d x^{3}}\\right)^{\\frac{3}{2}}=0\\)
    \nAnswer:
    \n\"Tamil<\/p>\n

    Question 28.
    \nSuppose the amount of milk sold daily at a milk booth is distributed with a minimum
    \nof 200 litres and a maximum of 600 litres with probability density function
    \n\\(f(x)=\\left\\{\\begin{array}{lr}
    \nk & 200 \\leq x \\leq 600 \\\\
    \n0 & \\text { otherwise }
    \n\\end{array}\\right.\\) Find the probability that daily sales will fall between 300 otherwise litres and 500 litres?
    \nAnswer:
    \nThe probability that daily sales will fall between 300 litres and 500 litres is
    \n\"Tamil<\/p>\n

    Question 29.
    \nLet A = \\(\\left(\\begin{array}{llll}
    \n1 & 0 & 1 & 0 \\\\
    \n0 & 1 & 0 & 1 \\\\
    \n1 & 0 & 0 & 1
    \n\\end{array}\\right)\\) ,B = \\(\\left(\\begin{array}{llll}
    \n0 & 1 & 0 & 1 \\\\
    \n1 & 0 & 1 & 0 \\\\
    \n1 & 0 & 0 & 1
    \n\\end{array}\\right)\\) , C = \\(\\left(\\begin{array}{llll}
    \n1 & 1 & 0 & 1 \\\\
    \n0 & 1 & 1 & 0 \\\\
    \n1 & 1 & 1 & 1
    \n\\end{array}\\right)\\) be any three boolean matrices of the same type. Find A V B
    \nAnswer:
    \n\"Tamil<\/p>\n

    Question 30.
    \nFind the general equation of a circle with centre (-3,- 4) and radius 3 units.
    \nAnswer:
    \nEquation of the circle in standard form is (x – h)2<\/sup> + (y – k)2<\/sup> = p2<\/sup>
    \n(x-(-3))2<\/sup> + (y-(-4))2<\/sup> = 32<\/sup> .
    \n(x + 3)2<\/sup> + (y + 4)2<\/sup> = 32<\/sup>
    \nx2<\/sup> + y2<\/sup> + 6x + 8y + 16 = 0.<\/p>\n

    Part – III<\/span><\/p>\n

    III. Answer any seven questions. Question No. 40 is compulsory. [7 x 3 = 21]<\/span><\/p>\n

    Question 31.
    \nIf A = \\(-\\left[\\begin{array}{lll}
    \n8 & 1 & 4 \\\\
    \n4 & 4 & 7 \\\\
    \n1 & 8 & 4
    \n\\end{array}\\right]\\) prove that A-1<\/sup> = AT<\/sup>.
    \nAnswer:
    \n\"Tamil
    \n\"Tamil<\/p>\n

    Question 32.
    \nIf z1<\/sub> = 2 + 5i, z2 <\/sub>= -3 – 4i, and z3<\/sub> = 1 + i, find the additive and multiplicative inverse of z1<\/sub>, z2<\/sub>, and z3<\/sub>.
    \nAnswer:
    \n\"Tamil<\/p>\n

    Question 33.
    \nSolve the equation 3x3<\/sup> – 26x2<\/sup> + 52x – 24 = 0 if its roots form a geometric progression
    \nAnswer:
    \nThe given equation is 3x3<\/sup> – 26x2<\/sup> + 52x – 24 = 0
    \n\"Tamil
    \nGiven that the root are GP
    \n\u2234 The roots are \\(\\frac{\\alpha}{r}\\) , \u03b1, \u03b1r
    \n\\(\\frac{\\alpha}{r}\\) x \u03b1 x \u03b1r = (-8)
    \n\u03b13<\/sup> = 8
    \n\u03b1 = 2
    \n\"Tamil<\/p>\n

    Question 34.
    \nFind the value of \\(\\cot \\left(\\sin ^{-1} \\frac{3}{5}+\\sin ^{-1} \\frac{4}{5}\\right)\\)
    \nAnswer:
    \n\"Tamil<\/p>\n

    Question 35.
    \nIf the normal at the point\u201c t1<\/sub>‘ on the parabola y2<\/sup> = 4ax meets the parabola again at the point \u2018t2<\/sub>\u2019, then prove that \\(t_{2}=-\\left(t_{1}+\\frac{2}{t_{1}}\\right)\\)
    \nAnswer:
    \nEquation of normal to y2<\/sup> = 4ax at ‘t\u2019 is y + xt = 2at + at3<\/sup>.
    \nSo equation of normal at \u2018t1<\/sub>,\u2019 is y + xt1<\/sub>, = 2at1<\/sub> + at1<\/sub>3<\/sup> ……….(1)
    \nThe normal meets the parabola y2<\/sup> = 4ax at \u201ct2<\/sub>‘ (ie.,) at (at2<\/sub>2<\/sup>, 2at2<\/sub>)
    \n\"Tamil<\/p>\n

    Question 36.
    \nFind the coordinates of the foot of the perpendicular drawn from the point (-1, 2, 3) to the straight line \\(\\vec{r}=(\\hat{i}-4 \\hat{j}+3 \\hat{k})+t(2 \\hat{i}+3 \\hat{j}+\\hat{k})\\). Also, find the shortest distance from the given point to the straight line.
    \nAnswer:
    \n\"Tamil
    \nComparing the given equation \\(\\vec{r}=(\\hat{i}-4 \\hat{j}+3 \\hat{k})+t(2 \\hat{i}+3 \\hat{j}+\\hat{k})\\) with \\(\\vec{r}=\\vec{a}+t \\vec{b}\\) we get \\(\\vec{a}=\\hat{i}-4 \\hat{j}+3 \\hat{k} \\hat{k}\\) and \\(\\vec{b}=2 \\hat{i}+3 \\hat{j}+\\hat{k}\\). We denote the given point (-1,2,3) by D, and the point (1,-4,3) on the straight line by A . If F is the foot of the perpendicular from D to the straight line, then F is of the form (2t+1, 3t – 4, t + 3) and \\(\\overrightarrow{\\mathrm{DF}}=\\overrightarrow{\\mathrm{OF}}-\\overrightarrow{\\mathrm{OD}}=(2 t+2) \\hat{i}+(3 t-6) \\hat{j}+t \\hat{k}\\) .
    \nSince \\(\\vec{b}\\) is perpendicular to \\(\\overrightarrow{\\mathrm{DF}}\\), we have
    \nLine \\(\\vec{b} \\cdot \\overrightarrow{\\mathrm{DF}}\\) = 0 = 2(2t + 2) + 3(3-6) +1(t) = 0 = t=1
    \nTherefore, the coordinate of F is (3,-1, 4).
    \nNow, the perpendicular distance from the given point to the given line is
    \n\\(\\mathrm{DF}=|\\overrightarrow{\\mathrm{DF}}|=\\sqrt{4^{2}+(-3)^{2}+1^{2}}=\\sqrt{26}\\) units.<\/p>\n

    \"Tamil<\/p>\n

    Question 37.
    \nFind the asymptotes of the curve f(x) = \\(\\frac{x^{2}+6 x-4}{3 x-6}\\)
    \nAnswer:
    \nSince the numerator is of highest degree than the denominator. We have a slant
    \nasymptote to find it we have to divide the numerator by the denominator.
    \n\"Tamil
    \nSo the equation of asymptote is \\(y=\\frac{x}{3}+\\frac{8}{3}\\) and 3x – 6 = 0
    \nx = 2<\/p>\n

    Question 38.
    \nEvaluate: \\(\\int_{0}^{1} x(1-x)^{n} d x\\)
    \nAnswer:
    \n\"Tamil<\/p>\n

    Question 39.
    \nFor the distribution function given by \\(\\mathbf{F}(x)=\\left\\{\\begin{array}{ll}
    \n0 & x<0 \\\\ x^{2} & 0 \\leq x \\leq 1 \\\\ 1 & x>1
    \n\\end{array}\\right.\\), find the density function. Also evaluate
    \n(i) P(0.5<x<0.75), (ii) P(X \u2264 50.5) (iii) P(x > 0.75)
    \nAnswer:
    \n\"Tamil<\/p>\n

    Question 40.
    \nLet w (x, y) = \\(x y+\\frac{e^{y}}{y^{2}+1}\\) for all (x, y) \u2208 R2<\/sup>. Calculate \\(\\frac{\\partial^{2} w}{\\partial y \\partial x}\\) and \\(\\frac{\\partial^{2} w}{\\partial x \\partial y}\\)
    \nAnswer:
    \n\"Tamil<\/p>\n

    Part – IV<\/span><\/p>\n

    IV. Answer all the questions. [7 x 5 = 35]<\/span><\/p>\n

    Question 41.
    \n(a) An amount of \u20b9 65,000 is invested in three bonds at the rates of 6%, 8% and 10% per annum respectively. The total annual income is \u20b9 5,000. The income from the third bond is \u20b9 800 more than that from the second bond. Determine the price of each bond. (Use Gaussian elimination method.)
    \nAnswer:
    \nLet the amount invested in 6% bond be \u20b9 x
    \nand the amount invested in 8% bond be \u20b9y
    \nand the amount invested in 10% bond be \u20b9z
    \nNow x + y + z = 65000
    \n\"Tamil
    \n(i.e) 6x + 8y + 10z = 500000
    \n(\u00f7by 2) 3x + 4y + 5z = 250000
    \nAlso given that
    \n\\frac{10}{100} z-\\frac{8}{100} y = 800 (i.e) -8y + 10z = 80000
    \n(\u00f7 by 2) -4y + 5z = 40000
    \n\"Tamil<\/p>\n

    (i.e,) x + y + z\u00a0 = 65000…..(1)
    \ny + 2z = 55000…..(2)
    \n13z = 260000 …..(3)
    \nForm (3) z = \\(\\frac{260000}{13}\\)
    \nSubstituting z = \u20b9 20000 in (2) we get
    \ny + 40000 = 55000 = y \u21d2 55000 – 40000 = \u20b9 15000
    \nSubstituting z = \u20b9 20000, y = \u20b9 15000 in (1) we get
    \nx + 15000 + 20000 = 65000
    \nx = 65000 – 35000 = \u20b9 30000
    \nSo the amount invested in
    \n6% bond x = \u20b9 30000
    \n8% bond y = \u20b9 15000
    \nand 10% bond z = \u20b9 20000<\/p>\n

    [OR]<\/p>\n

    (b) Find the equation of the curve whose slope is \\(\\frac{y-1}{x^{2}+x}\\) and which passes through the point (1, 0).
    \nAnswer:
    \n\"Tamil<\/p>\n

    Question 42.
    \n(a) If arg (z – 1) = \\(\\frac{\\pi}{6}\\) and arg (z + 1) = 2\\(\\frac{\\pi}{3}\\) , then prove that |z| = 1.
    \nAnswer:
    \nGiven arg (z-1) = \\(\\frac{\\pi}{6}\\) = 30\u00b0 and arg (z + 1) = \\(\\frac{2 \\pi}{3}\\) = 120\u00b0
    \nSo, arg (z + 1) – arg (z – 1) = 120\u00b0 – 30\u00b0 = 90\u00b0
    \nie, ang \\(\\frac{z+1}{z-1}=90^{\\circ}=\\frac{\\pi}{2}\\)
    \n\\(\\operatorname{Re} \\frac{z+1}{z-1}=0\\) [\u2234arg = \\(\\tan \\theta=\\frac{I P}{R P}\\)<\/p>\n

    \"Tamil
    \nTherefore x= -1 and x=1 are vertical asymptotes.
    \nThe rough sketch of the curve is shown on the right side.
    \n\"Tamil<\/p>\n

    \"Tamil<\/p>\n

    Question 43.
    \n(a) Solve the equation x3<\/sup> – 9x2<\/sup> + 14x + 24 = 0 if it is given that two of its roots are in the ratio 3:2.
    \nAnswer:
    \nThe given equation is x3<\/sup> – 9x2<\/sup> +14x + 24 = 0…
    \nSince, the two roots are in the ratio 3:2.
    \nThe roots are \u03b1 , 3\u03bb, 2\u03bb
    \n\u03b1 + 3\u03bb + 2\u03bb = -b = 9
    \n= a + 5\u03bb = 9 (a) ….. (1)
    \n\u03b1 (3\u03bb) (2\u03bb) = -24
    \n6\u03bb2<\/sup>a = -24 => \u03bb2<\/sup>\u03b1 = -4 …….(2)
    \n\u03b1 = 9 – 5\u03bb
    \n\u03bb2<\/sup> (9 – 52) = -4
    \n9\u03bb2<\/sup> – 5\u03bb3<\/sup> + 4 = 0
    \n5\u03bb3<\/sup> – 9\u03bb2<\/sup> – 4 = 0
    \n(\u03bb – 2) (5\u03bb2<\/sup> + \u03bb + 2) = 0
    \n\u03bb = 2,5\u03bb2<\/sup> + \u03bb+ 2 = 0 has only Imaginary roots \u2206 < 0
    \n\u2234 when \u03bb = 2, \u03b1 = 9-5 (2) = 9 – 10 = -1
    \n\u2234 The roots are \u03b1, 3\u03bb, 2\u03bb i.e., -1, 6, 4
    \n\"Tamil<\/p>\n

    [OR]<\/p>\n

    (b) Show that the lines \\(\\frac{x+3}{-3}=\\frac{y-1}{1}=\\frac{z-5}{5}\\) and \\(\\frac{x+1}{-1}=\\frac{y-2}{2}=\\frac{z-5}{5}\\)
    \nare coplanar. Also -3 .. 1 5 -1 2 find the equation of the plane containing these two lines.
    \nAnswer:
    \nFrom the lines we have,
    \n(x1<\/sub>, y1<\/sub>, z1<\/sub>) = (-3, 1, 5); (b1<\/sub>, b2<\/sub>, b3<\/sub>) = (-3, 1, 5)
    \n(x2<\/sub>, y2<\/sub>, z2<\/sub>) = (-1, 2, 5); (d1<\/sub>, d2<\/sub>, d3<\/sub>) = (-1, 2,5)
    \nConditions for coplanar
    \n\"Tamil
    \n\\(\\left|\\begin{array}{ccc}
    \n2 & 1 & 0 \\\\
    \n-3 & 1 & 5 \\\\
    \n-1 & 2 & 5
    \n\\end{array}\\right|\\) = 2(5 – 10) – 1(-15 + 5) + ((-6 + 1) = -10 + 10 = 0
    \nGiven two lines are coplanar
    \n\"Tamil
    \n(x + 3)(5 – 10) – (y – 1)[-15 + 5) + (2 – 5)(-6 + 1] = 0
    \n5(x + 3) + 10(y – 1) – 5(z – 5) = 0
    \n(\u00f7 by -5) \u21d2 (x + 3) – 2 (y – 1) + (z – 5) = 0
    \nx + 3 – 2y + 2 + 2 – 5 = 0 .
    \nor x – 2y + z = 0
    \n\"Tamil
    \n(x + 1)(5 – 10) – (y – 2)(-15 + 5) + (z – 5)(-6 + 1) = 0
    \n-5(x + 1) + 10(y – 2) – 5(2 – 5) = 0
    \n(x + 1) – 2(y – 2) + (2 – 5) = 0
    \nx + 1 – 2y + 4 + 2 – 5 = 0
    \nx – 2y + z = 0<\/p>\n

    Question 44.
    \n(a) Find the area of the region bounded by the parabola y2<\/sup> = x and the line y = x -2.
    \nAnswer:
    \nTo find the points of intersection solve the two equations y2<\/sup> = x and y=x-2
    \n\"Tamil<\/p>\n

    (b) Solve \\(\\frac{d y}{d x}+\\frac{y}{x \\log x}=\\frac{\\sin 2 x}{\\log x}\\)
    \nAnswer:
    \n\"Tamil<\/p>\n

    Question 45.
    \n(a) Solve \\(\\cos \\left(\\sin ^{-1}\\left(\\frac{x}{\\sqrt{1+x^{2}}}\\right)\\right)=\\sin \\left\\{\\cot ^{-1}\\left(\\frac{3}{4}\\right)\\right\\}\\)
    \nAnswer:
    \n\"Tamil
    \n\"Tamil<\/p>\n

    (b) Find the mean and variance of a random variable X, whose probability density x for x 20 function is \\(f(x)=\\left\\{\\begin{array}{cc}
    \n\\lambda e^{-\\lambda x} & \\text { for } x \\geq 0 \\\\
    \n0 & \\text { otherwise }
    \n\\end{array}\\right.\\)
    \nAnswer:
    \nObserve that the given distribution is continuous
    \n\"Tamil<\/p>\n

    Question 46.
    \n(a) Show that the equation of the normal to the curve x= a cos3<\/sup>\u03b8, y = a sin3<\/sup> \u03b8 at ‘\u03b8’ is
    \nx cos \u03b8 – y sin \u03b8= a cos 2 \u03b8.
    \nAnswer:
    \nx = a cos3<\/sup>\u03b8
    \n\\(\\frac{d x}{d \\theta}\\) = a[ 3cos2<\/sup>\u03b8](-sin \u03b8) = -3acos2<\/sup>\u03b8 sin \u03b8
    \ny= a sin3<\/sup>\u03b8
    \n\"Tamil
    \nSlope of the normal = \\(\\frac{-1}{m}=\\frac{\\cos \\theta}{\\sin \\theta}\\)Point = (x1, yy) = (a cos3<\/sup> \u03b8 , a sin3<\/sup> \u03b8 ) sin \u03b8
    \nSo, equation of the normal is
    \nY-a sin3<\/sup> \u03b8 = \\(\\frac{\\cos \\theta}{\\sin \\theta}\\)(x- a cos3<\/sup> \u03b8)
    \ni.e. y sin \u03b8 – a sin4<\/sup> \u03b8 = x cos \u03b8 – a cos4<\/sup> \u03b8
    \nx cos \u03b8 – sin \u03b8 = a cos4<\/sup> \u03b8 – a sin4<\/sup> \u03b8
    \ni.e x cos \u03b8 – y sin \u03b8 = a [cos2<\/sup>\u03b8+ sin2<\/sup>\u03b8] [cos2<\/sup>\u03b8 – sin2<\/sup>\u03b8]
    \n= a [cos 2 \u03b8]
    \nSo, the equation of the normal is x cos \u03b8 – y sin \u03b8 = a cos 2\u03b8<\/p>\n

    [OR]<\/p>\n

    (b) Verify (i) closure property (ii) commutative property (iii) associative property (iv) existence of identity and (v) existence of inverse for the operation +5<\/sub> on Z5<\/sub> using table corresponding to addition modulo 5.
    \nAnswer:
    \nIt is known that Z5<\/sub> = {[O], [1], [2], [3], [4]} . The table corresponding to addition modulo 5 is as follows: We take reminders {0,1,2,3,4} to represent the classes {[0],[1],[2],[3],[4]}.
    \n\"Tamil<\/p>\n

    (i) Since each box in the table is filled by exactly one element of Z5<\/sub>, the output a +5<\/sub>b is unique
    \nand hence +5<\/sub> is a binary operation.<\/p>\n

    (ii) The entries are symmetrically placed with respect to the main diagonal. So +5<\/sub> has commutative
    \nproperty.<\/p>\n

    (iii) The table cannot be used directly for the verification of the associative property. So it is to be verified as usual.
    \nFor instance, (2 +5<\/sub>3) +5<\/sub> 4 = 0 +5<\/sub> 4 = 4 (mod 5)
    \nand 2 +5<\/sub> (3 +5<\/sub> 4) = 2 +5<\/sub> 2 = 4 (mod 5). Hence (2 +5<\/sub>3) +5<\/sub> 4 = 2 +5<\/sub> (3 +5<\/sub>4).
    \nProceeding like this one can verify this for all possible triples and ultimately it can be shown that +5<\/sub> is associative.<\/p>\n

    (iv) The row headed by 0 and the column headed by 0 are identical. Hence the identity element is 0.<\/p>\n

    (v) The existence of inverse is guaranteed provided the identity O exists in each row and each column. From Table-A, it is clear that this property is true in this case. The method of finding the inverse of any one of the elements of Z5<\/sub> , say 2 is outlined below.<\/p>\n

    First find the position of the identity element 0 in the III row headed by 2. Move horizonta along the III row and after reaching 0, move vertically above 0 in the IV column, because 0 is in the III row and IV column. The element reached at the topmost position of IV column is 3. This element 3 is nothing but the inverse of 2, because, 2 +5<\/sub> 3 = 0 (mod 5). In this way, the inverse of each and every element of Z5<\/sub> can be obtained. Note that the inverse of 0 is 0, that of 1 is 4, that of 2 is 3, that of 3 is 2 , and that of 4 is 1.<\/p>\n

    Question 47.
    \n(a) Identify the type of conic and find centre, foci, vertices, and directrices the following:
    \n18x2<\/sup> + 12y2<\/sup> – 144x + 48y + 120 = 0
    \nAnswer:
    \n18x2<\/sup> + 12y2<\/sup> – 144x + 48y + 120 = 0
    \n(18x2<\/sup> – 144x) + (12y2<\/sup> + 48y) = -120
    \n18(x2<\/sup> – 8x) + 12 ( 12 + 4y) = -120
    \n18(x2<\/sup> – 8x + 16 – 16) + 12 (12+ 4y + 4 – 4) = -120
    \n18(x – 4)2<\/sup> – 288 + 12(y + 2)2<\/sup> – 48 = -120
    \n18(x – 4)2<\/sup> + 12(y + 2)2<\/sup> = -120 + 288 + 48 = 216
    \n\"Tamil
    \n\"Tamil<\/p>\n

    [OR]<\/p>\n

    (b) Prove that g(x, y) = x log \\(\\left(\\frac{y}{x}\\right)\\) is homogeneous; what is the degree? Verify Euler’s Theorem for g.
    \nAnswer:
    \ng (x, y) = x log(y\/x)
    \n8 (tx, ty) = (tx) log \\(\\left(\\frac{t y}{t x}\\right)\\)
    \n= txlog \\(\\left(\\frac{y}{x}\\right)\\)
    \ng(tx, ty) = t g(x, y)
    \n\u2234 ‘g’is a homogeneous function of degree 1. By Euler’s theorem, we have
    \n\"Tamil
    \n\u2234 Euler’s theorem verified.<\/p>\n","protected":false},"excerpt":{"rendered":"

    Students can Download Tamil Nadu 12th Maths Model Question Paper 5 English Medium Pdf, Tamil Nadu 12th Maths Model Question Papers\u00a0helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations. TN State Board 12th Maths Model Question Paper 5 English Medium Instructions: \u00a0The question paper comprises of […]<\/p>\n","protected":false},"author":2,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":"","footnotes":""},"categories":[8],"tags":[],"blocksy_meta":"","jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/samacheerguru.com\/wp-json\/wp\/v2\/posts\/13097"}],"collection":[{"href":"https:\/\/samacheerguru.com\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/samacheerguru.com\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/samacheerguru.com\/wp-json\/wp\/v2\/users\/2"}],"replies":[{"embeddable":true,"href":"https:\/\/samacheerguru.com\/wp-json\/wp\/v2\/comments?post=13097"}],"version-history":[{"count":1,"href":"https:\/\/samacheerguru.com\/wp-json\/wp\/v2\/posts\/13097\/revisions"}],"predecessor-version":[{"id":55524,"href":"https:\/\/samacheerguru.com\/wp-json\/wp\/v2\/posts\/13097\/revisions\/55524"}],"wp:attachment":[{"href":"https:\/\/samacheerguru.com\/wp-json\/wp\/v2\/media?parent=13097"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/samacheerguru.com\/wp-json\/wp\/v2\/categories?post=13097"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/samacheerguru.com\/wp-json\/wp\/v2\/tags?post=13097"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}