{"id":15599,"date":"2024-02-21T12:05:03","date_gmt":"2024-02-21T06:35:03","guid":{"rendered":"https:\/\/wordpress-505192-1602719.cloudwaysapps.com\/?p=15599"},"modified":"2024-02-22T17:04:49","modified_gmt":"2024-02-22T11:34:49","slug":"tamil-nadu-11th-maths-previous-year-question-paper-june-2019-english-medium","status":"publish","type":"post","link":"https:\/\/samacheerkalviguru.com\/tamil-nadu-11th-maths-previous-year-question-paper-june-2019-english-medium\/","title":{"rendered":"Tamil Nadu 11th Maths Previous Year Question Paper June 2019 English Medium"},"content":{"rendered":"

Students can Download Tamil Nadu 11th Maths Previous Year Question Paper June 2019 English Medium Pdf, Tamil Nadu 11th 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 11th Maths Previous Year Question Paper June 2019 English Medium<\/h2>\n

General Instructions:<\/p>\n

    \n
  1. The question paper comprises of four parts<\/span>.<\/li>\n
  2. You are to attempt all the parts. An internal choice of questions is provided wherever applicable.<\/li>\n
  3. All questions of Part I, II, III and IV are to be attempted separately.<\/li>\n
  4. Question numbers 1<\/span> to 20<\/span> in Part I<\/span> are Multiple Choice Questions of one mark<\/span> each.
    \nThese 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<\/span> to 30<\/span> in Part II<\/span> are two-mark<\/span> questions. These are to be answered in about one or two sentences.<\/li>\n
  6. Question numbers 31<\/span> to 40<\/span> in Part III<\/span> are three-mark<\/span> questions. These are to be answered in above three to five short sentences.<\/li>\n
  7. Question numbers 41<\/span> to 47<\/span> in Part IV<\/span> are five-mark<\/span> questions. These are to be answered in detail Draw diagrams wherever necessary.<\/li>\n<\/ol>\n

    Time: 2.30 Hours
    \nMaximum Marks: 90<\/p>\n

    PART – I<\/span><\/p>\n

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

    Question 1.
    \nThe range of the function \\(\\frac{1}{1-2sin x}\\) is………….
    \n(a) (-\u221e, -1) \u222a(\\(\\frac{1}{3}\\), \u221e)
    \n(b) (-1, \\(\\frac{1}{3}\\))
    \n(c) [-1, \\(\\frac{1}{3}\\)]
    \n(d) (-\u221e, -1] \u222a(\\(\\frac{1}{3}\\), \u221e)
    \nAnswer:
    \n(d) (-\u221e, -1] \u222a(\\(\\frac{1}{3}\\), \u221e)<\/p>\n

    Question 2.
    \nIf the function f : [-3, 3] \u2192 S defined by f(x) = x\u00b2 is onto, then S is…………
    \n(a) [-9, 9]
    \n(b) R
    \n(c) [-3, 3]
    \n(d) [0, 9]
    \nAnswer:
    \n(d) [0, 9]<\/p>\n

    Question 3.
    \nThe number of solutions of x\u00b2 + |x – 1| = 1 is ………..
    \n(a) 1
    \n(b) 0
    \n(c) 2
    \n(d) 3
    \nAnswer:
    \n(c) 2<\/p>\n

    \"Tamil<\/p>\n

    Question 4.
    \ncos 1\u00b0 + cos 2\u00b0 + cos 3\u00b0 + …. + cos 179\u00b0 =……………..
    \n(a) 0
    \n(b) 1
    \n(c) -1
    \n(d) 89
    \nAnswer:
    \n(a) 0<\/p>\n

    Question 5.
    \nIf tan \u03b1 and tan \u03b2 are the roots of x\u00b2 + ax + b = 0, then \\(\\frac{sin(\u03b1+\u03b2)}{sin \u03b1 sin \u03b2}\\) is equal to…………
    \n(a) \\(\\frac{b}{a}\\)
    \n(b) \\(\\frac{a}{b}\\)
    \n(c) –\\(\\frac{a}{b}\\)
    \n(d) –\\(\\frac{b}{a}\\)
    \nAnswer:
    \n(b) \\(\\frac{a}{b}\\)<\/p>\n

    Question 6.
    \nThe number of sides of a polygon having 44 diagonals is…………
    \n(a) 4
    \n(b) 4
    \n(c) 11
    \n(d) 22
    \nAnswer:
    \n(d) 22<\/p>\n

    Question 7.
    \nThe H.M. of two positive numbers whose A.M. and G.M. are 16, 8 respectively is………….
    \n(a) 10
    \n(b) 6
    \n(c) 5
    \n(d) 4
    \nAnswer:
    \n(d) 4<\/p>\n

    Question 8.
    \nThe nth term of the sequence \\(\\frac{1}{2}\\), \\(\\frac{3}{4}\\), \\(\\frac{7}{8}\\), \\(\\frac{15}{16}\\) is……………
    \n(a) 2n<\/sup> – n – 1
    \n(b) 1 – 2-n<\/sup>
    \n(c) 2-n<\/sup>+ n – 1
    \n(d) 2n-1<\/sup>
    \nAnswer:
    \n(b) 1 – 2-n<\/sup><\/p>\n

    Question 9.
    \nThe intercepts of the perpendicular bisector of the line segment joining (1,2) and (3,4) with coordinate axes are
    \n(a) 5, -5
    \n(b) 5, 5
    \n(c) 5, 3
    \n(d) 5, -4
    \nAnswer:
    \n(b) 5, 5<\/p>\n

    Question 10.
    \nThe image of the point (2, 3) in the line y = -x is
    \n(a) (-3, -2)
    \n(b) (-3, 2)
    \n(c) (-2, -3)
    \n(d) (3, 2)
    \nAnswer:
    \n(a) (-3, -2)<\/p>\n

    \"Tamil<\/p>\n

    Question 11.
    \nIf A = \\(\\left[\\begin{array}{cc} \\lambda & 1 \\\\ -1 & -\\lambda \\end{array}\\right]\\), then for what value of \u03bb, A\u00b2 = 0?………
    \n(a) 0
    \n(b) \u00b1 1
    \n(c) -1
    \n(d) 1
    \nAnswer:
    \n(b) \u00b1 1<\/p>\n

    Question 12.
    \nIf \\(\\vec{a}\\) and \\(\\vec{b}\\) are having same magnitute and angle between them is 60\u00b0 and their scalar product is \\(\\frac{1}{2}\\), then |\\(\\vec {a}\\)| is
    \n(a) 2
    \n(b) 3
    \n(c) 7
    \n(d) 1
    \nAnswer:
    \n(d) 1<\/p>\n

    Question 13.
    \n\\(\\lim _{x \\rightarrow \\infty} \\frac{a^{x}-b^{x}}{x}\\) = …………..
    \n(a) log ab
    \n(b) log (\\(\\frac{a}{b}\\))
    \n(c) log (\\(\\frac{b}{a}\\))
    \n(d) \\(\\frac{a}{b}\\)
    \nAnswer:
    \n(b) log (\\(\\frac{a}{b}\\))<\/p>\n

    Question 14.
    \nIf f(x) = \\(\\left\\{\\begin{array}{ccc}x, & x \\text { is irrational } \\\\ 1-x, & x \\text { is rational }\\end{array}\\right.\\) then f is………..
    \n(a) discontinuous at x = \\(\\frac{1}{2}\\)
    \n(b) continuous at x = \\(\\frac{1}{2}\\)
    \n(c) continuous everywhere
    \n(d) discontinuous everywhere
    \nAnswer:
    \n(b) continuous at x = \\(\\frac{1}{2}\\)<\/p>\n

    Question 15.
    \nThe derivative of f(x) = x |x| at x = -3 is…………..
    \n(a) 6
    \n(b) -6
    \n(c) does not exist
    \n(d) 0
    \nAnswer:
    \n(a) 6<\/p>\n

    Question 16.
    \nIf f(x) = x\u00b2 – 3x, then the points at which f(x) = f'(x) are…………..
    \n(a) both positive integers
    \n(b) both negative integers
    \n(c) both irrational
    \n(d) one rational and another irrational
    \nAnswer:
    \n(c) both irrational<\/p>\n

    \"Tamil<\/p>\n

    Question 17.
    \n\\(\\int \\tan ^{-1}(\\sqrt{\\frac{1-\\cos 2 x}{1+\\cos 2 x}})\\) dx ………….
    \n(a) x\u00b2 + c
    \n(b) 2x\u00b2 + c
    \n(c) \\(\\frac{x\u00b2}{2}\\) + c
    \n(d) –\\(\\frac{x\u00b2}{2}\\) + c
    \nAnswer:
    \n(c) \\(\\frac{x^2}{2}\\) + c<\/p>\n

    Question 18.
    \ne-7x<\/sup> sin 5x dx is…………
    \n(a) \\(\\frac{e^{-7x}}{74}\\) [-7 sin 5x – 5 cos 5x] + c
    \n(b) \\(\\frac{e^{-7x}}{74}\\) [7 sin 5x + 5 cos 5x] + c
    \n(c) \\(\\frac{e^{-7x}}{74}\\) [7 sin 5x – 5 cos 5x] + c
    \n(d) \\(\\frac{e^{-7x}}{74}\\) [-7 sin 5x + 5 cos 5x] + c
    \nAnswer:
    \n(a) \\(\\frac{e^{-7x}}{74}\\) [-7 sin 5x – 5 cos 5x] + c<\/p>\n

    Question 19.
    \nIf A and B are any two events then the probability that exactly one of them occur is
    \n(a) P(A \u222a\\(\\bar { B }\\)) + P(\\(\\bar { A }\\) \u222aB)
    \n(b) P(A \u2229\\(\\bar { B }\\)) + P(\\(\\bar { A }\\) \u2229B)
    \n(c) P(A) + P(B) – P(A \u2229 B)
    \n(d) P(A) + P(B) + 2P(A \u2229 B)
    \nAnswer:
    \n(a) P(A \u222a\\(\\bar { B }\\)) + P(\\(\\bar { A }\\) \u222aB)<\/p>\n

    Question 20.
    \nIn a certain college 4% of the boys and 1 % of the girls are taller than 1.8 meter. Further 60% of the students are girls. If a student is selected at random and is taller than 1.8 meters, then the probability that the student is a girl is ………….
    \n(a) \\(\\frac{2}{11}\\)
    \n(b) \\(\\frac{3}{11}\\)
    \n(c) \\(\\frac{5}{11}\\)
    \n(d) \\(\\frac{7}{11}\\)
    \nAnswer:
    \n(b) \\(\\frac{3}{11}\\)<\/p>\n

    PART- II<\/span><\/p>\n

    II. Answer any seven questions. Question No: 30 is compulsory. [7 \u00d7 2 = 14]<\/span><\/p>\n

    Question 21.
    \nResolve \\(\\frac{3x+1}{(x-2)(x+1)}\\) into partial fractions.
    \nAnswer:
    \n\"Tamil
    \nEquating numerator parts
    \n3x + 1 = A (x + 1) + B (x – 2)
    \nThis equation is true for any value of x.
    \nTo find A and B
    \nPut x = -1
    \n-3 + 1 = A (0) + B (-1 -2)
    \n-3 B = -2 \u21d2 B = 2\/3
    \nPut x = 2
    \n3(2) + 1 = A(2 + 1) + B (0)
    \n3A = 7 \u21d2 A = 7\/3
    \nHence \\(\\frac{3x+1}{(x-2)(x+1)}\\) = \\(\\frac{7}{3(x-2)}\\) + \\(\\frac{2}{3(x+1)}\\)<\/p>\n

    \"Tamil<\/p>\n

    Question 22.
    \nProve that
    \n\"Tamil
    \nAnswer:
    \ncot (180\u00b0 + \u03b8) = cot \u03b8
    \nsin (90\u00b0 – \u03b8) = cos \u03b8
    \ncos (- \u03b8) = cos \u03b8
    \nsin (270 + \u03b8) = – cos \u03b8
    \ntan (-\u03b8) = – tan \u03b8
    \ncosec (360\u00b0 + \u03b8) = cosec \u03b8
    \n\"Tamil<\/p>\n

    Question 23.
    \nIf cos \u03b8 = \\(\\frac{1}{2}\\) (a + \\(\\frac{1}{a}\\)), show that cos 3\u03b8 = \\(\\frac{1}{2}\\) (a\u00b3 + \\(\\frac{1}{a\u00b3}\\))
    \nAnswer:
    \n\"Tamil<\/p>\n

    Question 24.
    \nIf the letters of the word IITJEE are permuted in all possible ways and the strings thus formed are arranged in the lexicographic order, find the rank of the word IITJEE.
    \nAnswer:
    \nThe lexicographic order of the letters of given word is E, E, I, I, J, T. In the lexicographic order, the strings which begin with E come first. If we fill the first place with E, remaining 5 letters (E, I, I, J, T) can be arranged in \\(\\frac{5!}{2!}\\) ways. On proceeding like this we get,
    \nE – – – – = \\(\\frac{5!}{2!}\\) = 60 ways
    \nIIE – – – = 3! = 6 ways
    \nIIJ – – – = \\(\\frac{3!}{1!}\\) = 3 ways
    \nIITE – – =2! = 2 ways
    \nIITJEE = 1 way
    \nThe rank of the word IITJEE = 60 + 6 + 3 + 2 +1 = 72.<\/p>\n

    Question 25.
    \nProve that \\(\\frac{(2n)!}{n!}\\) = 2n<\/sup> (1, 3, 5 ……..(2n – 1)).
    \nAnswer:
    \n\"Tamil<\/p>\n

    Question 26.
    \nWrite the equation of the line passing through the point (1, -1) and parallel to the line x + 3y – 4 = 0
    \nAnswer:
    \nEquation of a line parallel to x + 3y – 4 = 0 will be of the form x + 3y + k = 0
    \nIt passes through (1,-1)
    \n\u21d2 (1) + 3(-1) + k = 0
    \n-2 + k = 0 \u21d2 k = 2
    \nSo the required equation is x + 3y + 2 = 0<\/p>\n

    Question 27.
    \nIf (k, 2), (2, 4) and (3, 2) are vertices of the triangle of area 4 square units then determine the value of k.
    \nAnswer:
    \nArea of \u0394 with vertices (k, 2) (2, 4) and (3, 2) = \\(\\frac{1}{2}\\) \\(\\left|\\begin{array}{lll} k & 2 & 1 \\\\ 2 & 4 & 1 \\\\ 3 & 2 & 1 \\end{array}\\right|\\) = 4 (given)
    \n\u21d2 \\(\\left|\\begin{array}{lll} k & 2 & 1 \\\\ 2 & 4 & 1 \\\\ 3 & 2 & 1 \\end{array}\\right|\\) = 2(4) = 8
    \n(i.e.,) k (4 – 2) -2(2 – 3) + 1 (4 – 12) = \u00b18
    \n(i.e.,) 2k – 2(-1) + 1(-8) = \u00b1 8
    \n(i.e.,) 2k + 2 – 8 = 8
    \n(i.e.,) 2k = 8 + 8 – 2 = 14
    \nk = 14\/2 = 7
    \n\u2234 k = 7
    \nor
    \n2k + 2 – 8 = -8
    \n\u21d2 2k = -8 + 8 – 2
    \n2k = -2
    \nk = -1
    \nSo k = 7 (or) k = -1<\/p>\n

    \"Tamil<\/p>\n

    Question 28.
    \nFind \u03bb, when the projection of \\(\\vec {a}\\) = \u03bb\\(\\hat{j}\\) + \\(\\hat{j}\\) + 4\\(\\hat{k}\\) on \\(\\vec {b}\\) = 2\\(\\hat{i}\\) + 6\\(\\hat{j}\\) + 2\\(\\hat{k}\\) is 4 units.
    \nAnswer:
    \n\\(\\vec {a}\\) = \u03bb\\(\\hat{j}\\) + \\(\\hat{j}\\) + 4\\(\\hat{k}\\) and \\(\\vec {b}\\) = 2\\(\\hat{i}\\) + 6\\(\\hat{j}\\) + 2\\(\\hat{k}\\)
    \nNow \\(\\vec {a}\\) – \\(\\vec {b}\\) = (\u03bb) (2) + (1) (6) + (4) (3)
    \n= 2\u03bb + 6 + 12 = 2\u03bb +18
    \n|\\(\\vec {a}\\)| = \\(\\sqrt{4+36+9}\\) = \\(\\sqrt{49}\\) = 7
    \nHere \\(\\frac{2\u03bb+18}{7}\\) = 4
    \n\u21d2 2\u03bb+ 18 = 4 \u00d7 7 = 28
    \n2\u03bb = 28 – 18 = 10
    \n\u03bb = 10\/2 = 5<\/p>\n

    Question 29.
    \nFind \\(\\frac{dy}{dx}\\) if y = ex<\/sup> sin x
    \nAnswer:
    \ny = ex<\/sup> sin x
    \n\u21d2 y’ = uv’ + vu’
    \nNow u = ex<\/sup> \u21d2 u’ = \\(\\frac{du}{dx}\\) = ex<\/sup>
    \nv = sin x \u21d2 v’ = \\(\\frac{dv}{dx}\\) = cos x
    \ny’ = ex<\/sup> (cos x) + sin x (ex<\/sup>)
    \n= ex<\/sup> [sin x + cos x]<\/p>\n

    Question 30.
    \nFind \\(\\frac{dy}{dx}\\) if x\u00b2 + y\u00b2 = 1
    \nAnswer:
    \nWe differentiate both sides of the equation.
    \n\\(\\frac{d}{dx}\\) (x)\u00b2 + \\(\\frac{d}{dx}\\) (y)\u00b2 = \\(\\frac{d}{dx}\\) (1)
    \n2x + 2y \\(\\frac{dy}{dx}\\) = 0
    \nSolving for the derivative yields
    \n\\(\\frac{dy}{dx}\\) = –\\(\\frac{x}{y}\\)<\/p>\n

    PART – III<\/span><\/p>\n

    III. Answer any seven questions. Question No. 40 is compulsory. [7 \u00d7 3 = 21]<\/span><\/p>\n

    Question 31.
    \nFrom the curve y = sin x, draw y = sin |x|
    \nAnswer:
    \n\"Tamil
    \n\"Tamil<\/p>\n

    Question 32.
    \nIf one root of k (x – 1)\u00b2 = 5x – 7 is double the other root, show that k = 2 or -25
    \nAnswer:
    \nk(x – 1)\u00b2 = 5x – 7
    \n(i.e.,) k (x\u00b2 – 2x + 1) – 5x + 7 = 0
    \nx\u00b2 (k) + x(-2k – 5) + k + 7 = 0
    \nkx\u00b2 – x(2k + 5) + (k + 7) = 0
    \nHere it is given that one root is double the other.
    \nSo let the roots to \u03b1 and 2\u03b1
    \nSum of the roots = \u03b1 + 2\u03b1 = 3\u03b1 = \\(\\frac{2k+5}{k}\\) \u03b1 \\(\\frac{2k+5}{3k}\\) ……..(1)
    \nProduct of the roots = \u03b1(2\u03b1) = 2\u03b1\u00b2 = \\(\\frac{k+7}{k}\\)
    \n\u21d2 \u03b1\u00b2 = \\(\\frac{k+7}{2k}\\) = ……..(2)
    \nSubstituting a value from (1) in (2)
    \n\"Tamil
    \n2(4k\u00b2 + 25 + 20k) = 9k (k + 7)
    \n2(4k\u00b2 + 25 + 20k) = 9k\u00b2 + 63k
    \n8k\u00b2 + 50 + 40k – 9k\u00b2 – 65k = 0
    \n-k\u00b2 – 25k + 50 = 0
    \nk\u00b2 + 23k – 50 = 0
    \n(k + 25) (k – 2) = 0
    \nk = -25 or 2<\/p>\n

    \"Tamil<\/p>\n

    Question 33.
    \nIf \u03b8 + \u2205 = \u03b1 and tan \u03b8 = k tan \u2205, then prove that sin(\u03b8 – \u2205) =\\(\\frac{k-1}{k+1}\\)
    \nAnswer:
    \n\"Tamil<\/p>\n

    Question 34.
    \nIf a, b, c are in geometric progression, and if a\\(\\frac{1}{x}\\)<\/sup> = b\\(\\frac{1}{y}\\)<\/sup> = c\\(\\frac{1}{z}\\)<\/sup> then prove that x, y, z are in arithmetic progression.
    \nAnswer:
    \nGiven a, b, c are in G.P.
    \n\u21d2 b\u00b2 = ac
    \n\u21d2 log b\u00b2 = log ac
    \n(i.e.) 2 log b = log a + log c …(1)
    \nWe are given a\\(\\frac{1}{x}\\)<\/sup> = b\\(\\frac{1}{y}\\)<\/sup> = c\\(\\frac{1}{z}\\)<\/sup> = k (say)
    \n\u21d2 log ak<\/sup> = \\(\\frac{1}{x}\\) = bk<\/sup> \\(\\frac{1}{y}\\) = ck<\/sup> \\(\\frac{1}{z}\\)
    \n\u21d2 ak<\/sup> = \\(\\frac{1}{x}\\) \u21d2 x = log ka<\/sup>
    \nSimilarly y = log kb<\/sup>
    \nz = log kc<\/sup>
    \nSubstituting these values in equation (1) we get 2y = x + z \u21d2 x, y, z are in A.P.<\/p>\n

    Question 35.
    \nShow that the points (1,3), (2,1) and (\\(\\frac{1}{2}\\), 4) are collinear.
    \nAnswer:
    \nLet the given points be A (1, 3), B (2, 1), and C (\\(\\frac{1}{2}\\), 4)
    \nSlope of AB = \\(\\frac{1-3}{2-1}\\) = \\(\\frac{-2}{1}\\) = -2 = m1<\/sub>
    \nSlope of BC = \\(\\frac{4-1}{1\/2-1}\\) = \\(\\frac{3}{-3\/2}\\) = -2 = m2<\/sub>
    \nSlope of AB = Slope of BC \u21d2 AB parallel to BC but B is a common point.
    \n\u21d2 The points A, B, C are collinear.<\/p>\n

    Question 36.
    \nIf A and B are symmetric matrices of same order, prove that AB + BA is a symmetric matrix.
    \nAnswer:
    \nGiven A and B are symmetric matrices
    \n\u21d2 AT<\/sup> = A and BT<\/sup> = B
    \nTo prove AB + BA is a symmetric matrix.
    \nProof: Now(AB + BA)T<\/sup> = (AB)T<\/sup> + (BA)T<\/sup> = BT<\/sup>AT<\/sup> + AT<\/sup>BT<\/sup>
    \n= BA + AB = AB + BA
    \ni.e. (AB + BA)T<\/sup> = AB + BA
    \n\u21d2 (AB + BA) is a symmetric matrix.<\/p>\n

    Question 37.
    \nLet \\(\\vec {a}\\), \\(\\vec {b}\\), \\(\\vec {c}\\) be three vectors such that |\\(\\vec {a}\\)| = 3, |\\(\\vec {b}\\)| = 4, |\\(\\vec {c}\\)| = 5 and each one of them being perpendicular to the sum of the other two, find |\\(\\vec {a}\\) + \\(\\vec {b}\\) + \\(\\vec {c}\\)|.
    \nAnswer:
    \nGiven |\\(\\vec {a}\\)| = 3; |\\(\\vec {b}\\)| = 4; |\\(\\vec {c}\\)| = 5
    \nNow, (\\(\\vec {a}\\) + \\(\\vec {b}\\) + \\(\\vec {c}\\))\u00b2 = a-2<\/sup> +b -2<\/sup>+c-2<\/sup> + 2(\\(\\vec {a}\\).\\(\\vec {b}\\) + \\(\\vec {b}\\).\\(\\vec {c}\\) + \\(\\vec {a}\\).\\(\\vec {c}\\))
    \n= 3\u00b2 + 4\u00b2 + 5\u00b2 + 2 \\(\\vec {a}\\).\\(\\vec {b}\\) + 2\\(\\vec {b}\\).\\(\\vec {c}\\) + 2\\(\\vec {a}\\).\\(\\vec {c}\\)
    \n= 9 + 16 + 25 + \\(\\vec {a}\\).\\(\\vec {b}\\) + \\(\\vec {a}\\).\\(\\vec {b}\\) + \\(\\vec {b}\\).\\(\\vec {c}\\) + \\(\\vec {b}\\).\\(\\vec {c}\\) + \\(\\vec {a}\\).\\(\\vec {c}\\) + \\(\\vec {a}\\).\\(\\vec {c}\\)
    \n= 50 + \\(\\vec {a}\\).(\\(\\vec {b}\\) + \\(\\vec {c}\\)) + \\(\\vec {b}\\).(\\(\\vec {c}\\) + \\(\\vec {a}\\)) + \\(\\vec {c}\\).(\\(\\vec {a}\\) + \\(\\vec {b}\\))
    \n= 50 + 0 + 0 + 0 = 50
    \n(one vector is \u22a5r<\/sup> to the sum of other two vectors)
    \n|\\(\\vec {a}\\) + \\(\\vec {b}\\) + \\(\\vec {c}\\)| = \\(\\sqrt {50}\\) = \\(\\sqrt {25\u00d72}\\) = 5\u221a2<\/p>\n

    \"Tamil<\/p>\n

    Question 38.
    \nEvaluate \\(\\lim _{x \\rightarrow 0} \\frac{\\sqrt{1+x^{2}}-1}{x}\\)
    \nAnswer:
    \n\"Tamil<\/p>\n

    Question 39.
    \nFind the derivatives of the following functions y = xcos x<\/sup>
    \nAnswer:
    \ny = xcos x<\/sup>
    \nTaking log on both sides, we get
    \nlog y = log xcos x<\/sup> = cos x log x
    \nDifferentiating w.r.to x we get
    \n\"Tamil<\/p>\n

    Question 40.
    \nEvaluate
    \n\"Tamil
    \nAnswer:
    \n\"Tamil<\/p>\n

    PART – IV<\/span><\/p>\n

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

    Question 41 (a).
    \nLet f, g: R \u2192 R be defined as f(x) = 2x – |x| and g(x) = 2x + |x|. Find fog.
    \nAnswer:
    \n\"Tamil
    \nfog (x) = f(g(x))
    \n= f(x) = 3x
    \nFor x > 0
    \nfog (x) = f(g(x))
    \n= f(3x) = 3x<\/p>\n

    [OR]<\/p>\n

    (b) Prove that \\(\\lim _{\\theta \\rightarrow 0} \\frac{\\sin \\theta}{\\theta}\\) = 1
    \nAnswer:
    \nLet \u03b8 \u2192 0, thought positive values
    \n\u2234 Let 0 < \u03b8 < \\(\\frac{\u03c0}{2}\\)
    \nDraw a circule center O and radius unity and let
    \n\u2220AOB = \u03b8 radiAnswer:
    \nLet the tagent at A meet OB product in the piont P.
    \nJion AB. From the figure it is clear that
    \nArea of \u0394AOB < Area of sector AOB < Area of \u0394OAP
    \n\"Tamil
    \ni.e., \\(\\frac{1}{2}\\) OA. OB sin \u03b8 < 1 (radius)\u00b2 \u03b8 < \\(\\frac{1}{2}\\) OA. AP
    \nor sin \u03b8 < \u03b8 < tan \u03b8
    \n[ \u2235 \\(\\frac{AP}{OA}\\) = tan \u03b8 or AP = tan \u03b8 as OA = 1
    \nDividing by sin \u03b8, which is positive
    \n\"Tamil<\/p>\n

    Question 42 (a).
    \n(i) If A \u00d7 A has 16 elements, S = {(a, b) \u2208 A \u00d7 A: a < b} ; (-1, 2) and (0,1) are two elements of S, then find the remaining elements of S.
    \n(ii) Find the range of the function \\(\\frac{1}{2 cos x – 1}\\)
    \nAnswer:
    \n(i) n(A \u00d7 A) = 16
    \n\u21d2 n(A) = 4
    \nS = {(-1, 0), (-1, 1), (0, 2), (1, 2)}<\/p>\n

    (ii) The range of cos x is – 1 to 1
    \n-1 \u2264 cos x \u2264 1
    \n(\u00d7 by 2) -2 \u2264 2 cos x \u2264 2
    \nadding -1 throughout
    \n-2 -1 \u2264 2 cos x – 1 \u2264 2 – 1
    \n(i.e.,) -3 \u2264 2 cos x – 1 \u2264 1
    \nso 1 \u2264 \\(\\frac{1}{2 cos x – 1}\\) \u2264 \\(\\frac{-1}{3}\\)
    \nThe range is outside \\(\\frac{-1}{3}\\) and 1
    \ni.e., range is (-\u221e,\\(\\frac{-1}{3}\\)] \u222a [1, \u221e)<\/p>\n

    \"Tamil<\/p>\n

    [OR]<\/p>\n

    (b) Evaluate: \\(\\int \\frac{3 x+5}{x^{2}+4 x+7}\\) dx
    \nAnswer:
    \nLet I = \\(\\int \\frac{3 x+5}{x^{2}+4 x+7}\\) dx
    \n3x + 5 = A \\(\\frac{d}{dx}\\) (x\u00b2 + 4x + 7) + B
    \n3x + 5 = A(2x + 4) + B
    \nComparing the coefficients of like terms we got
    \n2A = 3 \u21d2 A = \\(\\frac{3}{2}\\); 4A + B = 5 \u21d2 B = -1
    \n\"Tamil<\/p>\n

    Question 43 (a).
    \nDerive cosine formula using the law of sines in a \u0394ABC.
    \nAnswer:
    \n\"Tamil<\/p>\n

    [OR]<\/p>\n

    (b) Find \\(\\sqrt[3]{65}\\) using binomial expansion upto two decimal places.
    \nAnswer:
    \nWe know that for |x| < 1,
    \n\"Tamil
    \n\u2248 4 + 0.02 ( Since \\(\\frac{1}{36864}\\) + …….. is very small)
    \n\\(\\sqrt[3]{65}\\) = 4.02 (approximately)<\/p>\n

    \"Tamil<\/p>\n

    Question 44 (a).
    \n(i) Do the limit of the function \\(\\frac{sin |x|}{x}\\) Li exist as x \u2192 0? State reason for the answer.
    \nAnswer:
    \n\"Tamil
    \nHence the limit does ndt exist. Since that f(0–<\/sup>) \u2260 f(0+<\/sup>)
    \n\"Tamil<\/p>\n

    [OR]<\/p>\n

    (b) \"Tamil
    \nAnswer:
    \n\"Tamil<\/p>\n

    Question 45.
    \n(i) Evaluate \\(\\int \\frac{x \\sin ^{-1} x}{\\sqrt{1-x^{2}}}\\) dx
    \nAnswer:
    \nLet sin-1<\/sup> x = t
    \n\u21d2 \\(\\frac{1}{\\sqrt{1-x^{2}}}\\) dx = dt
    \nNow sin-1<\/sup> x = t \u21d2 x = sin t
    \nSo I = \u222b(sin t) (t) dt = \u222b t sin t dt
    \nNow \u222bt sin t dt = \u222b t d (-cos t)
    \n= t(- cos t) – \u222b(-cos t) dt
    \n= -t cos t + \u222bcos t dt
    \n= – t cos t + sin t
    \n\"Tamil<\/p>\n

    [OR]<\/p>\n

    (b) Show that the points whose position vectors 4\\(\\hat{i}\\) + 5\\(\\hat{j}\\) + \\(\\hat{k}\\),- \\(\\hat{j}\\) – \\(\\hat{k}\\), 3\\(\\hat{i}\\) +9\\(\\hat{j}\\) + 4\\(\\hat{k}\\) and -4\\(\\hat{i}\\) + 4\\(\\hat{j}\\) + 4\\(\\hat{k}\\) are coplanar.
    \nAnswer:
    \nLet the given points be A, B, C and D. To prove that the point A, B, C, D are coplanar, we have to prove that the vectors \\(\\overrightarrow{\\mathrm{AB}}\\), \\(\\overrightarrow{\\mathrm{AC}}\\) and \\(\\overrightarrow{\\mathrm{AD}}\\) are coplanar.
    \n\"Tamil
    \nEquating \\(\\hat{i}\\), \\(\\hat{j}\\) and \\(\\hat{k}\\) components we get
    \n-4 = -m – 8n
    \n\u21d2 m + 8n = 4 ….(i)
    \n-6 = 4m – n
    \n\u21d2 4m – n = -6 ……. (ii)
    \n-2 = 3m + 3n
    \n\u21d2 3m + 3n = -2 …….. (iii)
    \nSolving (i) and (ii)
    \n\"Tamil
    \nSubstituting m = –\\(\\frac{4}{3}\\) in (i) we get,
    \n–\\(\\frac{4}{3}\\) + 8n = 4
    \n\"Tamil
    \nwe are able to write one vector as a linear combination of the other two vectors \u21d2 the given vectors \\(\\vec{a}\\), \\(\\vec{b}\\), \\(\\vec{c}\\) are coplanar.
    \n(i.e.,) The given points A, B, C, D are coplanar.<\/p>\n

    \"Tamil<\/p>\n

    Question 46 (a).
    \nIf Q is a point on the locus of x\u00b2 + y\u00b2 + 4x – 3x + 7 = 0, then find the equation of locus of P which divides segment OQ externally in the ratio 3 : 4, where O is the origin.
    \nAnswer:
    \nLet (h, k) be the moving-point O = (0, 0), Let PQ = (a, b) on x\u00b2 + y\u00b2 + 4x – 3y + 7 = 0
    \nP divides OQ externally in the ratio 3 : 4
    \n\"Tamil
    \nh\u00b2 + k\u00b2 – 12h + 9k + 63 = 0, Locus of (h, k) is x\u00b2 + y\u00b2 – 12x + 9y + 63 = 0<\/p>\n

    [OR]<\/p>\n

    (b) A consulting firm rents car from three agencies such that 50% from agency L, 30% from agency M and 20% from agency N. If 90% of the cars from L, 70% of cars from M and 60% of cars from N are in good conditions (i) what is the probability that the firm will get a car in good condition? (ii) if a car is in good condition, what is probability that it has come from agency N?
    \nAnswer:
    \nLet A1<\/sub>, A2<\/sub> and A3<\/sub> be the events that the cars are rented from the agencies X, Y and Z respectively.
    \nLet G be the event of getting a car in good condition.
    \nWe have to find
    \n\"Tamil
    \n(i) The total probability of event G that is, P(G).
    \n(ii) Find the conditional probability A3, given G that is, P(A3<\/sub>\/G) We have
    \nP(A1<\/sub>) = 0.50, P(G\/A1<\/sub>) = 0.90
    \nP(A2<\/sub>) = 0.30, P(G\/A2<\/sub>) = 0.70
    \nP(A3<\/sub>) = 0.20, P(G\/A3<\/sub>) = 0.60.<\/p>\n

    (i) Since A1<\/sub>, A2<\/sub> and A3<\/sub> are mutually exclusive and exhaustive events and G is an event in S, then the total probability of event G is P(G).
    \nP(G) = P(A1<\/sub>) P(G\/A1<\/sub>) + P(A2<\/sub>) P(G\/A2<\/sub>) + P(A3<\/sub>) P(G\/A3<\/sub>)
    \nP(G) = (0.50) (0.90) + (0.30) (0.70) + (0.20) (0.60)
    \nP(G) = 0.78.<\/p>\n

    (ii) The conditional probability A3<\/sub> given G is P(A3<\/sub>\/G)
    \nBy Bayes\u2019 theorem,
    \n\"Tamil<\/p>\n

    \"Tamil<\/p>\n

    Question 47 (a).
    \nUsing the Mathematical induction, show that for any natural number n,
    \n\"Tamil
    \nAnswer:
    \n\"Tamil
    \n\"Tamil
    \n\u2234 P(k + 1) is true
    \nThus P(k) is true \u21d2 P(k + 1) is true.
    \nHence by principle of mathematical induction,
    \nP(n) is true for all n \u2208 Z.<\/p>\n

    (b) If 7 = (cos-1<\/sup> x)\u00b2, prove that (1 – x\u00b2) \\(\\frac{d\u00b2y}{dx\u00b2}\\) -x\\(\\frac{dx}{dy}\\) -2 = 0. Hence find y2<\/sub>, when x = 0.
    \nAnswer:
    \n\"Tamil
    \nSquaring on both sides (1 – x\u00b2) (\\(y_{1}^{2}\\)) = 4(cos-1<\/sup> x)\u00b2 = 4y
    \n\u21d2 (1 – x\u00b2)(\\(y_{1}^{2}\\)) = 4y
    \nDifferentiating again w.r.to x we get
    \n(1 – x\u00b2) (2y1<\/sub>, y2<\/sub>) + (\\(y_{1}^{2}\\)) (-2x) = 4y1<\/sub>
    \n\u21d2 (1 – x\u00b2) (2y1<\/sub>, y2<\/sub>) = 4y1<\/sub> + 2xy\\(y_{1}^{2}\\)
    \n(i.e.,) (1 – x\u00b2) (2y1<\/sub>, y2<\/sub>) = 2y1<\/sub> (2 +xy1<\/sub>)
    \n(\u00f7 by 2y1<\/sub>)(1 – x\u00b2)y\u00b2 = 2 + xy1<\/sub>
    \nSo (1 – x\u00b2)y2<\/sub> – xy1<\/sub> – 2 = 0
    \nWhen x = 0
    \n(1 – 0) y2<\/sub> – 0y1<\/sub> – 2 = 0
    \ny2<\/sub> – 2 = 0
    \ny2<\/sub> = 2<\/p>\n

    \"Tamil<\/p>\n","protected":false},"excerpt":{"rendered":"

    Students can Download Tamil Nadu 11th Maths Previous Year Question Paper June 2019 English Medium Pdf, Tamil Nadu 11th 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 11th Maths Previous Year Question Paper June 2019 English Medium General Instructions: …<\/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":[7],"tags":[],"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/samacheerkalviguru.com\/wp-json\/wp\/v2\/posts\/15599"}],"collection":[{"href":"https:\/\/samacheerkalviguru.com\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/samacheerkalviguru.com\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/samacheerkalviguru.com\/wp-json\/wp\/v2\/users\/2"}],"replies":[{"embeddable":true,"href":"https:\/\/samacheerkalviguru.com\/wp-json\/wp\/v2\/comments?post=15599"}],"version-history":[{"count":1,"href":"https:\/\/samacheerkalviguru.com\/wp-json\/wp\/v2\/posts\/15599\/revisions"}],"predecessor-version":[{"id":60615,"href":"https:\/\/samacheerkalviguru.com\/wp-json\/wp\/v2\/posts\/15599\/revisions\/60615"}],"wp:attachment":[{"href":"https:\/\/samacheerkalviguru.com\/wp-json\/wp\/v2\/media?parent=15599"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/samacheerkalviguru.com\/wp-json\/wp\/v2\/categories?post=15599"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/samacheerkalviguru.com\/wp-json\/wp\/v2\/tags?post=15599"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}