{"id":21789,"date":"2020-01-29T11:11:02","date_gmt":"2020-01-29T05:41:02","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=21789"},"modified":"2021-07-02T15:26:22","modified_gmt":"2021-07-02T09:56:22","slug":"2nd-puc-basic-maths-previous-year-question-paper-june-2017","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/2nd-puc-basic-maths-previous-year-question-paper-june-2017\/","title":{"rendered":"2nd PUC Basic Maths Previous Year Question Paper June 2017"},"content":{"rendered":"<p>Students can Download 2nd PUC Basic Maths Previous Year Question Paper June 2017, Karnataka <a href=\"https:\/\/kseebsolutions.guru\/2nd-puc-basic-maths-model-question-papers-with-answers\/\">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<h2>Karnataka 2nd PUC Basic Maths Previous Year Question Paper June 2017<\/h2>\n<p>Time: 3.15 hours<br \/>\nMax Marks: 100<\/p>\n<p>Instructions:<\/p>\n<ul>\n<li>The question paper has 5 parts A, B, C, D and E. Answer all the pats.<\/li>\n<li>Part-A carries 10 marks), Parts &#8211; B carries 20 marks, Part &#8211; C carries 3 marks, Part &#8211; D carries 30 marks and Part &#8211; E carries 1 marks.<\/li>\n<li>Write the question number properly as indicated in the question paper.<\/li>\n<\/ul>\n<p style=\"text-align: center;\"><span style=\"color: #0000ff;\">Part &#8211; A<\/span><\/p>\n<p><span style=\"color: #0000ff;\">Answer all the ten questions: (10 \u00d7 1 = 10)<\/span><\/p>\n<p>Question 1.<br \/>\nEvaluate \\(\\left| \\begin{matrix} 3200 &amp; 3201 \\\\ 3202 &amp; 3203 \\end{matrix} \\right| \\)<br \/>\nAnswer:<br \/>\n<img decoding=\"async\" loading=\"lazy\" class=\"alignnone size-full wp-image-65809\" src=\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2020\/01\/2nd-PUC-Basic-Maths-Previous-Year-Question-Paper-June-2017-1.png\" alt=\"2nd PUC Basic Maths Previous Year Question Paper June 2017 - 1\" width=\"303\" height=\"52\" srcset=\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2020\/01\/2nd-PUC-Basic-Maths-Previous-Year-Question-Paper-June-2017-1.png 303w, https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2020\/01\/2nd-PUC-Basic-Maths-Previous-Year-Question-Paper-June-2017-1-300x51.png 300w\" sizes=\"(max-width: 303px) 100vw, 303px\" \/><\/p>\n<p>Question 2.<br \/>\nIn how many ways can 10 people seated around a table ?<br \/>\nAnswer:<br \/>\n(10 &#8211; 1)! = 9!<\/p>\n<p>Question 3.<br \/>\nWrite the verbal form of the compound proposition pvq where<br \/>\nP:x is an integer; Q: 5 is odd number.<br \/>\nAnswer:<br \/>\nPVQ = x is an integer or 5 is odd number.<\/p>\n<p>Question 4.<br \/>\nFind the triplicate ration of 3 : 5.<br \/>\nAnswer:<br \/>\n3<sup>3 <\/sup>: 5<sup>3<\/sup> = 27:125<\/p>\n<p><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2019\/11\/KSEEB-Solutions.png\" alt=\"KSEEB Solutions\" width=\"172\" height=\"16\" \/><\/p>\n<p>Question 5.<br \/>\nA bill was drawn for 3 months was legally due on 18-08-2012. Find the date of drawing of the bill.<br \/>\nAnswer:<br \/>\nDrawn date = L.D.D &#8211; B.P &#8211; 3 days = 15 &#8211; 5 &#8211; 12<\/p>\n<p>Question 6.<br \/>\nFind the value of cos<sup>3<\/sup> 20\u00b0 &#8211; 3cos 20\u00b0.<br \/>\nAnswer:<br \/>\nCos 3(20) = cos60 = \\(\\frac { 1 }{ 2 }\\)<\/p>\n<p>Question 7.<br \/>\nIf the length of the latus rectum of the parabola x<sup>2<\/sup> = 4k y is 8. Find the value of k.<br \/>\nAnswer:<br \/>\n4k = 8 \u21d2 k = 2<\/p>\n<p>Question 8.<br \/>\nEvaluate line \\(\\lim _{x \\rightarrow 3}\\left(\\frac{x^{2}-x}{x-2}\\right)\\)<br \/>\nAnswer:<br \/>\n\\(=\\frac{3^{2}-3}{3-2}=\\frac{9-3}{1}=6\\)<\/p>\n<p>Question 9.<br \/>\nDifferentiate : 5e<sup>x<\/sup> &#8211; log<sub>e<\/sub> x &#8211; 3\u221ax w.r.t.x.<br \/>\nAnswer:<br \/>\n\\(\\frac{d y}{d x}=5 e^{x}-\\frac{1}{x}-\\frac{3}{2 \\sqrt{x}}\\)<\/p>\n<p>Question 10.<br \/>\nEvaluate \u222bsec<sup>2<\/sup> (x &#8211; 5)dx.<br \/>\nAnswer:<br \/>\nTan (x &#8211; 5)<\/p>\n<p style=\"text-align: center;\"><span style=\"color: #0000ff;\">Part &#8211; B<\/span><\/p>\n<p><span style=\"color: #0000ff;\">Answer any ten questions. (10 \u00d7 2 = 20)<\/span><\/p>\n<p>Question 11.<br \/>\nIf A = \\(\\left[ \\begin{matrix} 2 \\\\ -1 \\\\ 3 \\end{matrix} \\right] \\), B = \\(\\left[ \\begin{matrix} 1 &amp; 4 &amp; 2 \\end{matrix} \\right] \\) find AB and BA.<br \/>\nAnswer:<br \/>\n<img decoding=\"async\" loading=\"lazy\" class=\"alignnone size-full wp-image-65808\" src=\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2020\/01\/2nd-PUC-Basic-Maths-Previous-Year-Question-Paper-June-2017-2.png\" alt=\"2nd PUC Basic Maths Previous Year Question Paper June 2017 - 2\" width=\"376\" height=\"161\" srcset=\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2020\/01\/2nd-PUC-Basic-Maths-Previous-Year-Question-Paper-June-2017-2.png 376w, https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2020\/01\/2nd-PUC-Basic-Maths-Previous-Year-Question-Paper-June-2017-2-300x128.png 300w\" sizes=\"(max-width: 376px) 100vw, 376px\" \/><\/p>\n<p><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2019\/11\/KSEEB-Solutions.png\" alt=\"KSEEB Solutions\" width=\"172\" height=\"16\" \/><\/p>\n<p>Question 12.<br \/>\nIn a party each person shakes hands with everyone else. If there are 25 members in the party, calculate the number the number of hand shakes.<br \/>\nAnswer:<br \/>\n\\(25 C_{2}=\\frac{25 \\times 24}{2 \\times 1}=25 \\times 12=300\\)<\/p>\n<p>Question 13.<br \/>\nA bag contains 6 white beads and 4 red beads. A bead is drawn at random, what is the probability that the bead drawn is white.<br \/>\nAnswer:<br \/>\nP(beaddrawniswhite) = \\(\\frac{10 C_{1}}{6 C_{1}} \\frac{10}{6}=\\frac{5}{3}\\)<\/p>\n<p>Question 14.<br \/>\nIf p,q, r are the propositions with truth values F, T and F repectively. Then find the truth value of the ; compound proposition p \u2192 (q \u2192 r).<br \/>\nAnswer:<br \/>\nP \u2192 (q \u2192 r)<br \/>\nF \u2192 (T \u2192 F)<br \/>\nF \u2192 F = T<\/p>\n<p>Question 15.<br \/>\nIf a : b = 4 : 5. Find \\(\\frac{3 a+2 b}{3 a-2 b}\\)<br \/>\nAnswer:<br \/>\n<img decoding=\"async\" loading=\"lazy\" class=\"alignnone size-full wp-image-65807\" src=\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2020\/01\/2nd-PUC-Basic-Maths-Previous-Year-Question-Paper-June-2017-3.png\" alt=\"2nd PUC Basic Maths Previous Year Question Paper June 2017 - 3\" width=\"392\" height=\"138\" srcset=\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2020\/01\/2nd-PUC-Basic-Maths-Previous-Year-Question-Paper-June-2017-3.png 392w, https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2020\/01\/2nd-PUC-Basic-Maths-Previous-Year-Question-Paper-June-2017-3-300x106.png 300w\" sizes=\"(max-width: 392px) 100vw, 392px\" \/><\/p>\n<p>Question 16.<br \/>\nBanker&#8217;s gain on a bill due after 6 months at 4% p.a. is 24. Find true discount and Banker&#8217;s discount,<br \/>\nAnswer:<br \/>\nGiven B.G = 24. T = 6 months, = \\(\\frac { 1 }{ 2 }\\) yr, r = 4% = 0.0<br \/>\nBG = TD + r<br \/>\n24 = TD \u00d7 \\(\\frac { 1 }{ 2 }\\) \u00d7 0.04<br \/>\n<img decoding=\"async\" loading=\"lazy\" class=\"alignnone size-full wp-image-65806\" src=\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2020\/01\/2nd-PUC-Basic-Maths-Previous-Year-Question-Paper-June-2017-4.png\" alt=\"2nd PUC Basic Maths Previous Year Question Paper June 2017 - 4\" width=\"213\" height=\"51\" \/><br \/>\nTD = 1200<br \/>\nBG = BD &#8211; TD \u21d2 BD = BG + TD = 24 + 1200 = 1224<\/p>\n<p>Question 17.<br \/>\nProve that \\(\\frac{\\sin 3 A}{1+2 \\cos 2 A}\\) = sin A.<br \/>\nAnswer:<br \/>\n<img decoding=\"async\" loading=\"lazy\" class=\"alignnone size-full wp-image-65805\" src=\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2020\/01\/2nd-PUC-Basic-Maths-Previous-Year-Question-Paper-June-2017-5.png\" alt=\"2nd PUC Basic Maths Previous Year Question Paper June 2017 - 5\" width=\"360\" height=\"50\" srcset=\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2020\/01\/2nd-PUC-Basic-Maths-Previous-Year-Question-Paper-June-2017-5.png 360w, https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2020\/01\/2nd-PUC-Basic-Maths-Previous-Year-Question-Paper-June-2017-5-300x42.png 300w\" sizes=\"(max-width: 360px) 100vw, 360px\" \/><br \/>\n= Sin A = R.H.S<\/p>\n<p>Question 18.<br \/>\nIf tan A = \\(\\frac { 5 }{ 6 }\\) tan B = \\(\\frac { 1 }{ 11 }\\) Show that A + B = \\(\\frac{\\pi}{4}\\)<br \/>\nAnswer:<br \/>\nConsider tan (A+B) = \\(\\frac{\\tan A+\\tan B}{1+\\tan A \\cdot \\tan B}\\)<br \/>\n<img decoding=\"async\" loading=\"lazy\" class=\"alignnone size-full wp-image-65804\" src=\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2020\/01\/2nd-PUC-Basic-Maths-Previous-Year-Question-Paper-June-2017-6.png\" alt=\"2nd PUC Basic Maths Previous Year Question Paper June 2017 - 6\" width=\"181\" height=\"88\" \/><br \/>\nTan (A + B) = 1 \u21d2 A + B = \\(\\frac{\\pi}{4}\\)<\/p>\n<p><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2019\/11\/KSEEB-Solutions.png\" alt=\"KSEEB Solutions\" width=\"172\" height=\"16\" \/><\/p>\n<p>Question 19.<br \/>\nFind the equation of the circle whose centre is at (4,-2) and passing through the origin (0,0)<br \/>\nAnswer:<br \/>\nC = (4,-2) &amp; r = \\(\\sqrt{4^{2}+(-2)^{2}}=\\sqrt{16+4}=\\sqrt{20}\\)<br \/>\nEquation of the circle (x + 4)<sup>2<\/sup> + (y + 2)<sup>2<\/sup> = (\u221a 20)<br \/>\n\u21d2 x<sup>2<\/sup> + 16 &#8211; 8x + y<sup>2<\/sup> + y + 4y &#8211; 20 = 0<br \/>\n\u21d2 x<sup>2<\/sup> + y<sup>2<\/sup> &#8211; 8x + 4y = 0<\/p>\n<p>Question 20.<br \/>\n<img decoding=\"async\" loading=\"lazy\" class=\"alignnone size-full wp-image-65803\" src=\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2020\/01\/2nd-PUC-Basic-Maths-Previous-Year-Question-Paper-June-2017-7.png\" alt=\"2nd PUC Basic Maths Previous Year Question Paper June 2017 - 7\" width=\"550\" height=\"61\" srcset=\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2020\/01\/2nd-PUC-Basic-Maths-Previous-Year-Question-Paper-June-2017-7.png 550w, https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2020\/01\/2nd-PUC-Basic-Maths-Previous-Year-Question-Paper-June-2017-7-300x33.png 300w\" sizes=\"(max-width: 550px) 100vw, 550px\" \/><br \/>\nAnswer:<br \/>\n\\(\\lim _{x \\rightarrow 0}\\) f(x) = \\(\\lim _{x \\rightarrow 0}\\) (1 + 3x)<sup>1\/3<\/sup> = e<sup>3<\/sup> = f(0)<br \/>\n\u2234 f(x) is continuous at x = 0<\/p>\n<p>Question 21.<br \/>\nIf siny = x sin(a+y), prove that \\(\\frac{d y}{d x}=\\frac{\\sin ^{2}(a+y)}{\\sin a}\\)<br \/>\nAnswer:<br \/>\n<img decoding=\"async\" loading=\"lazy\" class=\"alignnone size-full wp-image-65802\" src=\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2020\/01\/2nd-PUC-Basic-Maths-Previous-Year-Question-Paper-June-2017-8.png\" alt=\"2nd PUC Basic Maths Previous Year Question Paper June 2017 - 8\" width=\"496\" height=\"162\" srcset=\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2020\/01\/2nd-PUC-Basic-Maths-Previous-Year-Question-Paper-June-2017-8.png 496w, https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2020\/01\/2nd-PUC-Basic-Maths-Previous-Year-Question-Paper-June-2017-8-300x98.png 300w\" sizes=\"(max-width: 496px) 100vw, 496px\" \/><\/p>\n<p>Question 22.<br \/>\nIf s = at<sup>3<\/sup> + bt, find a and b given that when t = 3, velocity is 0 and the acceleration is 14 units.<br \/>\nAnswer:<br \/>\nIf s = at<sup>3<\/sup> + bt velocity = \\(\\frac{\\mathrm{ds}}{\\mathrm{dt}}\\) = 3at<sup>2<\/sup> + b = 0<br \/>\nAcceleration = \\(\\frac{\\mathrm{dv}}{\\mathrm{dt}}\\) = 6at = 14<br \/>\nAt t = 3sec, 6a \u00d7 3 = 14 \u21d2 a = \\(\\frac{14}{18}=\\frac{7}{9}\\)<br \/>\n<img decoding=\"async\" loading=\"lazy\" class=\"alignnone size-full wp-image-65801\" src=\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2020\/01\/2nd-PUC-Basic-Maths-Previous-Year-Question-Paper-June-2017-9.png\" alt=\"2nd PUC Basic Maths Previous Year Question Paper June 2017 - 9\" width=\"249\" height=\"51\" \/><\/p>\n<p>Question 23.<br \/>\nEvaluate \\(\\int \\frac{3^{x} \\log 3}{\\left(3^{x}+1\\right)} d x\\)<br \/>\nAnswer:<br \/>\n= log(3<sup>x<\/sup> + 1) + C \u2234 \\(\\int \\frac{f^{1}(x)}{f(x)} d x\\) = log(3<sup>x<\/sup> +1)<\/p>\n<p>Question 24.<br \/>\nEvaluate \\(\\int_{1}^{2} \\frac{1}{(2 x+3)} d x\\)<br \/>\nAnswer:<br \/>\n<img decoding=\"async\" loading=\"lazy\" class=\"alignnone size-full wp-image-65800\" src=\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2020\/01\/2nd-PUC-Basic-Maths-Previous-Year-Question-Paper-June-2017-10.png\" alt=\"2nd PUC Basic Maths Previous Year Question Paper June 2017 - 10\" width=\"529\" height=\"57\" srcset=\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2020\/01\/2nd-PUC-Basic-Maths-Previous-Year-Question-Paper-June-2017-10.png 529w, https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2020\/01\/2nd-PUC-Basic-Maths-Previous-Year-Question-Paper-June-2017-10-300x32.png 300w\" sizes=\"(max-width: 529px) 100vw, 529px\" \/><\/p>\n<p style=\"text-align: center;\"><span style=\"color: #0000ff;\">Part &#8211; C<\/span><\/p>\n<p><span style=\"color: #0000ff;\">Answer any ten questions: (10 \u00d7 3 = 30)<\/span><\/p>\n<p>Question 25.<br \/>\nIf 2A + B = \\(\\left[ \\begin{matrix} 3 &amp; -1 \\\\ -2 &amp; 5 \\end{matrix} \\right] \\) and A &#8211; 2B = \\(\\left[ \\begin{matrix} 4 &amp; 2 \\\\ -1 &amp; 5 \\end{matrix} \\right] \\) then find A and B.<\/p>\n<p>Question 26.<br \/>\nSolve \\(\\left| \\begin{matrix} 2+x &amp; 3 &amp; -4 \\\\ 2 &amp; 3+x &amp; -4 \\\\ 2 &amp; 3 &amp; -4+x \\end{matrix} \\right| =0\\)<br \/>\nAnswer:<br \/>\n<img decoding=\"async\" loading=\"lazy\" class=\"alignnone size-full wp-image-65856\" src=\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2020\/01\/2nd-PUC-Basic-Maths-Previous-Year-Question-Paper-June-2017-1-1.png\" alt=\"2nd PUC Basic Maths Previous Year Question Paper June 2017 - 1\" width=\"472\" height=\"322\" srcset=\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2020\/01\/2nd-PUC-Basic-Maths-Previous-Year-Question-Paper-June-2017-1-1.png 472w, https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2020\/01\/2nd-PUC-Basic-Maths-Previous-Year-Question-Paper-June-2017-1-1-300x205.png 300w\" sizes=\"(max-width: 472px) 100vw, 472px\" \/><\/p>\n<p><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2019\/11\/KSEEB-Solutions.png\" alt=\"KSEEB Solutions\" width=\"172\" height=\"16\" \/><\/p>\n<p>Question 27.<br \/>\nFind the number of permutations of the letters of the word &#8216;ENGINEERING&#8217; How many of these<br \/>\ni. Begin with E and end with E<br \/>\nii. Have all the 3E&#8217;s together?<br \/>\nAnswer:<br \/>\nTotal-11, N &#8211; 3, G &#8211; 2, 1 &#8211; 2, E = 3<br \/>\nTotal permutations \\(\\frac{11 !}{3 ! \\times 3 ! \\times 2 ! \\times 2 !}\\)=<br \/>\nI. Number of permutations which begin with E &amp; end with E is = \\(\\frac{9 !}{3 ! \\times 2 ! \\times 2 !}\\)<br \/>\nII. Number of Permutations in Which all 3E&#8217;s are together (Total 11 &#8211; 3 + 1 = 9&#8242;) = \\(\\frac{9 !}{3 ! \\times 2 ! \\times 2 !}\\)<\/p>\n<p>Question 28.<br \/>\nA die is thrown, If E is the event &#8220;the number appearing is a multiple of 3&#8221; and F be the event &#8220;the number appearing is even&#8221;. Then find whether E are Fare independent.<br \/>\nAnswer:<br \/>\nS = {1,2,3,4,5,6}<br \/>\nE = {3,6), F = {2,4,6}<br \/>\nN(S) = 6, n(E) = 2 n(F) = 3<br \/>\nP(E) = \\(\\frac{2}{5}=\\frac{1}{3}\\) P(E) = \\(\\frac{3}{6}=\\frac{1}{3}\\)<br \/>\nThey are not independent<\/p>\n<p>Question 29.<br \/>\nTwo tap can separately fill a tank in 12 minutes and 15 minutes respectively. The tank when full can be emptied by a drain pipe in 20 minutes. When the tank was empty all the three were opened simultaneously in what time will the tank be filled up?<br \/>\nAnswer:<br \/>\nA fills tank in 12 mins \u2192 In 1 min A will fill 1\/12 of tank.<br \/>\nB fills tank in 15 mins \u2192 In 1 min B will fill 1\/15 of tank.<br \/>\nLets say C can alone empty the tank in x mins \u2192 In 1 min C will empty 1\/x of tank.<br \/>\nAccording to question, if all taps are opened tank will fill in 20 mins.<br \/>\nIn each min 1\/20 tank is filled which is due to A, B and C. Now A and B fill the tank will C empty the tank.<\/p>\n<p>1\/12 + 1\/15 &#8211; 1\/x = 1\/20<\/p>\n<p>1\/12 + 1\/15 &#8211; 1\/20 = 1\/x<\/p>\n<p>(5 + 4 &#8211; 3)\/60 = 1\/x<\/p>\n<p>6\/60 = 1\/x<\/p>\n<p>x = 10<\/p>\n<p>Question 30.<br \/>\nBharath bought a shirt for \u20b9 336 including 12% sales tax and a neck tie for 110 including 10% sales tax. Find the printed price of shirt and neck tie together.<br \/>\nAnswer:<br \/>\nGiven S.P = \u20b9 336, S.P (neck tie) = 110\/<br \/>\nLet C.P of shirt = x, &amp; C.P of neck tie = y<br \/>\nHere, setting price of shirt = C.P + 12% of C.P<br \/>\n336 = x + \\(\\frac{12 x}{100}\\)<br \/>\n336 = \\(=\\frac{112 x}{100}\\) \u21d2 x = \\(\\frac{33600}{112}\\)<br \/>\nSelling price of neck tie = C.P + 10% of C.P<br \/>\n110 = y + \\(\\frac{10 y}{100}\\)<br \/>\n110 = \\(\\frac{10 y}{100}\\) \u21d2 \\(\\frac{1100}{11}\\) = 100<br \/>\n\u2234 Printed price of shirt &amp; neck tie = 300 + 100 = \u20b9 400<\/p>\n<p>Question 31.<br \/>\nA Banker pays \u20b9 4,250 on a bill of \u20b9 5,000. 146 days before the legally due date. Find the rate of discount charged by the Banker.<br \/>\nAnswer:<br \/>\nGiven F = \u20b9 5000, Discounted Value = \u20b9 4,520<br \/>\nt = 146 days = \\(\\frac{146}{365}\\) year,r = ?<br \/>\nBD = F &#8211; Discounted value = 5000 &#8211; 4520 = \u20b9 480<br \/>\nBD = F\/r \u21d2 480 = \\(\\frac{5000 \\times 146 \\times r}{365}\\)<br \/>\n\u2234 r = \\(\\frac{480 \\times 365}{5000 \\times 146}\\) = 0.24 \u00d7 100=24<\/p>\n<p>Question 32.<br \/>\nRakshith decides to invest in TCS shares which are selling at 2020 per share. How many money is required to purchase 10 shares if the brokerage is 0.5%?<br \/>\nAnswer:<br \/>\nSelling price of 10 shares at 2020 per share = 20,200<br \/>\n0.5 Brokerage = \\(\\frac{0.5}{100}\\) \u00d7 20,200 = \u20b9 101<br \/>\n100 Amount required to purchase = 20,200 + 101 = 20,301<\/p>\n<p>Question 33.<br \/>\nFind the equation of the parabola if its focus is (4,0) and directrix is x = -4. Also find the equation of the tangent at the vertex.<br \/>\nAnswer:<br \/>\nGiven Directrix As x = -4 &amp; Focus = (4,0) \u21d2 a = 4<br \/>\nThe standard equation is y<sup>2<\/sup> = 4ax \u21d2 y<sup>2<\/sup> = 4 \u00d7 4 x X = 16x<br \/>\n\u2234 Equation of the parabola = 16x<\/p>\n<p><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2019\/11\/KSEEB-Solutions.png\" alt=\"KSEEB Solutions\" width=\"172\" height=\"16\" \/><\/p>\n<p>Question 34.<br \/>\nIf x = a cos<sup>4<\/sup>\u03b8, y = a sin<sup>4<\/sup>6 show that \\(\\frac{d y}{d x}-\\tan ^{2} \\theta\\)<br \/>\nAnswer:<br \/>\nx = a cost<sup>4<\/sup>\u03b8 &amp; y = a sin<sup>4<\/sup><br \/>\nDifferentiate both w.r.t \u03b8 we get<br \/>\n\\(\\frac{d y}{d \\theta}\\) = a.4cos<sup>3<\/sup>\u03b8.sin \u03b8, \\(\\frac{d y}{d \\theta}\\) = 2.4 sin<sup>3<\/sup>\u03b8. cos\u03b8<br \/>\n<img decoding=\"async\" loading=\"lazy\" class=\"alignnone size-full wp-image-65799\" src=\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2020\/01\/2nd-PUC-Basic-Maths-Previous-Year-Question-Paper-June-2017-11.png\" alt=\"2nd PUC Basic Maths Previous Year Question Paper June 2017 - 11\" width=\"319\" height=\"81\" srcset=\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2020\/01\/2nd-PUC-Basic-Maths-Previous-Year-Question-Paper-June-2017-11.png 319w, https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2020\/01\/2nd-PUC-Basic-Maths-Previous-Year-Question-Paper-June-2017-11-300x76.png 300w\" sizes=\"(max-width: 319px) 100vw, 319px\" \/><\/p>\n<p>Question 35.<br \/>\nThe side of an equilateral triangle is increasing at the rate of \u221a3 cm\/s. Find the rate at which its area is including when its side is 200 cms.<br \/>\nAnswer:<br \/>\nGiven \\(\\frac{\\mathrm{dx}}{\\mathrm{dt}}\\) = \\(\\sqrt{3}\\) cm\/sec., x = 2 meters, \\(\\frac{\\mathrm{dA}}{\\mathrm{dt}}\\) = ?<br \/>\nArea of equilateral \u0394<sup>le<\/sup> = A = \\(\\frac{\\sqrt{3}}{4}\\) x<sup>2<\/sup><br \/>\n\\(\\frac{d A}{d t}=\\frac{\\sqrt{3}}{4} \\cdot 2 x \\frac{d x}{d t}\\)<br \/>\n= \\(\\frac{\\sqrt{3}}{4}\\) . 2. 200. \\(\\sqrt{3}\\) = 300cm<sup>2<\/sup> \/ sec.<\/p>\n<p>Question 36.<br \/>\nFind two positive numbers whose sum is 14 and the sum of whose square is minimum.<br \/>\nAnswer:<br \/>\nLet the two numbers be x and y<br \/>\nGiven x + y = 14 &amp; S = x<sup>2<\/sup> + y<sup>2<\/sup> where y = 14 &#8211; x<br \/>\n\u2234 S = x<sup>2<\/sup> + (14 &#8211; x)<sup>2<\/sup> = x<sup>2<\/sup> + 14<sup>2<\/sup> + x<sup>2<\/sup> &#8211; 28x = 2x<sup>2<\/sup> &#8211; 28x + 14<sup>2<\/sup><br \/>\n\\(\\frac{d s}{d x}\\) = 4x &#8211; 28 \u2192 (1) \\(\\frac{d s}{d x}\\) = 0 \u21d2 4x &#8211; 28 = 0 \u21d2 x = 7<br \/>\n\\(\\frac{\\mathrm{d}^{2} \\mathrm{s}}{\\mathrm{d} \\mathrm{x}^{2}}\\) = 4 &gt; 0, sum is minimum.<br \/>\n\u2234 y = 14 &#8211; x = 14 &#8211; 7 = 7<br \/>\n\u2234 the two positive number are 7 &amp; 7.<\/p>\n<p>Question 37.<br \/>\nEvaluate \\(\\int \\frac{1+\\cos 2 x}{1-\\cos 2 x} d x\\)<br \/>\nAnswer:<br \/>\n= \\(=\\int \\frac{2 \\cos ^{2} x}{2 \\cos } d x\\) = \u222bcos<sup>2<\/sup> dx = \u222b(cosec<sup>2<\/sup>x &#8211; 1)dx = &#8211; cos x &#8211; x + c<\/p>\n<p>Question 38.<br \/>\nEvaluate \\(\\int_{0}^{1}(6 x+1) \\sqrt{3 x^{2}+x+5 d x}\\)<br \/>\nAnswer:<br \/>\nPut 3x<sup>2<\/sup> + x + 5 = t.<br \/>\n(6x + 1) dx = dt<br \/>\nWhen x = 0 then t = 5<br \/>\nWhen x = 1 then t = 9<br \/>\n\\(\\left.\\int_{0}^{1}(6 x+1)(\\sqrt{3 x^{2}+x+5}) d x=\\int_{0}^{1} \\sqrt{t} d t=\\frac{2}{3} t^{\\frac{3}{2}}\\right]_{5}^{7}\\)<br \/>\n\\(=\\frac{2}{3}\\left(9^{\\frac{3}{2}}-5^{\\frac{3}{2}}\\right)=\\frac{2}{3}(27-5 \\sqrt{5})\\)<\/p>\n<p><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2019\/11\/KSEEB-Solutions.png\" alt=\"KSEEB Solutions\" width=\"172\" height=\"16\" \/><\/p>\n<p style=\"text-align: center;\"><span style=\"color: #0000ff;\">Part-D<\/span><\/p>\n<p><span style=\"color: #0000ff;\">Answer any six questions: (6 \u00d7 5 = 30)<\/span><\/p>\n<p>Question 39.<br \/>\nSimplify (\u221a2 + 1)<sup>6<\/sup> &#8211; (\u221a2 &#8211; 1)<sup>6<\/sup> using Binomial theorem.<br \/>\nAnswer:<br \/>\n<img decoding=\"async\" loading=\"lazy\" class=\"alignnone size-full wp-image-65857\" src=\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2020\/01\/2nd-PUC-Basic-Maths-Previous-Year-Question-Paper-June-2017-2-1.png\" alt=\"2nd PUC Basic Maths Previous Year Question Paper June 2017 - 2\" width=\"661\" height=\"172\" srcset=\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2020\/01\/2nd-PUC-Basic-Maths-Previous-Year-Question-Paper-June-2017-2-1.png 661w, https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2020\/01\/2nd-PUC-Basic-Maths-Previous-Year-Question-Paper-June-2017-2-1-300x78.png 300w\" sizes=\"(max-width: 661px) 100vw, 661px\" \/><\/p>\n<p>Question 40.<br \/>\nResolve \\(\\frac{2 x^{2}-4 x+1}{(x-2)(x-3)^{2}}\\) into partial fractions.<br \/>\nAnswer:<br \/>\n2x<sup>2<\/sup> &#8211; 4x + 1 = A[x &#8211; 3)<sup>2<\/sup> + B(x &#8211; 2) (x-3) + C (x &#8211; 2)<br \/>\nPut x = 2, 8-8 + 1 = A(-1)<sup>2<\/sup> = A = 1<br \/>\nPut x = 3, 18 \u2013 12 + 1 = A(0) + B(0) + C(3 &#8211; 2) = 7 = C<br \/>\nComparing coefficient of x<sup>2<\/sup> = 2 = A + B = B = 2-A = 2 -1 =1<br \/>\n\u2234 \\(\\frac{2 x^{2}-4 x+1}{(x-2)(x-3)^{2}}=\\frac{1}{x-2}+\\frac{1}{x-3}+\\frac{7}{(x-3)^{2}}\\)<\/p>\n<p>Question 41.<br \/>\nProve that (p \u2228q)\u2227(~P\u2227q~) is a contradiction.<br \/>\nAnswer:<br \/>\n<img decoding=\"async\" loading=\"lazy\" class=\"alignnone size-full wp-image-65798\" src=\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2020\/01\/2nd-PUC-Basic-Maths-Previous-Year-Question-Paper-June-2017-12.png\" alt=\"2nd PUC Basic Maths Previous Year Question Paper June 2017 - 12\" width=\"520\" height=\"156\" srcset=\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2020\/01\/2nd-PUC-Basic-Maths-Previous-Year-Question-Paper-June-2017-12.png 520w, https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2020\/01\/2nd-PUC-Basic-Maths-Previous-Year-Question-Paper-June-2017-12-300x90.png 300w\" sizes=\"(max-width: 520px) 100vw, 520px\" \/><\/p>\n<p>Question 42.<br \/>\nDivide \u20b9 1,647 into three parts such that \\(\\frac { 3 }{ 7 }\\) th of the first, \\(\\frac { 2 }{ 3 }\\) rd of second and \\(\\frac { 4 }{ 5 }\\) th of the third are equal..<br \/>\nAnswer:<br \/>\nLet the 3 parts be A,B &amp; C<br \/>\nGiven \\(\\frac{3}{7} A=\\frac{2}{3} B \\Rightarrow \\frac{A}{B}=\\frac{2}{7} \\times \\frac{7}{3}=\\frac{14}{9}\\)<br \/>\n\u2234 A : B = 14 : 9<br \/>\nAlso given \\(\\frac{2}{3} B=\\frac{4}{5} \\Rightarrow \\frac{B}{C}=\\frac{4}{5} \\times \\frac{3}{2}=\\frac{12}{10}=\\frac{6}{5}\\)<br \/>\n\u2234 B : C = 6 : 5<br \/>\nMultiple (1) by 2 and (2) by 3 we get A : B = 28 : 18 and B : C = 18 : 5<br \/>\n\u2234 A : B : C = 28 : 18 : 15<br \/>\n\u2234 28x + 18x + 15x = 1647<br \/>\n\u21d2 61x = 1647 \u21d2 x = \\(\\frac{1647}{61}\\) = 27<br \/>\n\u2234 A receives = 28 \u00d7 27 = \u20b9 3 756<br \/>\nB receives = 18 \u00d7 27 = \u20b9 486<br \/>\nCreceives = 15 \u00d7 27 = \u20b9 405<\/p>\n<p><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2019\/11\/KSEEB-Solutions.png\" alt=\"KSEEB Solutions\" width=\"172\" height=\"16\" \/><\/p>\n<p>Question 43.<br \/>\nA company requires 100 hours to produce the first 10 units at 15\/hour. The learning effect is 80%. Find the total labour cost to produce a total of 160 units.<br \/>\nAnswer:<br \/>\nGiven<br \/>\na = 100 hrs. 10 units = 1 lot<br \/>\n\u2234 160 units = 16 lots<br \/>\n<img decoding=\"async\" loading=\"lazy\" class=\"alignnone size-full wp-image-65858\" src=\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2020\/01\/2nd-PUC-Basic-Maths-Previous-Year-Question-Paper-June-2017-3-1.png\" alt=\"2nd PUC Basic Maths Previous Year Question Paper June 2017 - 3\" width=\"524\" height=\"60\" srcset=\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2020\/01\/2nd-PUC-Basic-Maths-Previous-Year-Question-Paper-June-2017-3-1.png 524w, https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2020\/01\/2nd-PUC-Basic-Maths-Previous-Year-Question-Paper-June-2017-3-1-300x34.png 300w\" sizes=\"(max-width: 524px) 100vw, 524px\" \/><br \/>\ny = ax<sup>b<\/sup><br \/>\nlog y = log a + b log x<br \/>\n= log 100 \u2013 0.3219 log 16<br \/>\n= 2 &#8211; 0.3219 1.2041 = 2 &#8211; 0.3875 = 1.625<br \/>\ny = A(1.6125) = 40.98 hours\/unit<br \/>\n\u2234 Total time required for 160 units or 16 lots = 40.98 16 = 655.68 hrs.<br \/>\n\u2234 Total cost to produce 16 units at \u20b9 15\/hr. = 655.68 \u00d7 15 = \u20b9 9835.20<br \/>\nTable method<br \/>\n<img decoding=\"async\" loading=\"lazy\" class=\"alignnone size-full wp-image-65859\" src=\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2020\/01\/2nd-PUC-Basic-Maths-Previous-Year-Question-Paper-June-2017-4-1.png\" alt=\"2nd PUC Basic Maths Previous Year Question Paper June 2017 - 4\" width=\"568\" height=\"176\" srcset=\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2020\/01\/2nd-PUC-Basic-Maths-Previous-Year-Question-Paper-June-2017-4-1.png 568w, https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2020\/01\/2nd-PUC-Basic-Maths-Previous-Year-Question-Paper-June-2017-4-1-300x93.png 300w\" sizes=\"(max-width: 568px) 100vw, 568px\" \/><br \/>\nTotal time for 16 lots = 655.36 hrs.<br \/>\n\u2234 Total cost to produce 16 lots at \u20b9 15\/hr. = 655.36 \u00d7 15 = \u20b9 19830.4.<\/p>\n<p>Question 44.<br \/>\nSolve the following L.P.P. graphically maximize z = 60x + 15y subject to the constraints x + y \u2264 50, 3x + y \u2264 90, x \u2265 0, y \u2265 0.<br \/>\nAnswer:<br \/>\nConsider line x + y = 50<br \/>\nPut x = 0, y = 50, (0,50)<br \/>\nPut y = 0, x = 50 (50,0)<br \/>\nConsider the line 3x + y =- 90<br \/>\nPut x = 0, y = 90 (0,90)<br \/>\nPut y = 0, x = 30 (30,0)<br \/>\nPlot the above lines on the graph we get OABC is the solution region<br \/>\n<img decoding=\"async\" loading=\"lazy\" class=\"alignnone size-full wp-image-65860\" src=\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2020\/01\/2nd-PUC-Basic-Maths-Previous-Year-Question-Paper-June-2017-5-1.png\" alt=\"2nd PUC Basic Maths Previous Year Question Paper June 2017 - 5\" width=\"307\" height=\"151\" srcset=\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2020\/01\/2nd-PUC-Basic-Maths-Previous-Year-Question-Paper-June-2017-5-1.png 307w, https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2020\/01\/2nd-PUC-Basic-Maths-Previous-Year-Question-Paper-June-2017-5-1-300x148.png 300w\" sizes=\"(max-width: 307px) 100vw, 307px\" \/><br \/>\n\u2265 is maximum at c (30,0)<br \/>\nz<sub>max<\/sub> = 1800<br \/>\n<img decoding=\"async\" loading=\"lazy\" class=\"alignnone size-full wp-image-65861\" src=\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2020\/01\/2nd-PUC-Basic-Maths-Previous-Year-Question-Paper-June-2017-6-1.png\" alt=\"2nd PUC Basic Maths Previous Year Question Paper June 2017 - 6\" width=\"282\" height=\"420\" srcset=\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2020\/01\/2nd-PUC-Basic-Maths-Previous-Year-Question-Paper-June-2017-6-1.png 282w, https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2020\/01\/2nd-PUC-Basic-Maths-Previous-Year-Question-Paper-June-2017-6-1-201x300.png 201w\" sizes=\"(max-width: 282px) 100vw, 282px\" \/><\/p>\n<p><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2019\/11\/KSEEB-Solutions.png\" alt=\"KSEEB Solutions\" width=\"172\" height=\"16\" \/><\/p>\n<p>Question 45.<br \/>\nProve that sin 20\u00b0 sin 40\u00b0 sin 60\u00b0 sin 80\u00b0= \\(\\frac { 3 }{ 16 }\\)<br \/>\nAnswer:<br \/>\nL.H.S=sin 60\u00b0 (sin 40\u00b0 sin 20\u00b0 )sin 80\u00b0<br \/>\n<img decoding=\"async\" loading=\"lazy\" class=\"alignnone size-full wp-image-65862\" src=\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2020\/01\/2nd-PUC-Basic-Maths-Previous-Year-Question-Paper-June-2017-7-1.png\" alt=\"2nd PUC Basic Maths Previous Year Question Paper June 2017 - 7\" width=\"436\" height=\"265\" srcset=\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2020\/01\/2nd-PUC-Basic-Maths-Previous-Year-Question-Paper-June-2017-7-1.png 436w, https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2020\/01\/2nd-PUC-Basic-Maths-Previous-Year-Question-Paper-June-2017-7-1-300x182.png 300w\" sizes=\"(max-width: 436px) 100vw, 436px\" \/><\/p>\n<p>Question 46.<br \/>\nFind the equation of the circle passing through the points (5,1) and (3,4) and has its centre on the x &#8211; axis.<br \/>\nAnswer:<br \/>\nLet the required equation of the circle is x<sup>2<\/sup> + y<sup>2<\/sup> + 2gx + 2fy + c = 0<br \/>\nIt passes through (5, 1) &amp; (3, 4)<br \/>\n(5, 1) (5)<sup>2<\/sup> + (1)<sup>2<\/sup> + 2g(5) + 2f (1) +C = 0<br \/>\n10g + 2f + C + 26 = 0 &#8230;..(1)<br \/>\n(3,4) 3<sup>2<\/sup> + 4<sup>2<\/sup> + 2g (3) + 2f (4) +c= 0<br \/>\n68 + 8f + C + 25 = 0 &#8230;..(2)<br \/>\n&amp; the centre (- g, &#8211; f) lies on x-axis \u21d2 f = 0 &#8230;.(3)<br \/>\nSolving 1 &amp; 2 we get<br \/>\n<img decoding=\"async\" loading=\"lazy\" class=\"alignnone size-full wp-image-65864\" src=\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2020\/01\/2nd-PUC-Basic-Maths-Previous-Year-Question-Paper-June-2017-9-1.png\" alt=\"2nd PUC Basic Maths Previous Year Question Paper June 2017 - 9\" width=\"584\" height=\"236\" srcset=\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2020\/01\/2nd-PUC-Basic-Maths-Previous-Year-Question-Paper-June-2017-9-1.png 584w, https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2020\/01\/2nd-PUC-Basic-Maths-Previous-Year-Question-Paper-June-2017-9-1-300x121.png 300w\" sizes=\"(max-width: 584px) 100vw, 584px\" \/><br \/>\n\u21d2 2x<sup>2<\/sup> + 2y<sup>2<\/sup> &#8211; x &#8211; 47 = 0.<\/p>\n<p>Question 47.<br \/>\nIf y = \\((x+\\sqrt{x^{2}+1})\\) show that(x<sup>2<\/sup> &#8211; 1}y<sub>2<\/sub> + xy<sub>1<\/sub> &#8211; y = 0.<br \/>\nAnswer:<br \/>\ny = x + \\(\\sqrt{x^{2} &#8211; 1}\\) differentiate w.r.t x<br \/>\n<img decoding=\"async\" loading=\"lazy\" class=\"alignnone size-full wp-image-65865\" src=\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2020\/01\/2nd-PUC-Basic-Maths-Previous-Year-Question-Paper-June-2017-10-1.png\" alt=\"2nd PUC Basic Maths Previous Year Question Paper June 2017 - 10\" width=\"314\" height=\"56\" srcset=\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2020\/01\/2nd-PUC-Basic-Maths-Previous-Year-Question-Paper-June-2017-10-1.png 314w, https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2020\/01\/2nd-PUC-Basic-Maths-Previous-Year-Question-Paper-June-2017-10-1-300x54.png 300w\" sizes=\"(max-width: 314px) 100vw, 314px\" \/><br \/>\n\u2234 (x<sup>2<\/sup> &#8211; 1) y<sub>1<\/sub><sup>2<\/sup> = y<sup>2<\/sup> Differentiate agin w.r.t x<br \/>\n(x<sup>2<\/sup> &#8211; 1)2y<sub>1<\/sub>y<sub>2<\/sub> + y<sub>1<\/sub><sup>2<\/sup> (2x) = 2y .y<sub>1<\/sub> (\u00f7 2y<sub>1<\/sub> we get)<br \/>\n(x<sup>2<\/sup> &#8211; 1)y<sub>2<\/sub> + xy<sub>1<\/sub> &#8211; y = 0 Hence proved<\/p>\n<p>Question 48.<br \/>\nFind the area of the region between the parabolas y<sup>2<\/sup> = 4x and x<sup>2<\/sup> = 4y.<br \/>\nAnswer:<br \/>\n<img decoding=\"async\" loading=\"lazy\" class=\"alignnone size-full wp-image-65795\" src=\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2020\/01\/2nd-PUC-Basic-Maths-Previous-Year-Question-Paper-June-2017-22.png\" alt=\"2nd PUC Basic Maths Previous Year Question Paper June 2017 - 22\" width=\"292\" height=\"391\" srcset=\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2020\/01\/2nd-PUC-Basic-Maths-Previous-Year-Question-Paper-June-2017-22.png 292w, https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2020\/01\/2nd-PUC-Basic-Maths-Previous-Year-Question-Paper-June-2017-22-224x300.png 224w\" sizes=\"(max-width: 292px) 100vw, 292px\" \/><br \/>\n<img decoding=\"async\" loading=\"lazy\" class=\"alignnone size-full wp-image-65794\" src=\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2020\/01\/2nd-PUC-Basic-Maths-Previous-Year-Question-Paper-June-2017-23.png\" alt=\"2nd PUC Basic Maths Previous Year Question Paper June 2017 - 23\" width=\"388\" height=\"720\" srcset=\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2020\/01\/2nd-PUC-Basic-Maths-Previous-Year-Question-Paper-June-2017-23.png 388w, https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2020\/01\/2nd-PUC-Basic-Maths-Previous-Year-Question-Paper-June-2017-23-162x300.png 162w\" sizes=\"(max-width: 388px) 100vw, 388px\" \/><\/p>\n<p><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2019\/11\/KSEEB-Solutions.png\" alt=\"KSEEB Solutions\" width=\"172\" height=\"16\" \/><\/p>\n<p style=\"text-align: center;\"><span style=\"color: #0000ff;\">Part &#8211; E<\/span><\/p>\n<p><span style=\"color: #0000ff;\">Answer any one question: (1 \u00d7 10 = 10)<\/span><\/p>\n<p>Question 49.<br \/>\n(a) A school wants to award its students for the values of punctuality, good behavior and hardwork with a total cash award \u20b9 6,000. Three times the award money for hardwork added to that given for punctuality amounts to \u20b9 11,000. The award money for punctuality and hardwork together is double the one given for good behavior.<\/p>\n<p>Represent the above situation algebraically and also find the award money for each value using matrix method.<\/p>\n<p>(b) The total revenue function is given by R = 400x &#8211; 2x<sup>2<\/sup> and the total cost function given by C = 2x<sup>2<\/sup> + 40x + 4000 find<br \/>\ni. The marginal revenue and marginal cost function.<br \/>\nii. The average revenue and average cost.<br \/>\nAnswer:<br \/>\n(a) Let the values of punctuality, Good Behaviour and Hard work be denoted by x,y &amp; z receptively<br \/>\nGiven x + y +z = 6000 &#8212;&#8212;&#8212;1<br \/>\nX + 3z = 11000 &#8212;&#8212;&#8212;-2<br \/>\nX + z = 2y \u21d2 x &#8211; 2y + z = 0 &#8212;&#8212;- 3<br \/>\nSolve these equations by matrix method<br \/>\n<img decoding=\"async\" loading=\"lazy\" class=\"alignnone size-full wp-image-65796\" src=\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2020\/01\/2nd-PUC-Basic-Maths-Previous-Year-Question-Paper-June-2017-14.png\" alt=\"2nd PUC Basic Maths Previous Year Question Paper June 2017 - 14\" width=\"419\" height=\"79\" srcset=\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2020\/01\/2nd-PUC-Basic-Maths-Previous-Year-Question-Paper-June-2017-14.png 419w, https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2020\/01\/2nd-PUC-Basic-Maths-Previous-Year-Question-Paper-June-2017-14-300x57.png 300w\" sizes=\"(max-width: 419px) 100vw, 419px\" \/><br \/>\nThus x = \u20b9 500, y = \u20b9 2000 &amp; z = \u20b9 3500<\/p>\n<p>(b) (i) MR = \\(\\frac{d}{d x}\\) (TR) = \\(\\frac{d}{d x}\\) (400x &#8211; 2x<sup>2<\/sup>) = 400 &#8211; 4x<br \/>\nMC = \\(\\frac{d}{d x}\\) (TC) = \\(\\frac{d}{d x}\\)(2x<sup>2<\/sup> + 40x + 400 )= 4x &#8211; 40<\/p>\n<p>(ii) \\(\\mathrm{AR}=\\frac{\\mathrm{TR}}{\\mathrm{x}}=\\frac{400 \\mathrm{x}-2 \\mathrm{x}^{2}}{\\mathrm{x}}=400-2 \\mathrm{x}\\)<br \/>\n\\(A C=\\frac{T R}{x}=\\frac{2 x^{2}+40 x+4000}{x}=2 x+40+4000\\)<\/p>\n<p>Question 50.<br \/>\n(a) Prove that if \\(\\lim _{x \\rightarrow a}\\left(\\frac{x^{n}-a^{n}}{x-a}\\right)\\) = na<sup>n-1<\/sup> for all values of&#8217;n&#8217;.<br \/>\nAnswer:<\/p>\n<p>Case 1: Let n be a positive integer.<br \/>\nx<sup>n<\/sup> &#8211; a<sup>n<\/sup> = (x-a)(x<sup>n-1<\/sup> + x<sup>n-2<\/sup>. a + x<sup>n-3<\/sup>. a<sup>2<\/sup> + &#8212;&#8212;&#8212;- + 2n-1)<br \/>\n<img decoding=\"async\" loading=\"lazy\" class=\"alignnone size-full wp-image-65868\" src=\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2020\/01\/2nd-PUC-Basic-Maths-Previous-Year-Question-Paper-June-2017-13-1.png\" alt=\"2nd PUC Basic Maths Previous Year Question Paper June 2017 - 13\" width=\"513\" height=\"311\" srcset=\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2020\/01\/2nd-PUC-Basic-Maths-Previous-Year-Question-Paper-June-2017-13-1.png 513w, https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2020\/01\/2nd-PUC-Basic-Maths-Previous-Year-Question-Paper-June-2017-13-1-300x182.png 300w\" sizes=\"(max-width: 513px) 100vw, 513px\" \/><br \/>\n<img decoding=\"async\" loading=\"lazy\" class=\"alignnone size-full wp-image-65867\" src=\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2020\/01\/2nd-PUC-Basic-Maths-Previous-Year-Question-Paper-June-2017-12-1.png\" alt=\"2nd PUC Basic Maths Previous Year Question Paper June 2017 - 12\" width=\"208\" height=\"59\" \/><\/p>\n<p>Case 2: Let n be a negative integer<br \/>\nput n = -m, m &gt; 0<br \/>\n<img decoding=\"async\" loading=\"lazy\" class=\"alignnone size-full wp-image-65868\" src=\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2020\/01\/2nd-PUC-Basic-Maths-Previous-Year-Question-Paper-June-2017-13-1.png\" alt=\"2nd PUC Basic Maths Previous Year Question Paper June 2017 - 13\" width=\"513\" height=\"311\" srcset=\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2020\/01\/2nd-PUC-Basic-Maths-Previous-Year-Question-Paper-June-2017-13-1.png 513w, https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2020\/01\/2nd-PUC-Basic-Maths-Previous-Year-Question-Paper-June-2017-13-1-300x182.png 300w\" sizes=\"(max-width: 513px) 100vw, 513px\" \/><\/p>\n<p><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2019\/11\/KSEEB-Solutions.png\" alt=\"KSEEB Solutions\" width=\"172\" height=\"16\" \/><\/p>\n<p>Case 3: Let n = p\/q where p and q are integers and q \u2260 0<br \/>\n<img decoding=\"async\" loading=\"lazy\" class=\"alignnone size-full wp-image-65869\" src=\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2020\/01\/2nd-PUC-Basic-Maths-Previous-Year-Question-Paper-June-2017-14-1.png\" alt=\"2nd PUC Basic Maths Previous Year Question Paper June 2017 - 14\" width=\"650\" height=\"442\" srcset=\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2020\/01\/2nd-PUC-Basic-Maths-Previous-Year-Question-Paper-June-2017-14-1.png 650w, https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2020\/01\/2nd-PUC-Basic-Maths-Previous-Year-Question-Paper-June-2017-14-1-300x204.png 300w\" sizes=\"(max-width: 650px) 100vw, 650px\" \/><br \/>\nHence it proved for al the values of n<\/p>\n<p>(b) A person is at the top of a tower 75 feet high. From there he observes a vertical pole and finds the angles of depressions of top and bottom of the pole which are 30\u00b0 and 60\u00b0 respectively. Find the height of the pole.<br \/>\nAnswer:<br \/>\n<img decoding=\"async\" loading=\"lazy\" class=\"alignnone size-full wp-image-65870\" src=\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2020\/01\/2nd-PUC-Basic-Maths-Previous-Year-Question-Paper-June-2017-15.png\" alt=\"2nd PUC Basic Maths Previous Year Question Paper June 2017 - 15\" width=\"278\" height=\"194\" \/><br \/>\nwidth=&#8221;278&#8243; height=&#8221;194&#8243; \/&gt;<br \/>\nLet AB = Tower, CD = Pole = h<br \/>\nFrom triangle ABC we get<br \/>\ntan 60\u00b0 = \\(\\frac{A B}{A C} \\Rightarrow \\frac{75}{A C}=\\sqrt{3}\\)<br \/>\n\u21d2 AC = \\(\\frac{75}{\\sqrt{3}} &#8230;.. (1)\\) &#8230;. (1)<\/p>\n<p>From triangle BDE<br \/>\n<img decoding=\"async\" loading=\"lazy\" class=\"alignnone size-full wp-image-65871\" src=\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2020\/01\/2nd-PUC-Basic-Maths-Previous-Year-Question-Paper-June-2017-16.png\" alt=\"2nd PUC Basic Maths Previous Year Question Paper June 2017 - 16\" width=\"288\" height=\"51\" \/><br \/>\n\u21d2 AC = \\(\\sqrt{3}(75-h)\\)<\/p>\n<p>From 1 and 2 we get<br \/>\n\\(\\frac{75}{\\sqrt{3}}=\\sqrt{3}(75-h) \\Rightarrow 75=(\\sqrt{3})^{2}(75-h)\\)<br \/>\n75 = 225 &#8211; 3h 3h \u21d2 150 \u21d2 h = 50<br \/>\n\u2234 Height of the pole -50ft<\/p>\n","protected":false},"excerpt":{"rendered":"<p>Students can Download 2nd PUC Basic Maths Previous Year Question Paper June 2017, 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 Basic Maths Previous Year Question Paper June 2017 Time: 3.15 hours Max Marks: &hellip;<\/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\/21789"}],"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=21789"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/21789\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=21789"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=21789"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=21789"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}