{"id":19763,"date":"2020-01-14T16:22:02","date_gmt":"2020-01-14T10:52:02","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=19763"},"modified":"2021-07-02T15:28:59","modified_gmt":"2021-07-02T09:58:59","slug":"2nd-puc-maths-question-bank-chapter-6-ex-6-5","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/2nd-puc-maths-question-bank-chapter-6-ex-6-5\/","title":{"rendered":"2nd PUC Maths Question Bank Chapter 6 Application of Derivatives Ex 6.5"},"content":{"rendered":"

Students can Download Maths Chapter 6 Application of Derivatives Ex 6.5 Questions and Answers, Notes Pdf, 2nd PUC Maths Question Bank with Answers<\/a>\u00a0helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.<\/p>\n

Karnataka 2nd PUC Maths Question Bank Chapter 6 Application of Derivatives Ex 6.5<\/h2>\n

2nd PUC Maths Application of Derivatives NCERT Text Book Questions and Answers Ex 6.5<\/h3>\n

Question 1.
\nFind the maximum and minimum values, if any, of the following functions given by<\/p>\n

(i) f(x) = (2x – 1)2<\/sup> + 3
\nAnswer:
\nf (x) = (2x – 1)2<\/sup> + 3
\nFor all values of x, f (x) > 3 (2x – 1)2<\/sup> + 3 > 3
\n\u2234 minimum value is 3 when 2x -1 = 0 is x = \\(\\frac{1}{2}\\) however the function has no maximum values as f (x) \u2192 \u221e as |x| \u221e<\/p>\n

(ii) f (x) = 9x2<\/sup> + 12x + 2
\nAnswer:
\nf (x) = 9x2<\/sup> + 12x + 2
\n\"2nd<\/p>\n

\"KSEEB<\/p>\n

(iii) f (x) = – (x – 1)2<\/sup> + 10
\nAnswer:
\nf (x) = – (x – 1)2<\/sup> + 10
\nf(x)= 10 – (x – 1)2
\n<\/sup>10 – (x – 1)2<\/sup> < 10
\nf(x) has a maximum value when x – 1 = 0,x= 1.
\nhow ever f (x) has no minimum value.<\/p>\n

(iv) g(x) = x3<\/sup> + 1
\nAnswer:
\ng(x) = x3<\/sup> + 1 as x \u2192 \u221e g (x) \u2192 \u221e and\u00a0a
\ns -+ – \u221e, g (x) \u21d2 – \u221e
\n\u2234 g (x) has neither minimum value nor maximum value.<\/p>\n

Question 2.
\nFind the maximum and minimum values, if any, of the following functions given by<\/p>\n

(i) f (x) = |x + 2| – 1
\nAnswer:
\nf (x) = |x + 2| – 1
\nf (x) = |x + 2| – 1 \u2265 -1
\nminimum value is – 1 when x + 2 = 0, x = -2
\nhowever it has no maximum value.<\/p>\n

(ii) g(x) = – | x + 1| + 3
\nAnswer:
\ng(x) = – | x + 1| + 3 = 3,-1 x + 1|
\ng (x) < 3 v x + 1 = 0
\n\u2234 max. value is 3
\nwhen x = – 1
\nhow ever no minimum value.<\/p>\n

(iii) h(x) = sin (2x) + 5
\nAnswer:
\nh(x) = sin (2x) + 5
\nmaximum value of sin 2x = 1 and minimum value is -1
\nf(x) = 1 + 5 is 1 + 5
\n\u2234 max. h (x) = 6 and
\nmin. h (x) = 4.<\/p>\n

(iv) f (x) = |sin 4x + 3|
\nAnswer:
\nf (x) = |sin 4x + 3|
\n-1 < sin 4x < 1 \u21d2 3 -1 < sin 4x + 3 < + 1 + 3 + 2′< sin 4 x + 3 < 4
\n2 < | sin 4x + 3 | < 4
\nf (x) > 2 and f (x) < 4
\nmin. f (x) = 2 when sin 4x + 3 = 0
\nmax f (x) = 4 when sin 4x + 3 = 0
\n\u2234 minimum value is 2 at sin 4x = -1
\nmaximum value is 4 at sin 4x = 1<\/p>\n

\"KSEEB<\/p>\n

(v) h(x) = x + 1, x \u2208 (- 1, 1)
\nAns:
\nh(x) = x + 1, x \u2208 (-1,1)
\ngiven that x \u2208 (-1, 1)
\nie. -1 < x < 1
\n-1 + 1< x + 1 < 1 + 10 < x + 1< 2
\n\u2234 x + 1 > 0 or x + 1 < 2
\nx + 1 > 0 so no minimum value
\nx + 1 < 2, so no maximum value.<\/p>\n

Question 3.
\nFind the local maxima and local minima, if any, of the following functions. Find also the local maximum and the local minimum values, as the case may be:
\n(i) f (x) = x2
\n<\/sup>Answer:
\nf (x) = x2
\n<\/sup>f’ (x) = 2x, f’ (x) = 0
\n\u21d2 2x = 0, x = 0
\nf'(x)= 2
\nf'(x)> 0, hence f (x) has minimum value at x = 0
\nand the minimum value is (0)2<\/sup>= 0.<\/p>\n

(ii) g(x) = x3<\/sup> – 3x
\nAnswer:
\n\"2nd<\/p>\n

(iii) h(x) = sin x + cos x, o < x< \\(\\frac{\\pi}{2}\\)
\nAnswer:
\n\"2nd<\/p>\n

\"KSEEB<\/p>\n

(iv) f (x) = sin x – cos x, 0 < x < 2\u03c0
\nAnswer:
\n\"2nd
\n\"2nd
\n\"2nd<\/p>\n

(v) f (x) = x3<\/sup> – 6x2<\/sup> + 9x + 15
\nAnswer:
\n\"2nd<\/p>\n

\"KSEEB<\/p>\n

(vi) \\(g(x)=\\frac{x}{2}+\\frac{2}{x}, x>0\\)
\nAnswer:
\n\"2nd<\/p>\n

(vii) \\(g(x)=\\frac{1}{x^{2}+2}\\)
\nAnswer:
\n\"2nd<\/p>\n

(viii) \\(f(x)=x \\sqrt{1-x}, x>0\\)
\nAnswer:
\n\"2nd
\n\"2nd<\/p>\n

Question 4.
\nProve that the following functions do not have maxima or minima :<\/p>\n

(i) f (x) = ex
\n<\/sup>Answer:
\nf (x) = ex<\/sup>
\nf’ (x) = ex
\n<\/sup>f'(x) > 0 \u2200 x \u2208 R
\nhence function has no critical point There is no point at which the function is maximum or minimum.<\/p>\n

\"KSEEB<\/p>\n

(ii) g(x) = log x, x > 0
\nAnswer:
\ng (x) = log x, x > 0
\ng'(x) = \\(\\frac{1}{x}\\), where x > 0
\nhence the function has no critical point
\n\u2234 There is no point at which the function is maximum or minimum.<\/p>\n

(iii) h (x) = x3<\/sup> + x1<\/sup> + x +1
\nAnswer:
\nh (x) = x3<\/sup>+ x2<\/sup>+ x +1
\nh’ (x) = 3x2<\/sup> + 2x + 1
\nh’ (x) = 0 \u21d2 3x2<\/sup> + 2x + 1 = 0,
\nx has no real value, hence there is no critical point.
\n\u2234 For no point the function has max. or min. value.<\/p>\n

Question 5.
\nFind the absolute maximum value and the absolute minimum value of the following functions in the given intervals:<\/p>\n

(i) f(x) = x3<\/sup>,x \u2208 [-2,2]
\nAnswer:
\nf (x) = x3<\/sup>,x \u2208 [- 2,2]
\nf’ (x) = 0 \u21d2 3x2<\/sup> = 0 \u21d2 x = 0 for finding the absolute maximum and absolute minimum, we have to evaluate f (0), f (2), f (-2)
\nF (0) = (0)3<\/sup> = 0, F (2) = (2)3<\/sup> = 8,
\nF (-2) = (-2)3<\/sup> = – 8
\nAbsolute maximum = 8 and
\nAbsolute minimum = -8
\n\u2234 maximum at 2 is 8 and minimum at -2 is – 8.<\/p>\n

\"KSEEB<\/p>\n

(ii) f (x) = sin x + cos x , x \u2208 [0, \u03c0]
\nAnswer:
\n\"2nd<\/p>\n

(iii) \\(f(x)=4 x-\\frac{1}{2} x^{2}, x \\in[-2,9 \/ 2]\\)
\nAnswer:
\n\"2nd<\/p>\n

\"KSEEB<\/p>\n

(iv) f (x) = (x – 1)2<\/sup> + 3, x \u2208 [-3,1]
\nAnswer:
\nf (x) = (x – 1)2<\/sup> + 3, x \u2208 [-3,1]
\nf'(x) = 2 (x – 1)
\nf’ (x) = 0 \u21d2 (x – 1) = 0
\n\u21d2 x = 1 we will evaluate f (-3) and f (1)
\nf(-3) = (-3 – 1)2<\/sup> + 3= 16 + 3 = 19
\nf (1) = (0)2<\/sup> + 3 = 3
\nAbsolute maximum at x = -3 is 19 and
\nAbsolute minimum x = 1 is 3.<\/p>\n

Question 6.
\nFind the maximum profit that a company can make, if the profit function is given by
\np(x) = 41+24x – 18x2 <\/sup>
\nAnswer:
\n\"2nd<\/p>\n

Question 7.
\nFind both the maximum value and the minimum value of
\n3x4<\/sup> – 8x3<\/sup> + 12x2<\/sup> – 48x + 25 on the interval [0, 3].
\nAnswer:
\nf (x) = 3x4<\/sup> – 8x3<\/sup> + 12x2<\/sup> – 48 x + 25
\nf’ (x) = 12x3<\/sup> – 24x2<\/sup> + 24x – 48
\nf<\/sub>‘(x) = 0 \u21d2 12 (x3<\/sup> – 2x2<\/sup> + 2x -4) = 0
\n\u21d2 12 (x2<\/sup> (x – 2) + 2 (x – 2))
\n\u21d2 12 ( (x – 2) (x2<\/sup> + 2) ) = 0 (x – 2)(x2<\/sup> + 2) = 0
\n\u21d2 x = 2 but x2<\/sup> + 2 \u2260 0
\nThe points are f (0), f (2), f (3)
\nf (x) = 3x4<\/sup> – 8x3<\/sup> + 12x2<\/sup> – 48x + 25
\nf (0) = 25
\nf (2) = 3 (16) – 8 (8) + 12 (4) – 48 (2) + 25 = 48 – 64 + 48 – 96 + 25 = -39
\nf(3) = 3 (34<\/sup>) – 8 (33<\/sup>) + 12 (32<\/sup>) – 48 (3) + 25 = 243 – 216 + 108 – 144 + 25
\n376 – 360 = 16
\n\u2234 maximum of f (x) at x = 0 is 25
\nminimum of f (x) at x = 2 is – 39.<\/p>\n

\"KSEEB<\/p>\n

Question 8.
\nAt what points in the interval [0,2\u03c0] does the function sin 2x attain its maximum value?
\nAnswer:
\n\"2nd<\/p>\n

Question 9.
\nWhat is the maximum value of the function sin x + cos x on \\(\\left[0, \\frac{2}{\\pi}\\right]\\)
\nAnswer:
\n\"2nd
\n\"2nd<\/p>\n

Question 10.
\nFind the maximum value of 2x3<\/sup> – 24x + 107 in the interval [1, 3]. Find the maximum value of the same function in [-3, -1]
\nAnswer:
\nf (x) = 2x3<\/sup> – 24x +107
\nf’ (x) = 6x2<\/sup> – 24 \u21d2 x2<\/sup> = 4, x = \u00b1 2 + 2 s [1,3]
\n\u2234 we have to evaluate
\nf (1). f (2), f (3)
\nf (1) = 2 (1)3<\/sup> – 24 x 1 + 107 = 85
\nf (2) = 2 (2)3<\/sup>– 24 x 2+ 107 = 75
\nf(3) = 2(3)3<\/sup>-24 x 3 + 107 = 89
\nmaximum at x = 3 and miximum value is 89 minimum at x = 2, is 75
\nNow the interval is [-3, -1]
\n\u2234 points are -3, -2, -1
\nf (-3) = 2 (-3)3<\/sup> – 24 (-3) + 107 = 125
\nf (-2) = 2 (-2)3<\/sup> – 24 (-2) + 107
\n= 2 (-8+ 48 + 107 = 139 = 139
\nf (-1) = 2 (-1) 3-24 (-1) + 107 = 129
\nmaximum at -2 is 139
\nminimum at (-3) is 125.<\/p>\n

\"KSEEB<\/p>\n

Question 11.
\nIt is given that at x = 1, the function x4<\/sup> – 62x2<\/sup> + ax + 9 attains its maximum value, on the interval [0,2]. Find the value of a.
\nAnswer:
\nf (x) = x4<\/sup> – 62x2<\/sup> + ax + 9
\nf'(x) = 4x3<\/sup> – 124x + a
\nat x = 1, the function has maximum value
\n\u2234 f'(1) = 0 \u21d2 4 (1)2<\/sup> -124 (1) + a = 0, a = 120
\n\u2234 when a = 120, the function attains maximum value.<\/p>\n

Question 12.
\nFind the maximum and minimum values of x + sin 2x on [0,2\u03c0]
\nAnswer:
\n\"2nd<\/p>\n

Question 13.
\nFind two numbers whose sum is 24 and whose product is as large as possible.
\nAnswer:
\n\"2nd<\/p>\n

\"KSEEB<\/p>\n

Question 14.
\nFind two positive numbers x and y such that x + y = 60 and xy3<\/sup> is maximum.
\nAnswer:
\nx + y = 60 \u21d2 y = 60 – x
\n\"2nd
\n\"2nd<\/p>\n

Question 15.
\nFind two positive numbers x and y such that their sum is 35 and the product x2<\/sup> y5 <\/sup>is a maximum.
\nAnswer:
\n\"2nd
\n\"2nd<\/p>\n

\"KSEEB<\/p>\n

Question 16.
\nFind two positive numbers whose sum is 16 and the sum of whose cubes is minimum.
\nAnswer:
\n\"2nd<\/p>\n

Question 17.
\nA square piece of tin of side 18 cm \u00a1s to be made into a box without top, by cutting a square from each corner and folding up the flaps to form the box. What should be the side of the square to be cut off so that the volume of the box is the maximum possible.
\nAnswer:
\n\"2nd
\n\"2nd
\nx = 9 is not possible, it so side = 0
\nvolume is maximum when the sides are 12,12,3
\nvolume =12 x 12 x 3 = 432 cm3
\n<\/sup>The volume is maximum when the side of the square to be cut off is 3cm.<\/p>\n

Question 18.
\nA rectangular sheet of tin 45 cm by 24 cm is to be made into a box without top, by cutting off square from each corner and folding up the flaps. What should be the side of the square to be cut off so that the volume of the box is maximum ?
\nAnswer:
\nThe sides of the box are x, 45 – 24, 45 – 2x
\n\"2nd
\n\"2nd<\/p>\n

\"KSEEB<\/p>\n

Question 19.
\nShow that of all the rectangles inscribed in a given fixed circle, the square has the maximum area.
\nAnswer:
\nLet the sides of rectangle be x and y and radius of the circle is r
\n\"2nd
\nhence rectangle becomes square.
\n\u2234 The area is maximum when the rectangle is a square.<\/p>\n

Question 20.
\nShow that the right circular cylinder of given surface and maximum volume is such that its height is equal to the diameter of the base.
\nAnswer:
\n\"2nd<\/p>\n

Question 21.
\nOf all the closed cylindrical cans (right circular), of a given volume of 100 cubic centimetres, find the dimensions of the can which has the minimum surface area?
\nAnswer:
\n\"2nd
\n\"2nd<\/p>\n

\"KSEEB<\/p>\n

Question 22.
\nA wire of length 28 m is to be cut into two pieces. One of the pieces is to be made into a square and the other into a circle. What should be the length of the two pieces so that the combined area of the square and the circle is minimum?
\nAnswer:
\n\"2nd
\n\"2nd<\/p>\n

\"KSEEB<\/p>\n

Question 23.
\nProve that the volume of the largest cone that can be inscribed in a sphere of radius
\nR is \\(\\frac{8}{27} \\)of the volume of the sphere.
\nAnswer:
\n\"2nd
\n\"2nd<\/p>\n

Question 24.
\nShow that the right circular cone of least curved surface and given volume has an altitude equal to \\(\\sqrt{2}\\) time the radius of the base.
\nAnswer:
\n\"2nd
\n\"2nd
\n\"2nd<\/p>\n

Question 25.
\nShow that the semi-vertical angle of the cone of the maximum volume and of given slant height is \\(\\tan ^{-1} \\sqrt{2}\\).
\nAnswer:
\n\"2nd
\n\"2nd<\/p>\n

\"KSEEB<\/p>\n

Question 26.
\nShow that semi-vertical angle of right circular cone of given surface area and maximum volume is \\(\\sin ^{-1}\\left(\\frac{1}{3}\\right)\\)
\nAnswer:
\n\"2nd
\n\"2nd<\/p>\n

Choose the correct answer in the Exercises 27 and 29.<\/p>\n

Question 27.
\nThe point on the curve x2<\/sup> = 2y which is nearest to the point (0, 5) is
\n(A) (2,\\(\\sqrt{2}\\),4)
\n(B) (2, \\(\\sqrt{2}\\), 0)
\n(C) (0,0)
\n(D) (2,2)
\nAnswer:
\n\"2nd
\n\"2nd<\/p>\n

\"KSEEB<\/p>\n

Question 28.
\nFor all real values of x, the minimum value of \\(\\frac{1-x+x^{2}}{1+x+x^{2}}\\) is
\n(A) 0
\n(B) 1
\n(C) 3
\n(D) 1\/3
\nAnswer:
\n\"2nd<\/p>\n

Question 29.
\nThe maximum value of [x(x – 1) + 1]1\/3<\/sup>< x < 1 is 0 \u2264 x \u2264 1 is
\nAnswer:
\n\"2nd
\n\"2nd<\/p>\n","protected":false},"excerpt":{"rendered":"

Students can Download Maths Chapter 6 Application of Derivatives Ex 6.5 Questions and Answers, Notes Pdf, 2nd PUC Maths Question Bank with Answers\u00a0helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations. Karnataka 2nd PUC Maths Question Bank Chapter 6 Application of Derivatives Ex 6.5 2nd PUC Maths …<\/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\/19763"}],"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=19763"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/19763\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=19763"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=19763"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=19763"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}