{"id":172797,"date":"2022-03-14T10:33:23","date_gmt":"2022-03-14T05:03:23","guid":{"rendered":"https:\/\/www.cbselabs.com\/?p=172797"},"modified":"2022-03-14T10:33:23","modified_gmt":"2022-03-14T05:03:23","slug":"cbse-sample-papers-for-class-12-maths-term-2-set-4","status":"publish","type":"post","link":"https:\/\/www.cbselabs.com\/cbse-sample-papers-for-class-12-maths-term-2-set-4\/","title":{"rendered":"CBSE Sample Papers for Class 12 Maths Term 2 Set 4 with Solutions"},"content":{"rendered":"<p>Students can access the <a href=\"https:\/\/www.cbselabs.com\/cbse-sample-papers-for-class-12-maths\/\">CBSE Sample Papers for Class 12 Maths with Solutions<\/a> and marking scheme Term 2 Set 4 will help students in understanding the difficulty level of the exam.<\/p>\n<h2>CBSE Sample Papers for Class 12 Maths Term 2 Set 4 with Solutions<\/h2>\n<p>Maximum Marks : 40<br \/>\nTime : 2 Hours<\/p>\n<p><span style=\"color: #0000ff;\">General Instructions:<\/span><\/p>\n<ul>\n<li>This question paper contains three sections &#8211; A, B and C. Each part is compulsory.<\/li>\n<li>Section &#8211; A has 6 short answer type (SA1) questions of 2 marks each.<\/li>\n<li>Section &#8211; B has 4 short answer type (SA2) questions of 3 marks each.<\/li>\n<li>Section &#8211; C has 4 long answer type questions (LA) of 4 marks each.<\/li>\n<li>There is an internal choice in some of the questions.<\/li>\n<li>Q14 is a case-based problem having 2 sub parts of 2 marks each.<\/li>\n<\/ul>\n<p style=\"text-align: center;\"><span style=\"color: #0000ff;\">Section &#8211; A<\/span><\/p>\n<p>Question 1.<br \/>\nDetermine order and degree of given differential equation:<br \/>\n\\(\\left(\\frac{d \\mathrm{~S}}{d t}\\right)^{4}\\) + 3S\\(\\frac{d^{2} S}{d t^{2}}\\) = 0<br \/>\nAnswer:<br \/>\nThe given differential equation<br \/>\n\\(\\left(\\frac{d \\mathrm{~S}}{d t}\\right)^{4}\\) + 3S\\(\\frac{d^{2} S}{d t^{2}}\\) = 0<\/p>\n<p>The highest order derivative presents in the given differential equation is \\(\\frac{d^{2} S}{d x^{2}}\\) Therefore, its order is two. It is a polynomial equation in \\(\\frac{d^{2} S}{d t^{2}}\\) and \\(\\frac{d \\mathrm{~S}}{d t}\\). The power raised to \\(\\frac{d^{2} S}{d t^{2}}\\) is 1. Hence, its degree is one.<\/p>\n<p>Question 2.<br \/>\nFind the coordinates of the foot of the perpendicular from 0(0, 0, 0) to the plane 3x &#8211; 4y + 5z &#8211; 10 = 0.<br \/>\nAnswer:<br \/>\nLet the coordinates of foot of perpendicular be V(x<sub>1<\/sub>, y<sub>1<\/sub>, z<sub>1<\/sub>) from point 0(0, 0, 0). Equation of plane in normal from is given by<br \/>\n\\(\\frac{3}{\\sqrt{50}}\\)x &#8211; \\(\\frac{4}{\\sqrt{50}}\\)y + \\(\\frac{5}{\\sqrt{50}}\\)z = \u221a2<br \/>\n[Dividing through by \\(\\sqrt{3^{2}+(-4)^{2}+5^{2}}=\\sqrt{50}\\)]<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-172844\" src=\"https:\/\/i2.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2022\/03\/CBSE-Sample-Papers-for-Class-12-Maths-Term-2-Set-4-with-Solutions-1.png?resize=241%2C196&#038;ssl=1\" alt=\"CBSE Sample Papers for Class 12 Maths Term 2 Set 4 with Solutions 1\" width=\"241\" height=\"196\" data-recalc-dims=\"1\" \/><\/p>\n<p>The direction cosines of the normal drawn from the origin to the given plane are<br \/>\nl = \\(\\frac{3}{5 \\sqrt{2}}\\)<br \/>\nm = \\(\\frac{-4}{5 \\sqrt{2}}\\)<br \/>\nn = \\(\\frac{5}{5 \\sqrt{2}}\\)<br \/>\nas OP = \u221a2<\/p>\n<p>So coordinate of foot of perpendicular P(\\(\\frac{3}{5}, \\frac{-4}{5}\\), 1)<\/p>\n<p><img loading=\"lazy\" src=\"https:\/\/i0.wp.com\/ncertmcq.com\/wp-content\/uploads\/2022\/03\/CBSE-Labs-1.png?resize=110%2C13&#038;ssl=1\" alt=\"CBSE Sample Papers for Class 12 Maths Term 2 Set 4 with Solutions\" width=\"110\" height=\"13\" data-recalc-dims=\"1\" \/><\/p>\n<p>Question 3.<br \/>\nFind the angle between unit vectors \\(\\widehat{\\mathrm{a}}\\) and \\(\\widehat{\\mathrm{b}}\\) so that \\(\\sqrt{3a}\\) &#8211; b is also a unit vector.<br \/>\nAnswer:<br \/>\na and b are unit vectors and \\(\\sqrt{3a}\\) -b is also unit vector<br \/>\nTo find angle between a and b.<\/p>\n<p>Suppose angle between a and b is \u03b8.<br \/>\n\\(\\vec{a} \\cdot \\vec{b}=|\\vec{a}| \\cdot|\\vec{b}|\\) cos \u03b8 (Dot product of two vectors)<br \/>\n\\(\\vec{a} \\cdot \\vec{b}\\) = cos \u03b8<\/p>\n<p>As \\(\\vec{a}\\) and \\(\\vec{b}\\) are unit vector so, \\(|\\vec{a}|=|\\vec{b}|\\) = 1<br \/>\n\\(\\sqrt{3a}\\) &#8211; b is also unit vector i.e. \\(|\\sqrt{3} \\vec{a}-\\vec{b}|\\) = 1<\/p>\n<p>Squaring both sides, we get<br \/>\n(\\( \\sqrt{3} \\vec{a}-\\vec{b}\\))<sup>2<\/sup> = 1<br \/>\n\u21d2 (\u221a3)2 \\(|\\vec{a}|^{2}+|\\vec{b}|^{2}-2 \\sqrt{3} \\vec{a} \\cdot \\vec{b}\\) = 1 [since, \\(\\vec{a} \\cdot \\vec{b}\\) = cos \u03b8]<br \/>\n\u21d2 3.1 + 1 &#8211; 2.\u221a3.cos\u03b8 = 1<br \/>\n\u21d2 4 &#8211; 2\u221a3.cos\u03b8 = 1<br \/>\n\u21d2 2\u221a3. cos \u03b8 = 3<br \/>\n\u21d2 cos \u03b8 = \\(\\frac{\\sqrt{3}}{2}\\)<br \/>\n\u21d2 \u03b8 = \\(\\frac{\\pi}{6}\\)<br \/>\nTherefore, the angle between the two unit vectors is \\(\\frac{\\pi}{6}\\)<\/p>\n<p>Question 4.<br \/>\nTen cards numbered 1 to 10 are placed in box, mixed up throughly and then one card is drawn randomly. If it is known that the number on the drawn card is more than 3, what is the probability that it is an even number?<br \/>\nAnswer:<br \/>\nLet A be the event the number on the card drawn is even and B be the event the number on the card drawn is greater than 3.<br \/>\nNow, the sample space of the experiment is<br \/>\nS = {1, 2, 3, 4, 5, 6, 7, 8, 9,10}<br \/>\nThen, A = {2,4, 6, 8,10}, B = {4, 5, 6, 7, 8, 9,10}<br \/>\nand A \u2229 B = {4, 6, 8,10}<br \/>\nP(A) = \\(\\frac{5}{10}\\) P(B) = \\(\\frac{7}{10}\\) = and P(A \u2229 B) = \\(\\frac{4}{10}\\)<br \/>\nThen, P(A\/B) = \\(\\frac{P(A \\cap B)}{P(B)}=\\frac{4 \/ 10}{7 \/ 10}=\\frac{4}{7}\\)<\/p>\n<p>Question 5.<br \/>\nFind \u222b\\(\\frac{\\sin x-\\cos x}{\\sqrt{1+\\sin 2 x}}\\) dx, 0 &lt; x &lt; \u03c0\/2.<br \/>\nOR<br \/>\nFind \u222b(log x)<sup>2<\/sup>dx<br \/>\nAnswer:<br \/>\nAccording to given<br \/>\nLet I = \u222b\\(\\frac{\\sin x-\\cos x}{\\sqrt{1+\\sin 2 x}}\\) dx, 0 &lt; x &lt; \u03c0\/2.<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-172843\" src=\"https:\/\/i0.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2022\/03\/CBSE-Sample-Papers-for-Class-12-Maths-Term-2-Set-4-with-Solutions-2.png?resize=267%2C163&#038;ssl=1\" alt=\"CBSE Sample Papers for Class 12 Maths Term 2 Set 4 with Solutions 2\" width=\"267\" height=\"163\" data-recalc-dims=\"1\" \/><br \/>\nLet sin x + cos x = t<br \/>\n\u21d2 (cos x &#8211; sin x) dx = dt<br \/>\n\u2234 I = \u222b\\(\\frac{-1}{t}\\)dt<br \/>\n= -log t + C<br \/>\n= log\\(\\left(\\frac{1}{t}\\right)\\) + C<br \/>\n\u21d2 I = log\\(\\left(\\frac{1}{\\sin x+\\cos x}\\right)\\) + C<\/p>\n<p>OR<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-172842\" src=\"https:\/\/i2.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2022\/03\/CBSE-Sample-Papers-for-Class-12-Maths-Term-2-Set-4-with-Solutions-3.png?resize=255%2C85&#038;ssl=1\" alt=\"CBSE Sample Papers for Class 12 Maths Term 2 Set 4 with Solutions 3\" width=\"255\" height=\"85\" data-recalc-dims=\"1\" \/><br \/>\n\u21d2 I = x(logx)<sup>2<\/sup> &#8211; \u222b\\(\\frac{2 x \\log }{x}\\) dx<br \/>\n\u21d2 I= x.(logx)<sup>2<\/sup> &#8211; I<sub>1<\/sub> + C<sub>1<\/sub> &#8230;&#8230;..(1)<br \/>\n\u21d2 I<sub>1<\/sub> = \u222b2.log x dx<br \/>\n\u21d2 I<sub>1<\/sub> =2x.log &#8211; 2\u222b\\(\\frac{x}{x}\\) dx<br \/>\n\u21d2 I<sub>1<\/sub> = 2x.log &#8211; 2x + C<sub>2<\/sub><br \/>\n\u21d2 I = x.(logx)<sup>2<\/sup> &#8211; 2x.log x + 2x + C<sub>1<\/sub> &#8211; C<sub>2<\/sub><br \/>\n\u21d2 I = x.(log x)<sup>2<\/sup> &#8211; 2x. Iogx + 2x + C (where C = C<sub>1<\/sub> &#8211; C<sub>2<\/sub>)<\/p>\n<p><img loading=\"lazy\" src=\"https:\/\/i0.wp.com\/ncertmcq.com\/wp-content\/uploads\/2022\/03\/CBSE-Labs-1.png?resize=110%2C13&#038;ssl=1\" alt=\"CBSE Sample Papers for Class 12 Maths Term 2 Set 4 with Solutions\" width=\"110\" height=\"13\" data-recalc-dims=\"1\" \/><\/p>\n<p>Question 6.<br \/>\nA bag contains 8 black and 5 blue balls three balls are drawn at random without replacements. What is the probability that all drawn balls are black colours?<br \/>\nAnswer:<br \/>\n1st black ball drawn at random the probability = \\(\\frac{8}{13}\\)<br \/>\n2nd black ball drawn at random the probability = \\(\\frac{7}{12}\\)<br \/>\n3rd black ball drawn at random the probability = \\(\\frac{6}{11}\\)<br \/>\nP = \\(\\frac{8}{13} \\times \\frac{7}{12} \\times \\frac{6}{11}\\)<br \/>\nSo, the total probability = \\(\\frac{28}{13 \\times 11}\\) = 0.195<\/p>\n<p style=\"text-align: center;\"><span style=\"color: #0000ff;\">Section &#8211; B<\/span><\/p>\n<p>Question 7.<br \/>\nFind the scalar components of a unit vector which is perpendicular to each of the vectors \u00ee +2\u0135 &#8211; k\u0302 and 3\u00ee &#8211; \u0135 &#8211; 2k\u0302.<br \/>\nAnswer:<br \/>\nLet, \\(\\vec{a}\\) = \u00ee +2\u0135 &#8211; k\u0302<br \/>\nand \\(\\vec{b}\\) = 3\u00ee &#8211; \u0135 &#8211; 2k\u0302<\/p>\n<p>We know that a unit vector perpendicular to each of vector \\(\\vec{a}\\) and \\(\\vec{b}\\) is given as \\(\\left[\\frac{\\vec{a} \\times \\vec{b}}{|\\vec{a} \\times \\vec{b}|}\\right]\\)<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-172841\" src=\"https:\/\/i0.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2022\/03\/CBSE-Sample-Papers-for-Class-12-Maths-Term-2-Set-4-with-Solutions-4.png?resize=407%2C367&#038;ssl=1\" alt=\"CBSE Sample Papers for Class 12 Maths Term 2 Set 4 with Solutions 4\" width=\"407\" height=\"367\" srcset=\"https:\/\/i0.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2022\/03\/CBSE-Sample-Papers-for-Class-12-Maths-Term-2-Set-4-with-Solutions-4.png?w=407&amp;ssl=1 407w, https:\/\/i0.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2022\/03\/CBSE-Sample-Papers-for-Class-12-Maths-Term-2-Set-4-with-Solutions-4.png?resize=300%2C271&amp;ssl=1 300w\" sizes=\"(max-width: 407px) 100vw, 407px\" data-recalc-dims=\"1\" \/><br \/>\n\u2234 Sclar components are \\(\\frac{3}{\\sqrt{83}}, \\frac{-5}{\\sqrt{83}}, \\frac{-7}{\\sqrt{83}}\\)<\/p>\n<p>Question 8.<br \/>\nFind the shortest distance between the lines and l2 whose vector equations are \\(\\vec{r}\\) = \u00ee + j\u0302 + \u03bb(2i\u0302 &#8211; j\u0302 + k\u0302) and \\(\\vec{r}\\) = 2i\u0302 + j\u0302 &#8211; k\u0302 + \u03bc(3i\u0302 &#8211; 5j\u0302 + 2k\u0302).<br \/>\nOR<br \/>\nFind cartesian equation of a line where the direction ratios of the parallel vector to it are (2, -1, 3) and passes through (5, -2, 4).<br \/>\nAnswer:<br \/>\nWe have given two vector equations<br \/>\n\\(\\vec{r}\\) = \u00ee + j\u0302 + \u03bb(2i\u0302 &#8211; j\u0302 + k\u0302) &#8230;&#8230;.(1)<br \/>\nand \\(\\vec{r}\\) = 2i\u0302 + j\u0302 &#8211; k\u0302 + \u03bc(3i\u0302 &#8211; 5j\u0302 + 2k\u0302)<\/p>\n<p>Comparing (1) and (2), with \\(\\vec{r}=\\overrightarrow{a_{1}}+\\lambda \\overrightarrow{b_{1}}\\) and \\(\\vec{r}=\\overrightarrow{a_{2}}+\\mu \\overrightarrow{b_{2}}\\) respectively,<\/p>\n<p>We get, \\(\\overrightarrow{a_{1}}\\) = \u00ee + j\u0302 , \\(\\overrightarrow{b_{1}}\\) = 2\u00ee &#8211; j\u0302 + k\u0302<br \/>\n\\(\\overrightarrow{a_{2}}\\) = 2\u00ee + j\u0302 &#8211; k\u0302 and \\(\\overrightarrow{b_{1}}\\) = 3\u00ee &#8211; 5j\u0302 + 2k\u0302<\/p>\n<p>Therefore, \\(\\overrightarrow{a_{2}}-\\overrightarrow{a_{1}}\\) = \u00ee &#8211; k\u0302<br \/>\nand \\(\\overrightarrow{b_{1}} \\times \\overrightarrow{b_{2}}\\) = (2i\u0302 &#8211; j\u0302 + k\u0302) \u00d7 (3i\u0302 &#8211; 5j\u0302 + 2k\u0302)<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-172840\" src=\"https:\/\/i2.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2022\/03\/CBSE-Sample-Papers-for-Class-12-Maths-Term-2-Set-4-with-Solutions-5.png?resize=503%2C365&#038;ssl=1\" alt=\"CBSE Sample Papers for Class 12 Maths Term 2 Set 4 with Solutions 5\" width=\"503\" height=\"365\" srcset=\"https:\/\/i2.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2022\/03\/CBSE-Sample-Papers-for-Class-12-Maths-Term-2-Set-4-with-Solutions-5.png?w=503&amp;ssl=1 503w, https:\/\/i2.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2022\/03\/CBSE-Sample-Papers-for-Class-12-Maths-Term-2-Set-4-with-Solutions-5.png?resize=300%2C218&amp;ssl=1 300w\" sizes=\"(max-width: 503px) 100vw, 503px\" data-recalc-dims=\"1\" \/><\/p>\n<p>OR<\/p>\n<p>Given, direction ratios of the parallel vector = (2, -1, 3)<br \/>\n\u2234 The direction ratios of required line are (2k\u0302 &#8211; k\u0302, 3k\u0302)<br \/>\nAlso, the given point is (x<sub>1<\/sub>, y<sub>1<\/sub>, z<sub>1<\/sub>) = (5, -2, 4)<br \/>\nCartesian equation of line passes through (x<sub>1<\/sub>, y<sub>1<\/sub>, z<sub>1<\/sub>) and with direction ratios a, b, c is given by<br \/>\n\\(\\frac{x-x_{1}}{a}=\\frac{y-y_{1}}{b}=\\frac{z-z_{1}}{c}\\)<\/p>\n<p>\u2234 The equation of required lines is,<br \/>\n\\(\\frac{x-5}{2 k}=\\frac{y+2}{-k}=\\frac{z-4}{3 k}\\)<br \/>\ni.e, \\(\\frac{x-5}{2}=\\frac{y+2}{-1}=\\frac{z-4}{3}\\)<\/p>\n<p><img loading=\"lazy\" src=\"https:\/\/i0.wp.com\/ncertmcq.com\/wp-content\/uploads\/2022\/03\/CBSE-Labs-1.png?resize=110%2C13&#038;ssl=1\" alt=\"CBSE Sample Papers for Class 12 Maths Term 2 Set 4 with Solutions\" width=\"110\" height=\"13\" data-recalc-dims=\"1\" \/><\/p>\n<p>Question 9.<br \/>\nFind the general solution of the following differential equation :<br \/>\nxdy &#8211; (y + 2x<sup>2<\/sup>)dx = 0<br \/>\nOR<br \/>\nFind the particular solution of the differential equation e<sup>x<\/sup> tan y dx + (2 &#8211; e<sup>x<\/sup>) sec<sup>2<\/sup> y dy = 0, given that y = \u03c0\/4 when x = 0.<br \/>\nAnswer:<br \/>\nxdy &#8211; (y + 2x<sup>2<\/sup>)dx = 0<br \/>\n\u21d2 \\(\\frac{d y}{d x}=\\frac{y+2 x^{2}}{x}\\)<br \/>\n\u21d2 \\(\\frac{d y}{d x}-\\frac{y}{x}\\) = 2x<\/p>\n<p>Comparing \\(\\frac{d y}{d x}\\) +Py = Q<br \/>\nP = \\(\\frac{-1}{x}\\), Q = 2x<br \/>\nI.F = e<sup>\u222bPdx<\/sup> = e<sup>-\u222b1\/xdx<\/sup> = e<sup>-logx<\/sup><br \/>\n= x<sup>-1<\/sup> = \\(\\frac{1}{x}\\)<\/p>\n<p>Solution is given by<br \/>\ny(I.F) = \u222bQ(I.F) dx<br \/>\n\u21d2 \\(\\frac{y}{x}\\) = \u222b(2x \u00d7 \\(\\frac{1}{x}\\))dx<br \/>\n\u21d2 \\(\\frac{y}{x}\\) = 2x +C<br \/>\n\u21d2 y = 2x<sup>2<\/sup> +Cx<\/p>\n<p>OR<\/p>\n<p>e<sup>x<\/sup> tan y dx + (2 &#8211; e<sup>x<\/sup>) sec<sup>2<\/sup>y dy = 0<br \/>\n\u21d2 e<sup>x<\/sup> tan y dx &#8211; (e<sup>x<\/sup> &#8211; 2)sec<sup>2<\/sup>y dy<br \/>\n\u2234 \u222b\\(\\frac{e^{x}}{e^{x}-2}\\)dx = \u222b\\(\\frac{\\sec ^{2} y}{\\tan y}\\)dx<br \/>\n\u21d2 log |e<sup>x<\/sup> &#8211; 2| = log |tan y| + log C<br \/>\n\u21d2 log e<sup>x<\/sup> &#8211; 2|= log (C + tan y)<br \/>\n\u21d2 e<sup>x<\/sup> &#8211; 2 = C tany<\/p>\n<p>Given, x = 0, y = \\(\\frac{\\pi}{4}\\)<br \/>\n\u2234 e<sup>0<\/sup> &#8211; 2 = C tan\\(\\left(\\frac{\\pi}{4}\\right)\\)<br \/>\n\u21d2 1 &#8211; 2 = C tan\\(\\left(\\frac{\\pi}{4}\\right)\\)<br \/>\n\u21d2 -1 = C \u00d7 1<br \/>\n\u21d2 C = -1<br \/>\n\u2234 e<sup>x<\/sup> &#8211; 2 = -tany<br \/>\nor<br \/>\ne<sup>x<\/sup> &#8211; 2 + tany = 0.<\/p>\n<p>Question 10.<br \/>\nFind: \u222b\\(\\frac{x^{2}+1}{\\left(x^{2}+2\\right)\\left(x^{2}+3\\right)}\\)dx<br \/>\nAnswer:<br \/>\nI = \u222b\\(\\frac{x^{2}+1}{\\left(x^{2}+2\\right)\\left(x^{2}+3\\right)}\\)dx<br \/>\nLet x<sup>2<\/sup> = y<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-172839\" src=\"https:\/\/i1.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2022\/03\/CBSE-Sample-Papers-for-Class-12-Maths-Term-2-Set-4-with-Solutions-6.png?resize=587%2C453&#038;ssl=1\" alt=\"CBSE Sample Papers for Class 12 Maths Term 2 Set 4 with Solutions 6\" width=\"587\" height=\"453\" srcset=\"https:\/\/i1.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2022\/03\/CBSE-Sample-Papers-for-Class-12-Maths-Term-2-Set-4-with-Solutions-6.png?w=587&amp;ssl=1 587w, https:\/\/i1.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2022\/03\/CBSE-Sample-Papers-for-Class-12-Maths-Term-2-Set-4-with-Solutions-6.png?resize=300%2C232&amp;ssl=1 300w\" sizes=\"(max-width: 587px) 100vw, 587px\" data-recalc-dims=\"1\" \/><\/p>\n<p style=\"text-align: center;\"><span style=\"color: #0000ff;\">Section &#8211; C<\/span><\/p>\n<p>Question 11.<br \/>\nFind the coordinates of foot of perpendicular drawn from the point (0, 2, 3) on the line = \\(\\frac{x+3}{5}=\\frac{y-1}{2}=\\frac{z+4}{3}\\). Also, find the length of perpendicular.<br \/>\nAnswer:<br \/>\nGiven equation of line is<br \/>\n\\(\\frac{x+3}{5}=\\frac{y-1}{2}=\\frac{z+4}{3}\\)<\/p>\n<p>Any random point T on the given line is calculated as follows<br \/>\n\\(\\frac{x+3}{5}=\\frac{y-1}{2}=\\frac{z+4}{3}\\) = \u03bb(say)<\/p>\n<p>\u2234 coordinates of point T are (5\u03bb -3, 2\u03bb + 1, 3\u03bb &#8211; 4).<br \/>\nLet point is P (0, 2, 3)<\/p>\n<p>Now, Direction ratios of line PT = (5\u03bb, &#8211; 3 &#8211; 0, 2\u03bb + 1 &#8211; 2, 3\u03bb &#8211; 4 &#8211; 3)<br \/>\n= (5\u03bb &#8211; 3, 2\u03bb &#8211; 1, 3\u03bb &#8211; 7)<br \/>\n\u2235 PT is perpendicular to the given line.<br \/>\n\u21d2 a<sub>1<\/sub>a<sub>2<\/sub> + b<sub>1<\/sub>b<sub>2<\/sub> + c<sub>1<\/sub>c<sub>2<\/sub> = 0<br \/>\na<sub>1<\/sub> = 5\u03bb- 3, b<sub>1<\/sub> = 2\u03bb &#8211; 1, c<sub>1<\/sub> = 3\u03bb &#8211; 7<br \/>\na<sub>2<\/sub> = 5, b<sub>2<\/sub> = 2, c<sub>2<\/sub> = 3<br \/>\n\u21d2 5(5\u03bb &#8211; 3) + 2 (2\u03bb &#8211; 1) + 3 (3\u03bb &#8211; 7) = 0<br \/>\n\u21d2 25\u03bb &#8211; 15 + 4\u03bb &#8211; 2 + 9\u03bb &#8211; 21 =0<br \/>\n\u21d2 38\u03bb &#8211; 38 = 0<br \/>\n\u21d2 \u03bb = 1<br \/>\n\u2234 The foot of perpendicular T = (5L- 3,2X +1,3X &#8211; 4)<br \/>\n= (2, 3, -1) (Put \u03bb = 1)<\/p>\n<p>Also, length of perpendicular, PT = Distance between point P and T<br \/>\nPT = \\(\\sqrt{(0-2)^{2}+(2-3)^{2}+(3+1)^{2}}\\)<br \/>\n= \\(\\sqrt{4+1+16}\\)<br \/>\n= \\(\\sqrt{21}\\) units<\/p>\n<p><img loading=\"lazy\" src=\"https:\/\/i0.wp.com\/ncertmcq.com\/wp-content\/uploads\/2022\/03\/CBSE-Labs-1.png?resize=110%2C13&#038;ssl=1\" alt=\"CBSE Sample Papers for Class 12 Maths Term 2 Set 4 with Solutions\" width=\"110\" height=\"13\" data-recalc-dims=\"1\" \/><\/p>\n<p>Question 12.<br \/>\nFind the area under the curve y = 2\u221ax and between the lines x = 0 and x = 1.<br \/>\nOR<br \/>\nFind the area of parabola y<sup>2<\/sup> = 4ax bounded by its latus rectum.<br \/>\nAnswer:<br \/>\nGiven y = 2<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-172838\" src=\"https:\/\/i0.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2022\/03\/CBSE-Sample-Papers-for-Class-12-Maths-Term-2-Set-4-with-Solutions-7.png?resize=261%2C280&#038;ssl=1\" alt=\"CBSE Sample Papers for Class 12 Maths Term 2 Set 4 with Solutions 7\" width=\"261\" height=\"280\" data-recalc-dims=\"1\" \/><br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-172837\" src=\"https:\/\/i1.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2022\/03\/CBSE-Sample-Papers-for-Class-12-Maths-Term-2-Set-4-with-Solutions-8.png?resize=121%2C209&#038;ssl=1\" alt=\"CBSE Sample Papers for Class 12 Maths Term 2 Set 4 with Solutions 8\" width=\"121\" height=\"209\" data-recalc-dims=\"1\" \/><\/p>\n<p>OR<\/p>\n<p>The graph of required parabola and its latus rectum is given alongside. The vertex of parabola y<sup>2<\/sup>= 4ax is at (O, O). The equation of latus rectum LSL&#8217; is x = a<br \/>\nAlso, parabola is symmetric about the X-axis.<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-172836\" src=\"https:\/\/i1.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2022\/03\/CBSE-Sample-Papers-for-Class-12-Maths-Term-2-Set-4-with-Solutions-9.png?resize=223%2C168&#038;ssl=1\" alt=\"CBSE Sample Papers for Class 12 Maths Term 2 Set 4 with Solutions 9\" width=\"223\" height=\"168\" data-recalc-dims=\"1\" \/><br \/>\n\u2234 The required area of the region OLL&#8217;O<br \/>\n= 2(area of region OLSO)<br \/>\n= 2\u222b<sup>a<\/sup><sub>0<\/sub>ydx = 2\u222b<sup>a<\/sup><sub>0<\/sub>\\(\\sqrt{4 a x}\\) dx<br \/>\n= 2 \u00d7 2\u221aa\u222b<sup>a<\/sup><sub>0<\/sub> \u221ax dx<br \/>\n= 4\u221aa\\(\\frac{2}{3}\\)[x<sup>3\/2<\/sup>]<sup>a<\/sup><sub>0<\/sub> = \\(\\frac{8}{3}\\)\u221aa(a<sup>3\/2<\/sup>)<br \/>\n= \\(\\frac{8}{3}\\) a<sup>2<\/sup> sq. units<\/p>\n<p>Question 13.<br \/>\nEvaluate \u222b\\(\\frac{x^{9}}{\\left(4 x^{2}+1\\right)^{6}}\\)dx<br \/>\nAnswer:<br \/>\nLet I = \u222b\\(\\frac{x^{9}}{\\left(4 x^{2}+1\\right)^{6}}\\)dx<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-172835\" src=\"https:\/\/i0.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2022\/03\/CBSE-Sample-Papers-for-Class-12-Maths-Term-2-Set-4-with-Solutions-10.png?resize=287%2C567&#038;ssl=1\" alt=\"CBSE Sample Papers for Class 12 Maths Term 2 Set 4 with Solutions 10\" width=\"287\" height=\"567\" srcset=\"https:\/\/i0.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2022\/03\/CBSE-Sample-Papers-for-Class-12-Maths-Term-2-Set-4-with-Solutions-10.png?w=287&amp;ssl=1 287w, https:\/\/i0.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2022\/03\/CBSE-Sample-Papers-for-Class-12-Maths-Term-2-Set-4-with-Solutions-10.png?resize=152%2C300&amp;ssl=1 152w\" sizes=\"(max-width: 287px) 100vw, 287px\" data-recalc-dims=\"1\" \/><\/p>\n<p><img loading=\"lazy\" src=\"https:\/\/i0.wp.com\/ncertmcq.com\/wp-content\/uploads\/2022\/03\/CBSE-Labs-1.png?resize=110%2C13&#038;ssl=1\" alt=\"CBSE Sample Papers for Class 12 Maths Term 2 Set 4 with Solutions\" width=\"110\" height=\"13\" data-recalc-dims=\"1\" \/><\/p>\n<p>Question 14.<br \/>\nA doctor is going to visit a patient. From the past experience, it is known that the probabilities that he will come by cab, metro, bike or by other means of transport are respectively 0.3, 0.2, 0.1 and 0.4. The probabilities that he will be late are 0.25, 0.3, 0.35 and 0.1, if he comes by cab, metro, bike and other means of transport respectively.<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-172834\" src=\"https:\/\/i1.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2022\/03\/CBSE-Sample-Papers-for-Class-12-Maths-Term-2-Set-4-with-Solutions-11.png?resize=239%2C308&#038;ssl=1\" alt=\"CBSE Sample Papers for Class 12 Maths Term 2 Set 4 with Solutions 11\" width=\"239\" height=\"308\" srcset=\"https:\/\/i1.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2022\/03\/CBSE-Sample-Papers-for-Class-12-Maths-Term-2-Set-4-with-Solutions-11.png?w=239&amp;ssl=1 239w, https:\/\/i1.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2022\/03\/CBSE-Sample-Papers-for-Class-12-Maths-Term-2-Set-4-with-Solutions-11.png?resize=233%2C300&amp;ssl=1 233w\" sizes=\"(max-width: 239px) 100vw, 239px\" data-recalc-dims=\"1\" \/><br \/>\nBased on the above information, answer the following questions :<br \/>\n(i) When the doctor arrives late, what is the probability that he comes by metro?<br \/>\n(ii) When the doctor arrives late, what is the probability that he comes by other means of transport?<br \/>\nAnswer:<br \/>\nLet E be the event that the doctor visit the patient late and let A<sub>1<\/sub>, A<sub>2<\/sub>, A<sub>3<\/sub>, A<sub>4<\/sub> be the events that the doctor comes by cab, metro, bike and other means of transport respectively.<br \/>\nP(A<sub>1<\/sub>) = 0.3, P(A<sub>2<\/sub>) = 0.2, P(A<sub>3<\/sub>) = 0.1, P(A<sub>4<\/sub>) = 0.4<br \/>\nP(E\/A<sub>1<\/sub>) = Probability that the doctor arriving late when he comes by cab = 0.25<br \/>\nSimilarly, P(E\/A<sub>2<\/sub>) = 0.3,<br \/>\nP(E\/A<sub>3<\/sub>) = 0.35<br \/>\nAnd P(E\/A<sub>4<\/sub>) = 0.1<\/p>\n<p>(i) P(A<sub>2<\/sub>\/E) = Probability that the doctor arriving late when he comes by metro<br \/>\nP(A<sub>2<\/sub>)P(E\/A<sub>2<\/sub>)<br \/>\n= \\(\\frac{\\mathrm{P}\\left(\\mathrm{A}_{2}\\right) \\mathrm{P}\\left(\\mathrm{E} \/ \\mathrm{A}_{2}\\right)}{\\Sigma \\mathrm{P}\\left(\\mathrm{A}_{i}\\right) \\mathrm{P}\\left(\\mathrm{E} \/ \\mathrm{A}_{i}\\right)}\\)<br \/>\n= \\(\\frac{(0.2)(0.3)}{(0.3)(0.25)+(0.2)(0.3)+(0.1)(0.35)+(0.4)(0.1)}\\)<br \/>\n= \\(\\frac{0.06}{0.075+0.06+0.035+0.04}\\)<br \/>\n= \\(\\frac{0.06}{0.21}\\)<br \/>\n= \\(\\frac{2}{7}\\)<br \/>\nThus, the probability that the doctor arriving late when he comes by metro is \\(\\frac{2}{7}\\).<\/p>\n<p>(ii) P(A<sub>4<\/sub>\/E) = Probability that the doctor arriving late when he comes by any means of transport<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-172833\" src=\"https:\/\/i2.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2022\/03\/CBSE-Sample-Papers-for-Class-12-Maths-Term-2-Set-4-with-Solutions-12.png?resize=343%2C276&#038;ssl=1\" alt=\"CBSE Sample Papers for Class 12 Maths Term 2 Set 4 with Solutions 12\" width=\"343\" height=\"276\" srcset=\"https:\/\/i2.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2022\/03\/CBSE-Sample-Papers-for-Class-12-Maths-Term-2-Set-4-with-Solutions-12.png?w=343&amp;ssl=1 343w, https:\/\/i2.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2022\/03\/CBSE-Sample-Papers-for-Class-12-Maths-Term-2-Set-4-with-Solutions-12.png?resize=300%2C241&amp;ssl=1 300w\" sizes=\"(max-width: 343px) 100vw, 343px\" data-recalc-dims=\"1\" \/><br \/>\nThus, the probability that the doctor arriving late when he comes by any means of transport is \\(\\frac{4}{21}\\).<\/p>\n","protected":false},"excerpt":{"rendered":"<p>Students can access the CBSE Sample Papers for Class 12 Maths with Solutions and marking scheme Term 2 Set 4 will help students in understanding the difficulty level of the exam. CBSE Sample Papers for Class 12 Maths Term 2 Set 4 with Solutions Maximum Marks : 40 Time : 2 Hours General Instructions: This &hellip;<\/p>\n","protected":false},"author":29,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[84914],"tags":[],"yoast_head":"<!-- This site is optimized with the Yoast SEO Premium plugin v17.2 (Yoast SEO v17.2) - https:\/\/yoast.com\/wordpress\/plugins\/seo\/ -->\n<title>CBSE Sample Papers for Class 12 Maths Term 2 Set 4 with Solutions - CBSE Labs<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/www.cbselabs.com\/cbse-sample-papers-for-class-12-maths-term-2-set-4\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"CBSE Sample Papers for Class 12 Maths Term 2 Set 4 with Solutions\" \/>\n<meta property=\"og:description\" content=\"Students can access the CBSE Sample Papers for Class 12 Maths with Solutions and marking scheme Term 2 Set 4 will help students in understanding the difficulty level of the exam. 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