**Answer is not given due to the change in present syllabus<\/span><\/span><\/p>\n Section – A<\/strong><\/p>\n Question 1. Question 2. Question 3. Question 4. Section – B<\/strong><\/p>\n Question 5. Question 6. Question 7. Question 8. Question 9. Question 10. Question 11. Question 12. Section – C<\/strong><\/p>\n Question 13. Question 14. Question 15. Question 16. Question 17. Question 18. Question 19. Question 20. Question 21. Question 22. Question 23. Section – D<\/strong><\/p>\n Question 24. Question 25. Question 26. Question 27. Question 28. Question 29. Note : Except for the following questions, all the remaining questions have been asked in previous set.<\/p>\n Section – B<\/strong><\/p>\n Question 12. Section – C<\/strong><\/p>\n Question 20. Question 21. Question 22. Question 23. Section – D<\/strong><\/p>\n Question 28. Question 29.CBSE Previous Year Question Papers Class 12 Maths 2017 Delhi Set I<\/h3>\n
\nIf A is a 3 \u00d7 3 invertible matrix, then what will be the value of k if det (A-1<\/sup>) = (det A)k<\/sup>. [1]
\nSolution:
\nGiven, A is an invertible matrix.
\n\u2234 A. A-1<\/sup> = I
\n\u21d2 det (A.A-1<\/sup>) = det (I)
\n\u21d2 det (A).det(A-1<\/sup>) = 1 [ \u2235 det(I) = 1]
\n\u21d2 det (A), (det A)k<\/sup> = 1 [\u2235 det(A-1<\/sup>)=(det A)k<\/sup>]
\n![\"CBSE](\"https:\/\/i0.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/CBSE-Previous-Year-Question-Papers-Class-12-Maths-2017-Delhi-1.png?resize=216%2C89&ssl=1\")
\non comparing both sides, we get
\nk = -1<\/p>\n
\nDetermine the value of the constant ‘k’ so that the function f(x) = \\(\\left\\{\\begin{array}{l}{\\frac{k x}{|x|}, \\text { if } x<0} \\\\ {3, \\text { if } x \\geq 0}\\end{array}\\right.\\) is continuous at x = 0. [1]
\nSolution:
\nGiven, that the function is continuous
\n![\"CBSE](\"https:\/\/i1.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/CBSE-Previous-Year-Question-Papers-Class-12-Maths-2017-Delhi-2.png?resize=364%2C94&ssl=1\")
\n![\"CBSE](\"https:\/\/i0.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/CBSE-Previous-Year-Question-Papers-Class-12-Maths-2017-Delhi-3.png?resize=297%2C195&ssl=1\")
<\/p>\n
\nEvaluate: \\(\\int_{2}^{3} 3^{x} d x\\). [1]
\nSolution:
\n![\"CBSE](\"https:\/\/i0.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/CBSE-Previous-Year-Question-Papers-Class-12-Maths-2017-Delhi-4.png?resize=361%2C250&ssl=1\")
<\/p>\n
\nIf a line makes angles 90\u00b0 and 60\u00b0 respectively with the positive directions of X and Y-axes, find the angle which it makes with the positive direction of Z-axis. [1]
\nSolution:
\nWe know that:
\nl2<\/sup> + m2<\/sup> + n2<\/sup> = 1 …(i)
\nand l = cos \u03b1, m = cos \u03b2, n = cos \u03b3
\nGiven, \u03b1 = 90\u00b0, \u03b2 = 60\u00b0
\n\u2234 cos \u03b1 = cos 90\u00b0 = 0 and cos \u03b2 = cos 60\u00b0 = \\(\\frac{1}{2}\\)
\n![\"CBSE](\"https:\/\/i1.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/CBSE-Previous-Year-Question-Papers-Class-12-Maths-2017-Delhi-5.png?resize=340%2C318&ssl=1\")
<\/p>\n
\nShow that all the diagonal elements of a skew symmetric matrix are zero. [2]
\nSolution:
\n![\"CBSE](\"https:\/\/i2.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/CBSE-Previous-Year-Question-Papers-Class-12-Maths-2017-Delhi-6.png?resize=343%2C210&ssl=1\")
\nHence, all the diagonal elements of a skew, symmetric matrix are zero (as diagonal elements are : a11<\/sub>; a22<\/sub>, …….. ann<\/sub>). Hence Proved.<\/p>\n
\nFind \\(\\frac{d y}{d x}\\) at x = 1, y = \\(\\frac{\\pi}{4}\\) if sin2<\/sup> y + cos xy = K. [2]
\nSolution:
\nGiven, sin2<\/sup> y + cos xy = K
\nDifferentiating both sides w.r.t x, we get
\n![\"CBSE](\"https:\/\/i0.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/CBSE-Previous-Year-Question-Papers-Class-12-Maths-2017-Delhi-7.png?resize=378%2C467&ssl=1\")
\n![\"CBSE](\"https:\/\/i0.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/CBSE-Previous-Year-Question-Papers-Class-12-Maths-2017-Delhi-8.png?resize=139%2C150&ssl=1\")
<\/p>\n
\nThe volume of a sphere is increasing at the rate of 3 cubic centimetre per second. Find the rate of increase of its surface area, when the radius is 2 cm. [2]
\nSolution:
\nLet V be the volume and r be the radius of sphere at any time t.
\n![\"CBSE](\"https:\/\/i0.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/CBSE-Previous-Year-Question-Papers-Class-12-Maths-2017-Delhi-9.png?resize=371%2C536&ssl=1\")
\n\u2234 Rate of increase of surface area of the sphere is 3 square centimetre per second.<\/p>\n
\nShow that the function f(x) = 4x3<\/sup> – 18x2<\/sup> + 27x – 7 is always increasing on R. [2]
\nSolution:
\nGiven, f(x)= 4x3<\/sup> – 18x2<\/sup> + 27x – 7
\nDifferentiating f(x) w.r.t. x, we get
\nf'(x) = 12x2<\/sup> – 36x + 27 = 3(4x2<\/sup> – 12x + 9)
\n= 3(2x – 3)2<\/sup> for any x \u03f5 R
\n3 > 0 and (2x – 3)2<\/sup> \u2265 0
\n\u2234 f'(x) \u2265 o
\n\u21d2 The fimction is always increasing on R.<\/p>\n
\nFind the vector equation of the line passing through the point A(1, 2, -1) and parallel to the line 5x – 25 = 14 – 7y = 35z. [2]
\nSolution:
\nGiven line is 5x – 25 = 14 – 7y = 35z
\n![\"CBSE](\"https:\/\/i1.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/CBSE-Previous-Year-Question-Papers-Class-12-Maths-2017-Delhi-10.png?resize=305%2C155&ssl=1\")
\n\u2234 Vector equation of the line which passes through the point A (1, 2, – 1) and its direction ratio are proportional to 7, – 5, 1 is
\n![\"CBSE](\"https:\/\/i1.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/CBSE-Previous-Year-Question-Papers-Class-12-Maths-2017-Delhi-11.png?resize=262%2C49&ssl=1\")
<\/p>\n
\nProve that if E and F are independent events, then the events E and F’ are also independent. [2]
\nSolution:
\nSince E and F are independent events :
\nP(E \u2229 F) = P (E) . P (F) …(i)
\nP(E \u2229 F’) = P (E) – P (E \u2229 F)
\n= P (E) – P (E) P (F) [From (i)]
\n= P(E)(1 – P(F))
\n= P (E) P (F’)
\n\u2234 E and F’ are also independent. Hence Proved.<\/p>\n
\nA small firm manufactures necklaces and bracelets. The total number of necklaces and bracelets that it can handle per day is at most 24. It takes one hour to make a bracelet and half an hour to make a necklace. The maximum number of hours available per day is 16. If the profit on a necklace is \u20b9 100 and that on a bracelet is \u20b9 300. Formulate an L.P.P. for finding how many of each should be produced daily to maximize the profit ? It is being given that at least one of each must be produced. [2]
\nSolution:
\nLet the manufacturer produces x pieces of necklaces and y pieces of bracelets.
\nSince total number of necklaces and bracelets that can be handle per day are 24.
\nso, x + y \u2264 24 …(i)
\nTo make bracelet one needs one hour and half an hour is need to make necklace and maximum time available is 16 hours
\nso, \\(\\frac{1}{2}\\)x + y \u2264 16 ….(ii)
\nNow, let Z be the profit and we have to maximize it, so our LPP will be
\nMaximize Z = 100x + 300y
\nSubject to constraints :
\nx + y \u2264 24
\n\\(\\frac{1}{2}\\) x + y \u2264 16
\nor x + 2y \u2264 32
\nand x, y \u2265 1<\/p>\n
\nFind: \\(\\int \\frac{d x}{x^{2}+4 x+8}\\). [2]
\nSolution:
\n![\"CBSE](\"https:\/\/i1.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/CBSE-Previous-Year-Question-Papers-Class-12-Maths-2017-Delhi-12.png?resize=169%2C55&ssl=1\")
\n![\"CBSE](\"https:\/\/i2.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/CBSE-Previous-Year-Question-Papers-Class-12-Maths-2017-Delhi-13.png?resize=300%2C259&ssl=1\")
<\/p>\n
\nProve that [4]
\n![\"CBSE](\"https:\/\/i0.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/CBSE-Previous-Year-Question-Papers-Class-12-Maths-2017-Delhi-14.png?resize=244%2C108&ssl=1\")
\nSolution:
\n![\"CBSE](\"https:\/\/i1.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/CBSE-Previous-Year-Question-Papers-Class-12-Maths-2017-Delhi-15.png?resize=333%2C454&ssl=1\")
\n![\"CBSE](\"https:\/\/i1.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/CBSE-Previous-Year-Question-Papers-Class-12-Maths-2017-Delhi-16.png?resize=214%2C182&ssl=1\")
<\/p>\n
\nUsing properties of determinants, prove that: [4]
\n![\"CBSE](\"https:\/\/i0.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/CBSE-Previous-Year-Question-Papers-Class-12-Maths-2017-Delhi-17.png?resize=299%2C78&ssl=1\")
\nSolution:
\n![\"CBSE](\"https:\/\/i2.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/CBSE-Previous-Year-Question-Papers-Class-12-Maths-2017-Delhi-18.png?resize=369%2C524&ssl=1\")
\nOR
\n![\"CBSE](\"https:\/\/i2.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/CBSE-Previous-Year-Question-Papers-Class-12-Maths-2017-Delhi-19.png?resize=320%2C57&ssl=1\")
\nfind the matrix D such that CD – AB = 0.
\nSolution:
\nLet D be the matrix of order 2 \u00d7 2,
\n![\"CBSE](\"https:\/\/i0.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/CBSE-Previous-Year-Question-Papers-Class-12-Maths-2017-Delhi-20.png?resize=371%2C469&ssl=1\")
\n![\"CBSE](\"https:\/\/i1.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/CBSE-Previous-Year-Question-Papers-Class-12-Maths-2017-Delhi-21.png?resize=338%2C497&ssl=1\")
\n![\"CBSE](\"https:\/\/i2.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/CBSE-Previous-Year-Question-Papers-Class-12-Maths-2017-Delhi-22.png?resize=315%2C185&ssl=1\")
<\/p>\n
\nDifferentiate the function (sin x)x<\/sup> + sin-1<\/sup> \\(\\sqrt{x}\\) with respect to x.
\nSolution:
\n![\"CBSE](\"https:\/\/i2.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/CBSE-Previous-Year-Question-Papers-Class-12-Maths-2017-Delhi-23.png?resize=375%2C401&ssl=1\")
\n![\"CBSE](\"https:\/\/i0.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/CBSE-Previous-Year-Question-Papers-Class-12-Maths-2017-Delhi-24.png?resize=374%2C250&ssl=1\")
\nOR
\nIF xm<\/sup>yn<\/sup> = (x + y)m+n<\/sup>, prove that \\(\\frac{d^{2} y}{d x^{2}}\\) = 0.
\nSolution:
\n![\"CBSE](\"https:\/\/i2.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/CBSE-Previous-Year-Question-Papers-Class-12-Maths-2017-Delhi-25.png?resize=365%2C481&ssl=1\")
\n![\"CBSE](\"https:\/\/i0.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/CBSE-Previous-Year-Question-Papers-Class-12-Maths-2017-Delhi-26.png?resize=368%2C210&ssl=1\")
<\/p>\n
\nFind [4]
\n![\"CBSE](\"https:\/\/i1.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/CBSE-Previous-Year-Question-Papers-Class-12-Maths-2017-Delhi-27.png?resize=162%2C48&ssl=1\")
\nSolution:
\n![\"CBSE](\"https:\/\/i1.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/CBSE-Previous-Year-Question-Papers-Class-12-Maths-2017-Delhi-28.png?resize=308%2C50&ssl=1\")
\n![\"CBSE](\"https:\/\/i1.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/CBSE-Previous-Year-Question-Papers-Class-12-Maths-2017-Delhi-29.png?resize=406%2C547&ssl=1\")
\n![\"CBSE](\"https:\/\/i0.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/CBSE-Previous-Year-Question-Papers-Class-12-Maths-2017-Delhi-30.png?resize=369%2C363&ssl=1\")
<\/p>\n
\nEvaluate: [4]
\n![\"CBSE](\"https:\/\/i2.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/CBSE-Previous-Year-Question-Papers-Class-12-Maths-2017-Delhi-31.png?resize=123%2C51&ssl=1\")
\nSolution:
\n![\"CBSE](\"https:\/\/i0.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/CBSE-Previous-Year-Question-Papers-Class-12-Maths-2017-Delhi-32.png?resize=285%2C171&ssl=1\")
\n![\"CBSE](\"https:\/\/i0.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/CBSE-Previous-Year-Question-Papers-Class-12-Maths-2017-Delhi-33.png?resize=373%2C523&ssl=1\")
\nOR
\nEvaluate: ![\"CBSE](\"https:\/\/i2.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/CBSE-Previous-Year-Question-Papers-Class-12-Maths-2017-Delhi-34.png?resize=137%2C37&ssl=1\")
\nSolution:
\n![\"CBSE](\"https:\/\/i1.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/CBSE-Previous-Year-Question-Papers-Class-12-Maths-2017-Delhi-35.png?resize=391%2C532&ssl=1\")
\n![\"CBSE](\"https:\/\/i0.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/CBSE-Previous-Year-Question-Papers-Class-12-Maths-2017-Delhi-36.png?resize=368%2C221&ssl=1\")
<\/p>\n
\nProve that x2<\/sup> – y2<\/sup> = c(x2<\/sup> +y2<\/sup>)2<\/sup> is the general solution of the differential equation (x3<\/sup> – 3xy2<\/sup>) dx = (y3<\/sup> – 3x2<\/sup>y) dy, where c is a parameter. [4]
\nSolution:
\nGiven differential equation is,
\n![\"CBSE](\"https:\/\/i0.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/CBSE-Previous-Year-Question-Papers-Class-12-Maths-2017-Delhi-37.png?resize=370%2C504&ssl=1\")
\n![\"CBSE](\"https:\/\/i0.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/CBSE-Previous-Year-Question-Papers-Class-12-Maths-2017-Delhi-38.png?resize=282%2C322&ssl=1\")
\n![\"CBSE](\"https:\/\/i0.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/CBSE-Previous-Year-Question-Papers-Class-12-Maths-2017-Delhi-39.png?resize=372%2C478&ssl=1\")
\n![\"CBSE](\"https:\/\/i2.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/CBSE-Previous-Year-Question-Papers-Class-12-Maths-2017-Delhi-40.png?resize=395%2C530&ssl=1\")
<\/p>\n
\nLet \\(\\vec{a}=\\hat{i}+\\hat{j}+\\hat{k}, \\vec{b}=\\hat{i} \\text { and } \\vec{c}=c_{1} \\hat{i}+c_{2} \\hat{j}+c_{3} \\hat{k}\\) then:
\n(a) Let c1<\/sub> = 1 and c2<\/sub> = 2, find c3<\/sub> which makes \\(\\vec{a}, \\vec{b} \\text { and } \\vec{c}\\) coplanar.
\n(b) If c2<\/sub> = – 1 and c3<\/sub> = 1, show that no value of c1<\/sub> can make \\(\\vec{a}, \\vec{b} \\text { and } \\vec{c}\\) coplanar. [4]
\nSolution:
\n![\"CBSE](\"https:\/\/i0.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/CBSE-Previous-Year-Question-Papers-Class-12-Maths-2017-Delhi-41.png?resize=366%2C516&ssl=1\")
\n![\"CBSE](\"https:\/\/i2.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/CBSE-Previous-Year-Question-Papers-Class-12-Maths-2017-Delhi-42.png?resize=368%2C251&ssl=1\")
<\/p>\n
\nIf \\(\\vec{a}, \\vec{b}, \\vec{c}\\) are mutually perpendicular vectors of equal magnitudes, show that the vector \\(\\vec{a}+\\vec{b}+\\vec{c}\\) is equally inclined to \\(\\vec{a}, \\vec{b} \\text { and } \\vec{c}\\). Also, find the angle which \\(\\vec{a}+\\vec{b}+\\vec{c}\\) makes with \\(\\vec{a} \\text { or } \\vec{b} \\text { or } \\vec{c}\\). [4]
\nSolution:
\n![\"CBSE](\"https:\/\/i2.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/CBSE-Previous-Year-Question-Papers-Class-12-Maths-2017-Delhi-43.png?resize=365%2C122&ssl=1\")
\n![\"CBSE](\"https:\/\/i0.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/CBSE-Previous-Year-Question-Papers-Class-12-Maths-2017-Delhi-44.png?resize=381%2C558&ssl=1\")
\n![\"CBSE](\"https:\/\/i0.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/CBSE-Previous-Year-Question-Papers-Class-12-Maths-2017-Delhi-45.png?resize=369%2C198&ssl=1\")
<\/p>\n
\nThe random variable X can take only the values 0, 1, 2, 3. Given that P (X = 0) = P (X = 1) = p and P(X = 2) = P(X = 3) such that \u03a3pi<\/sub>xi<\/sub>2<\/sup> = 2\u03a3pi<\/sub>xi<\/sub> , find the value of p. [4]
\nSolution:
\nGiven, P(X = 0) = P(X = 1) = p and P(X = 2) = P (X = 3)
\nLet P (X = 2) = P(X = 3) = k
\n![\"CBSE](\"https:\/\/i0.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/CBSE-Previous-Year-Question-Papers-Class-12-Maths-2017-Delhi-46.png?resize=364%2C208&ssl=1\")
\n![\"CBSE](\"https:\/\/i1.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/CBSE-Previous-Year-Question-Papers-Class-12-Maths-2017-Delhi-47.png?resize=366%2C222&ssl=1\")
<\/p>\n
\nOften it is taken that a truthful person commands, more respect in the society. A man is known to speak the truth 4 out of 5 times. He throws a die and reports that it is a six. Find the probability that it is actually a six.
\nDo you also agree that the value of truthfulness leads to more respect in the society ? [4]
\nSolution:
\nLet E1<\/sub>, E2<\/sub> and A be the events defined as follows:
\nE1<\/sub> = Six appears on throwing a die.
\nE2<\/sub> = Six does not appear on throwing a die.
\nand A = the man reports that it is a six We have,
\n![\"CBSE](\"https:\/\/i0.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/CBSE-Previous-Year-Question-Papers-Class-12-Maths-2017-Delhi-48.png?resize=93%2C95&ssl=1\")
\nNow P(A\/E1<\/sub>) = Probability that the man reports that there is a six on the die given that six has occurred on the die = 4\/5 (probability that the man speaks truth)
\nand P(A\/E2<\/sub>) = Probability that the man reports that there is a six on the die given that six has not occurred on the die (probability that the man does not speak truth).
\n![\"CBSE](\"https:\/\/i0.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/CBSE-Previous-Year-Question-Papers-Class-12-Maths-2017-Delhi-49.png?resize=345%2C273&ssl=1\")
\nYes, truthfulness always leads to more respect in the society as truth always wins.<\/p>\n
\nSolve the following L.P.P. graphically : Minimise Z = 5x + 10y Subject to constraints
\nx + 2y \u2264 120
\nx + y \u2265 60
\nx – 2y \u2265 0
\nand x, y \u2265 0 [4]
\nSolution:
\nWe have,
\nMinimise Z = 5x + 10y
\nSubject to the constraints :
\nx + 2y \u2264 120
\nx + y \u2265 60
\nx – 2y \u2265 0
\nand x, y \u2265 0
\nConverting the given inequalities into equations, we obtain the following equations :
\nx + 2y = 120
\nx + y = 60
\nx – 2y = 0
\nThen, x + 2y = 120
\n![\"CBSE](\"https:\/\/i2.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/CBSE-Previous-Year-Question-Papers-Class-12-Maths-2017-Delhi-50.png?resize=396%2C536&ssl=1\")
\nThe shaded region ABCD represented by the given constraints is the feasible region. Comer points of the common shaded region are A (40, 20), B (60, 30), C (120, 0) and D (60, 0). Value of Z at each corner point is given as :
\n![\"CBSE](\"https:\/\/i2.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/CBSE-Previous-Year-Question-Papers-Class-12-Maths-2017-Delhi-51.png?resize=365%2C140&ssl=1\")
\nHence, minimum value of Z is 300 at (60, 0).<\/p>\n
\nUse product \\(\\left[\\begin{array}{rrr}{1} & {-1} & {2} \\\\ {0} & {2} & {-3} \\\\ {3} & {-2} & {4}\\end{array}\\right]\\left[\\begin{array}{rrr}{-2} & {0} & {1} \\\\ {9} & {2} & {-3} \\\\ {6} & {1} & {-2}\\end{array}\\right]\\) to solve the system of equations x + 3z = 9, – x + 2y – 2z = 4, 2x – 3y + 4z = 3. [6]
\nSolution:
\nConsider,
\n![\"CBSE](\"https:\/\/i2.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/CBSE-Previous-Year-Question-Papers-Class-12-Maths-2017-Delhi-52.png?resize=374%2C575&ssl=1\")
\n![\"CBSE](\"https:\/\/i1.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/CBSE-Previous-Year-Question-Papers-Class-12-Maths-2017-Delhi-53.png?resize=213%2C91&ssl=1\")
\n![\"CBSE](\"https:\/\/i2.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/CBSE-Previous-Year-Question-Papers-Class-12-Maths-2017-Delhi-54.png?resize=320%2C384&ssl=1\")
<\/p>\n
\nConsider f : R+<\/sub> \u2192 [- 5, \u221e), given by f(x) = 9x2<\/sup> + 6x – 5. Show that f is invertible with f-1<\/sup>(y) = \\(\\left(\\frac{\\sqrt{y+6}-1}{3}\\right)\\). [6]
\nhence find:
\n(i) f-1<\/sup>(10)
\n(ii) y if f-1<\/sup>(y) = \\(\\frac{4}{3}\\),
\nwhere R+<\/sub> is the set of all non-negative real numbers.
\nSolution:
\nFor one-one :
\n![\"CBSE](\"https:\/\/i0.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/CBSE-Previous-Year-Question-Papers-Class-12-Maths-2017-Delhi-55.png?resize=371%2C481&ssl=1\")
\n![\"CBSE](\"https:\/\/i2.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/CBSE-Previous-Year-Question-Papers-Class-12-Maths-2017-Delhi-56.png?resize=372%2C440&ssl=1\")
\nOR
\nDiscuss the commutativity and associativity of binary operation defined on A = Q – {1} by the rule a * b = a – b + ab for all, a, b \u03f5 A. Also find the identity element of * in A and hence find the invertible elements of A.**<\/p>\n
\nIf the sum of lengths of the hypotenuse and a side of a right angled triangle is given, show that the area of the triangle is maximum, when the angle between them is \\(\\frac{\\pi}{3}\\).
\nSolution:
\nLet h, x and y be the length of hypotenuse and sides of the right triangle ABC.
\n![\"CBSE](\"https:\/\/i1.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/CBSE-Previous-Year-Question-Papers-Class-12-Maths-2017-Delhi-57.png?resize=317%2C422&ssl=1\")
\n![\"CBSE](\"https:\/\/i0.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/CBSE-Previous-Year-Question-Papers-Class-12-Maths-2017-Delhi-58.png?resize=373%2C574&ssl=1\")
\n![\"CBSE](\"https:\/\/i2.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/CBSE-Previous-Year-Question-Papers-Class-12-Maths-2017-Delhi-59.png?resize=302%2C535&ssl=1\")
\n![\"CBSE](\"https:\/\/i1.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/CBSE-Previous-Year-Question-Papers-Class-12-Maths-2017-Delhi-60.png?resize=366%2C45&ssl=1\")
<\/p>\n
\nUsing integration, find the area of region bounded by the triangle whose vertices are (- 2, 1), (0, 4) and (2, 3). [6]
\nSolution:
\nThe vertices of the \u2206ABC are A(- 2, 1), B(0, 4) and C(2, 3).
\nEquation of the side AB is
\n![\"CBSE](\"https:\/\/i2.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/CBSE-Previous-Year-Question-Papers-Class-12-Maths-2017-Delhi-61.png?resize=361%2C391&ssl=1\")
\n![\"CBSE](\"https:\/\/i2.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/CBSE-Previous-Year-Question-Papers-Class-12-Maths-2017-Delhi-62.png?resize=403%2C516&ssl=1\")
\nOR
\nFind the area bounded by the circle x2<\/sup> + y2<\/sup> = 16 and the line \\( \\sqrt{{3}} \\)y = x in the first quadrant, using integration.
\nSolution:
\nGiven, x = \\( \\sqrt{{3}} \\)y
\nand x2<\/sup> + y2<\/sup> = 16,
\n![\"CBSE](\"https:\/\/i0.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/CBSE-Previous-Year-Question-Papers-Class-12-Maths-2017-Delhi-63.png?resize=372%2C481&ssl=1\")
\n![\"CBSE](\"https:\/\/i0.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/CBSE-Previous-Year-Question-Papers-Class-12-Maths-2017-Delhi-64.png?resize=361%2C395&ssl=1\")
<\/p>\n
\nSolve the differential equation x\\(\\frac{d y}{d x}\\) + y = x cos x + sin x, given that y = 1 when x = \\(\\frac{\\pi}{2}\\).[6]
\nSolution:
\nGiven differential equation is :
\n![\"CBSE](\"https:\/\/i2.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/CBSE-Previous-Year-Question-Papers-Class-12-Maths-2017-Delhi-65.png?resize=199%2C48&ssl=1\")
\n![\"CBSE](\"https:\/\/i1.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/CBSE-Previous-Year-Question-Papers-Class-12-Maths-2017-Delhi-66.png?resize=375%2C496&ssl=1\")
\n![\"CBSE](\"https:\/\/i2.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/CBSE-Previous-Year-Question-Papers-Class-12-Maths-2017-Delhi-67.png?resize=370%2C272&ssl=1\")
<\/p>\n
\nFind the equation of the plane through the line of intersection of \\(\\vec{r} \\cdot(2 \\hat{i}-3 \\hat{j}+4 \\hat{k})=1\\) and \\(\\vec{r} \\cdot(\\hat{i}-\\hat{j})+4=0\\) and perpendicular to the plane \\(\\vec{r} \\cdot(2 \\hat{i}-\\hat{j}+\\hat{k})+8=0\\). Hence find whether the plane thus obtained contains the line x – 1 = 2y – 4 = 3z – 12. [6]
\nSolution:
\nThe equation of the plane passing through the line of intersection of the given planes is :
\n![\"CBSE](\"https:\/\/i0.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/CBSE-Previous-Year-Question-Papers-Class-12-Maths-2017-Delhi-68.png?resize=346%2C103&ssl=1\")
\n![\"CBSE](\"https:\/\/i2.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/CBSE-Previous-Year-Question-Papers-Class-12-Maths-2017-Delhi-69.png?resize=377%2C525&ssl=1\")
\n![\"CBSE](\"https:\/\/i0.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/CBSE-Previous-Year-Question-Papers-Class-12-Maths-2017-Delhi-70.png?resize=377%2C534&ssl=1\")
\nOR
\nFind the vector and Cartesian equations of a line passing through (1, 2, – 4) and perpendicular to the two lines \\(\\frac{x-8}{3}=\\frac{y+19}{-16}=\\frac{z-10}{7}\\) and \\(\\frac{x-15}{3}=\\frac{y-29}{8}=\\frac{z-5}{-5}\\).
\nSolution:
\nCartesian equation of the line passing through (1, 2, – 4) is
\n![\"CBSE](\"https:\/\/i2.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/CBSE-Previous-Year-Question-Papers-Class-12-Maths-2017-Delhi-71.png?resize=373%2C473&ssl=1\")
\n![\"CBSE](\"https:\/\/i0.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/CBSE-Previous-Year-Question-Papers-Class-12-Maths-2017-Delhi-72.png?resize=367%2C568&ssl=1\")
<\/p>\nCBSE Previous Year Question Papers Class 12 Maths 2017 Delhi Set II<\/h3>\n
\nFor the curve y = 5x – 2x3<\/sup>, if x increases at the rate of 2 units\/sec, then find the rate of change of the slope of the curve when 2 = 3. [2]
\nSolution:
\nGiven curve is,
\ny = 5x – 2x3<\/sup>
\n![\"CBSE](\"https:\/\/i0.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/CBSE-Previous-Year-Question-Papers-Class-12-Maths-2017-Delhi-73.png?resize=313%2C200&ssl=1\")
\n![\"CBSE](\"https:\/\/i2.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/CBSE-Previous-Year-Question-Papers-Class-12-Maths-2017-Delhi-74.png?resize=368%2C118&ssl=1\")
<\/p>\n
\nThe random variable X can take only the values 0, 1, 2, 3. Given that P(2) = P(3) = p and P(0) = 2P(1). If \u03a3pi<\/sub>xi<\/sub>2<\/sup> = 2\u03a3pi<\/sub>xi<\/sub>, find the value of p. [4]
\nSolution:
\nGiven, P(2) = P(3) = p
\nand P(0) = 2P(1)
\nLet P(1) = k
\n![\"CBSE](\"https:\/\/i2.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/CBSE-Previous-Year-Question-Papers-Class-12-Maths-2017-Delhi-75.png?resize=265%2C91&ssl=1\")
\n\u21d2 0 + k + 4p + 9p = 2(0 + k + 2p + 3p)
\n\u21d2 k + 13p = 2k + 10p
\nor 3p = k ….(i)
\nalso we know that
\n\u03a3pi<\/sub> = 1
\n\u21d2 2k + k + p + p = l
\n\u21d2 3k + 2p = 1
\n\u21d2 9p + 2p = 1 [using (i)]
\n\u21d2 11p = 1
\nor P = \\(\\frac{1}{11}\\)<\/p>\n
\nUsing vectors find the area of triangle ABC with vertices A(1, 2, 3), B(2, – 1, 4) and C (4, 5, – 1). [4]
\nSolution:
\nVertices of the given \u2206ABC are A(1, 2, 3) B(2, – 1, 4) and C(4, 5, – 1)
\n![\"CBSE](\"https:\/\/i0.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/CBSE-Previous-Year-Question-Papers-Class-12-Maths-2017-Delhi-76.png?resize=370%2C415&ssl=1\")
<\/p>\n
\nSolve the following L. P. P. graphically
\nMaximise Z = 4x + y
\nSubject to following constraints :
\nx + y \u2264 50,
\n3x + y \u2264 90,
\nx \u2265 10
\nx, y \u2265 0 [4]
\nSolution:
\nWe have,
\nMaximise Z = 4x + y
\nSubject to the constraints :
\nx + y \u2264 50
\n3x + y \u2264 90
\nx \u2265 10
\nx, y \u2265 0
\nConverting the given inequalities into equations, we obtain the following equations :
\nx + y = 50
\n3x + y = 90
\nx = 10
\n![\"CBSE](\"https:\/\/i0.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/CBSE-Previous-Year-Question-Papers-Class-12-Maths-2017-Delhi-77.png?resize=373%2C484&ssl=1\")
\nPlotting these points on the graph, we get the shaded feasible region i.e., ABCD.
\n![\"CBSE](\"https:\/\/i0.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/CBSE-Previous-Year-Question-Papers-Class-12-Maths-2017-Delhi-78.png?resize=365%2C163&ssl=1\")
<\/p>\n
\nFind: [4]
\n![\"CBSE](\"https:\/\/i2.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/CBSE-Previous-Year-Question-Papers-Class-12-Maths-2017-Delhi-79.png?resize=163%2C54&ssl=1\")
\nSolution:
\n![\"CBSE](\"https:\/\/i1.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/CBSE-Previous-Year-Question-Papers-Class-12-Maths-2017-Delhi-80.png?resize=362%2C295&ssl=1\")
\n![\"CBSE](\"https:\/\/i0.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/CBSE-Previous-Year-Question-Papers-Class-12-Maths-2017-Delhi-81.png?resize=395%2C497&ssl=1\")
\n![\"CBSE](\"https:\/\/i1.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/CBSE-Previous-Year-Question-Papers-Class-12-Maths-2017-Delhi-82.png?resize=394%2C298&ssl=1\")
<\/p>\n
\nA metal box with a square base and vertical sides is to contain 1024 cm3<\/sup>. The material for the top and bottom costs \u20b9 5 per cm2<\/sup> and the material for the sides costs \u20b92.50 per cm2<\/sup>. Find the least cost of the box. [6]
\nSolution:
\nLet the length, breadth and height of the metal box be x cm, x cm and y cm respectively. It is given that,
\n![\"CBSE](\"https:\/\/i1.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/CBSE-Previous-Year-Question-Papers-Class-12-Maths-2017-Delhi-83.png?resize=372%2C105&ssl=1\")
\n![\"CBSE](\"https:\/\/i2.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/CBSE-Previous-Year-Question-Papers-Class-12-Maths-2017-Delhi-84.png?resize=304%2C510&ssl=1\")
\n![\"CBSE](\"https:\/\/i2.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/CBSE-Previous-Year-Question-Papers-Class-12-Maths-2017-Delhi-85.png?resize=370%2C414&ssl=1\")
<\/p>\n
\nIf A = \\(\\left(\\begin{array}{rrr}{2} & {3} & {10} \\\\ {4} & {-6} & {5} \\\\ {6} & {9} & {-20}\\end{array}\\right)\\), find A-1<\/sup>. Using A-1<\/sup> solve the system of equations: [6]
\n![\"CBSE](\"https:\/\/i0.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/CBSE-Previous-Year-Question-Papers-Class-12-Maths-2017-Delhi-86.png?resize=136%2C109&ssl=1\")
\n![\"CBSE](\"https:\/\/i2.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/CBSE-Previous-Year-Question-Papers-Class-12-Maths-2017-Delhi-87.png?resize=142%2C55&ssl=1\")
\nSolution:
\n![\"CBSE](\"https:\/\/i1.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/CBSE-Previous-Year-Question-Papers-Class-12-Maths-2017-Delhi-88.png?resize=369%2C556&ssl=1\")
\n![\"CBSE](\"https:\/\/i2.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/CBSE-Previous-Year-Question-Papers-Class-12-Maths-2017-Delhi-89.png?resize=330%2C409&ssl=1\")
\n![\"CBSE](\"https:\/\/i0.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/CBSE-Previous-Year-Question-Papers-Class-12-Maths-2017-Delhi-90.png?resize=373%2C546&ssl=1\")
\n![\"CBSE](\"https:\/\/i2.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/CBSE-Previous-Year-Question-Papers-Class-12-Maths-2017-Delhi-91.png?resize=335%2C476&ssl=1\")
<\/p>\nCBSE Previous Year Question Papers Class 12 Maths 2017 Delhi Set III<\/h3>\n
![\"CBSE](\"https:\/\/i0.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/CBSE-Previous-Year-Question-Papers-Class-12-Maths-2017-Delhi-92.png?resize=339%2C76&ssl=1\")
![\"CBSE](\"https:\/\/i0.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/CBSE-Previous-Year-Question-Papers-Class-12-Maths-2017-Delhi-93.png?resize=354%2C295&ssl=1\")
<\/p>\n![\"CBSE](\"https:\/\/i1.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/CBSE-Previous-Year-Question-Papers-Class-12-Maths-2017-Delhi-94.png?resize=372%2C488&ssl=1\")
![\"CBSE](\"https:\/\/i1.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/CBSE-Previous-Year-Question-Papers-Class-12-Maths-2017-Delhi-95.png?resize=370%2C164&ssl=1\")
<\/p>\n![\"CBSE](\"https:\/\/i0.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/CBSE-Previous-Year-Question-Papers-Class-12-Maths-2017-Delhi-96.png?resize=368%2C248&ssl=1\")
![\"CBSE](\"https:\/\/i2.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/CBSE-Previous-Year-Question-Papers-Class-12-Maths-2017-Delhi-97.png?resize=371%2C420&ssl=1\")
<\/p>\n![\"CBSE](\"https:\/\/i1.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/CBSE-Previous-Year-Question-Papers-Class-12-Maths-2017-Delhi-98.png?resize=183%2C51&ssl=1\")
![\"CBSE](\"https:\/\/i0.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/CBSE-Previous-Year-Question-Papers-Class-12-Maths-2017-Delhi-99.png?resize=372%2C522&ssl=1\")
![\"CBSE](\"https:\/\/i2.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/CBSE-Previous-Year-Question-Papers-Class-12-Maths-2017-Delhi-100.png?resize=340%2C137&ssl=1\")
![\"CBSE](\"https:\/\/i1.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/CBSE-Previous-Year-Question-Papers-Class-12-Maths-2017-Delhi-101.png?resize=367%2C51&ssl=1\")
<\/p>\n![\"CBSE](\"https:\/\/i0.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/CBSE-Previous-Year-Question-Papers-Class-12-Maths-2017-Delhi-102.png?resize=369%2C322&ssl=1\")
<\/p>\n![\"CBSE](\"https:\/\/i2.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/CBSE-Previous-Year-Question-Papers-Class-12-Maths-2017-Delhi-103.png?resize=372%2C476&ssl=1\")
![\"CBSE](\"https:\/\/i0.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/CBSE-Previous-Year-Question-Papers-Class-12-Maths-2017-Delhi-104.png?resize=331%2C561&ssl=1\")
![\"CBSE](\"https:\/\/i1.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/CBSE-Previous-Year-Question-Papers-Class-12-Maths-2017-Delhi-105.png?resize=361%2C428&ssl=1\")
![\"CBSE](\"https:\/\/i2.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/CBSE-Previous-Year-Question-Papers-Class-12-Maths-2017-Delhi-106.png?resize=248%2C108&ssl=1\")
<\/p>\n![\"CBSE](\"https:\/\/i1.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/CBSE-Previous-Year-Question-Papers-Class-12-Maths-2017-Delhi-107.png?resize=402%2C542&ssl=1\")
![\"CBSE](\"https:\/\/i2.wp.com\/www.cbselabs.com\/wp-content\/uploads\/2019\/08\/CBSE-Previous-Year-Question-Papers-Class-12-Maths-2017-Delhi-108.png?resize=313%2C242&ssl=1\")
<\/p>\n