{"id":31479,"date":"2022-03-29T14:30:42","date_gmt":"2022-03-29T09:00:42","guid":{"rendered":"https:\/\/mcq-questions.com\/?p=31479"},"modified":"2022-03-29T15:06:35","modified_gmt":"2022-03-29T09:36:35","slug":"ncert-solutions-for-class-12-maths-chapter-6-ex-6-5","status":"publish","type":"post","link":"https:\/\/mcq-questions.com\/ncert-solutions-for-class-12-maths-chapter-6-ex-6-5\/","title":{"rendered":"NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Ex 6.5"},"content":{"rendered":"

These NCERT Solutions for Class 12 Maths<\/a> Chapter 6 Application of Derivatives Ex 6.5 Questions and Answers are prepared by our highly skilled subject experts. https:\/\/mcq-questions.com\/ncert-solutions-for-class-12-maths-chapter-6-ex-6-5\/<\/p>\n

NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Exercise 6.5<\/h2>\n

\"NCERT<\/p>\n

Ex 6.5 Class 12 Question 1.<\/strong>
\nFind the maximum and minimum values, if any, of the following functions given by
\n(i) f(x) = (2x – 1)\u00b2 + 3
\n(ii) f(x) = 9x\u00b2 + 12x + 2
\n(iii) f(x) = – (x – 1)\u00b2 + 10
\n(iv) g(x) = x3<\/sup> + 1
\nSolution:
\nMethod I:
\n(i) f(x) = (2x – 1)\u00b2 + 3
\n\u2234 f(x) \u2265 3 since (2x – 1)\u00b2 \u2265 0 for x \u2208 R
\nHence the minimum value of f is 3, when
\n2x – 1 or x = \\(\\frac { 1 }{ 2 }\\)
\nThere is no maximum value since
\nf(x) \u2192 \u221e as x \u2192 \u221e and x \u2192 – \u221e<\/p>\n

Method II:
\nf(x) = (2x – 1)\u00b2 + 3
\nThen f'(x) = 2(2x – 1)
\n\u2234 f'(x) = 0 \u21d2 2(2x – 1) = 0 \u21d2 \\(\\frac { 1 }{ 2 }\\)
\nSign of f'(x) = 2(2x – 1)
\nf'(x) = 0 \u21d2 6(3x + 2) = 0 \u21d2 x = \\(\\frac { – 2 }{ 3 }\\)
\nSign of f'(x) = 6(3x + 2)
\n\"Ex
\nf has no maximum value<\/p>\n

Method III
\nf(x) – 9x\u00b2 + 12x + 2
\nf'(x) = 18x + 12 and f”'(x) = 18
\nf'(x) = 0 \u21d2 18x + 12 = 0
\n\u21d2 6(3x + 2) = 0
\n\u21d2 x = \\(\\frac { – 2 }{ 3 }\\)
\n\"Exercise
\nf has no maximum value<\/p>\n

iii.
\nMethod I
\nf(x) = – (x – 1)\u00b2 + 10 = 10 – (x – 1)\u00b2
\n\u2234 f'(x) \u2264 10 since (x – 1)\u00b2 \u2265 0 for all x \u2208 R
\nThe maximum value of is 10, when x – 1 = 0
\nof x = 1
\nf has no minimum value as f(x) \u2192 – \u221e as x \u2192 \u221e and x \u2192 – \u221e.<\/p>\n

Method II
\nf(x) = – (x – 1)\u00b2 + 10, f'(x) = – 2(x – 1)
\n\"6.5
\nThere is no maximum value<\/p>\n

Method III (Second derivative test)
\n\"Ex6.5
\nThere is no maximum value<\/p>\n

ii.
\nMethod I
\nf(x) = 9x\u00b2 + 12x + 2 = (3x + 2)\u00b2 – 2
\nf(x) \u2265 – 2 since (3x + 2)\u00b2 \u2265 0 for all x \u2208 R
\nThe minimum value of f is – 2, when
\n(3x + 2) = 0 or when x = \\(\\frac { – 2 }{ 3 }\\)
\nf has no maximum value as f(x) \u2192 \u221e as x \u2192 \u221e and x \u2192 \u221e<\/p>\n

Method II
\nf(x) = 9x\u00b2 + 12x + 2
\nf'(x) = 18x + 12 = 6(3x + 2)
\nf'(x) = 0 \u21d2 – 2(x – 1) = 0 \u21d2 x = 1
\nSign of f'(x) = – 2(x – 1)
\n\"Ex
\n\u2234 f has a maximum at x – 1 and the maxi-mum value = f(1) = -(1 – 1)\u00b2 + 10 = 10
\nThere is no minimum value<\/p>\n

Method III
\nf(x) = – (x – 1)\u00b2 + 10
\nf'(x) = – 2 (x – 1) and f”(x) = – 2
\nf'(x) = 0 \u21d2 – 2 (x – 1) = 0 \u21d2 x = 1
\nf”(1) = – 2, is negative
\n\u2234 f has a maximum value at x = 1
\nThe maximum value = f(1) = – (1 – 1)\u00b2 + 10 = 10
\nf has no minimum value<\/p>\n

iv. Method I
\ng(x) = x\u00b3 + 1
\nWhen x \u2192 \u221e, g(x) \u2192 \u221e and
\nWhen x \u2192 – \u221e, g(x) \u2192 – \u221e
\n\u2234 g does not attain a maximum value or a minimum value.<\/p>\n

Method II
\ng(x) = x\u00b3 + 1 g'(x) = 3x\u00b2
\ng'(x) = 0 \u21d2 3x\u00b2 = 0 \u21d2 x = 0
\nSign of g'(x) = 3x\u00b2
\n\"Exercise
\ng does not attain a maximum value or a minimum value<\/p>\n

Method III
\ng(x) = x\u00b3 + 1
\ng'(x) = 3x\u00b2 and g”(x) = 6x
\ng'(x) = 0 \u21d2 3x\u00b2 = 0 \u21d2 x = 0
\ng'(0) = 6(0) = b
\nHence the second derivative test fails.
\nSo use the first derivative test (Method II).<\/p>\n

\"NCERT<\/p>\n

Exercise 6.5 Class 12 Question 2.<\/strong>
\nFind the maximum and minimum values, if any, of the following functions given by
\n(i) f(x) = |x + 2| – 1
\n(ii) g(x) = -|x + 1| + 3
\n(iii) h (x) = sin 2x + 5
\n(iv) f(x) = |sin(4x + 3)|
\n(v) h(x) = x + 1, x \u2208 (- 1, 1)
\nSolution:
\n(i) We have : f(x) = |x + 2 |-1 \u2200x\u2208R
\nNow |x + 2|\u22650\u2200x \u2208 R
\n|x + 2| – 1 \u2265 – 1 \u2200x \u2208 R ,
\nSo -1 is the min. value of f(x)
\nnow f(x) = – 1
\n\u21d2 |x + 2|- 1
\n\u21d2 |x + 2| = 0
\n\u21d2 x = – 2<\/p>\n

(ii) We have g(x) = – |x + 1| + 3 \u2200x \u2208 R
\nNow | x + 1| \u2265 0 \u2200x \u2208 R
\n-|x + 1| + 3 \u22643 \u2200x \u2208 R
\nSo 3 is the minimum value of f(x).
\nNow f(x) = 3
\n\u21d2 -|x + 1| + 3
\n\u21d2 |x + 1| = 0
\n\u21d2 x = – 1.<\/p>\n

(iii) Thus maximum value of f(x) is 6 and minimum value is 4.<\/p>\n

(iv) Let f(x) = |sin 4x + 3|
\nMaximum value of sin 4x is 1
\n\u2234 Maximum value of |sin(4x+3)| is |1+3| = 4
\nMinimum value of sin 4x is -1
\n\u2234 Minimum value of f(x) is |-1+3| = |2|= 2<\/p>\n

(v) h(x) = x + 1 x \u2208 (- 1, 1)
\ni.e., – 1 < x < 1
\n\u2234 – 1 + 1 < x + 1 < 2
\ni.e., 0 < x + 1 < 2
\ni.e., 0 < h(x) < 2
\nHence h(x) does not attain a maximum value or a minimum value
\nAnother Method
\nh(x) = x + 1
\n\u2234 h'(x) = 1
\nh'(x) \u2260 0 for x \u2208 (- 1, 1)
\n\u2234 h(x) does not attain a maximum value or a minimum value in ( – 1, 1)<\/p>\n

6.5 Class 12 Question 3.<\/strong>
\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<\/sup>
\n(ii) g(x) = x3<\/sup> – 3x
\n(iii) h(x) = sinx + cosx, 0 < x < \\(\\frac { \\pi }{ 2 }\\)
\n(iv) f(x) = sin x – cos x, 0 < x < 2\u03c0
\n(v) f(x) = x3\u00a0<\/sup> – 6x2<\/sup> + 9x + 15
\n(vi) g(x) = \\(\\frac { x }{ 2 } +\\frac { 2 }{ x }\\), x > 0
\n(vii) g(x) = \\(\\frac { 1 }{ { x }^{ 2 }+2 }\\)
\n(viii) f(x) = \\(x\\sqrt { 1-x } \\), x < 1
\nSolution:
\n(i) f(x) = x2<\/sup>
\nNow f'(x) = 0 \u21d2 2x = 0 i.e., x = 0
\nAt x = 0; When x is slightly < 0, f’ (x) is -ve When x is slightly > 0, f(x) is +ve
\n\u2234 f(x) changes sign from -ve to +ve as x increases through 0.
\n\u21d2 f’ (x) has a local minimum at x = 0 local minimum value f(0) = 0.<\/p>\n

(ii) g(x) = x3<\/sup> – 3x
\ng'(x) = 3x\u00b2 – 3 = 3(x\u00b2 – 1)
\ng'(x) = 0 \u21d2 3(x\u00b2 – 1) = 0
\n\u21d2 3(x + 1) (x – 1) = 0
\n\u21d2 x = – 1 or x = 1
\ng”(x) = 6x
\ng”(-1) = 6(- 1) = – 6 < 0 g”(1) = 6(1) = 6 > 0
\n\u2234 By the second derivative test,
\nx = – 1 is a point of local maximum,
\nx = 1 is a point of local minimum
\nThe local maximum value = g(-1)
\n= (-1)3<\/sup> – 3(- 1) = 2
\nThe local minimum value = g(1)
\n= (1)3<\/sup> – 3(1) = – 2<\/p>\n

(iii) h(x) = sinx + cosx, 0 < x < \\(\\frac { \\pi }{ 2 }\\)
\nh(x) = cos x – sin x
\nh(x) = 0 \u21d2 cos x – sin x = 0
\n\"Ex
\nTherefore, by second derivative test, x = \\(\\frac { \\pi }{ 4 }\\)
\nis a local maximum point
\nThe local maximum value = h\\(\\frac { \\pi }{ 4 }\\)
\n= \\(\\sin \\frac{\\pi}{4}+\\cos \\frac{\\pi}{4}\\)
\n= \\(=\\frac{1}{\\sqrt{2}}+\\frac{1}{\\sqrt{2}}=\\frac{2}{\\sqrt{2}}=\\sqrt{2}\\)<\/p>\n

(iv) f(x) = sin x – cos x, 0 < x < 2\u03c0
\nf'(x) = cosx + sinx
\nf”(x) = – sinx + cosx
\nf'(x) = 0 \u21d2 cosx + sinx 0
\n\u21d2 sinx = – cosx
\n\"Class<\/p>\n

(v) f(x) = x3\u00a0<\/sup> – 6x2<\/sup> + 9x + 15
\nf'(x) = 3x\u00b2 – 12x + 9 = 3(x\u00b2 – 4x + 3)
\nf”(x) = 6x – 12
\nf'(x) = 0 \u21d2 3(x\u00b2 – 4x + 3) = 0
\n\u21d2 3(x – 1) (x – 3) = 0
\n\u21d2 x = 1 or x = 3
\nAt x = 1, f”(x) = 6(1) – 12 = 6 – 12 < 0 At x = 3, f”(x) = 6(3) – 12 = 18 – 12 >0
\nf has a local maxima at x = 1 and a local minima at x = 3
\nThe local maximum value = f(1)
\n= (1)\u00b3 – 6 (1)\u00b2 + 9(1) + 15 = 19
\nThe local minimum value = f(3)
\n= (3)\u00b3 – 6(3)\u00b2 + 9(3) +15 = 15<\/p>\n

(vi) g(x) = \\(\\frac { x }{ 2 } +\\frac { 2 }{ x }\\), x > 0
\n\"6.5<\/p>\n

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

(viii) f(x) = \\(x\\sqrt { 1-x } \\), x < 1
\n\"6.5<\/p>\n

\"NCERT<\/p>\n

Ex6.5 Class 12 Question 4.<\/strong>
\nProve that the following functions do not have maxima or minima:
\n(i) f(x) = ex<\/sup>
\n(ii) f(x) = log x
\n(iii) h(x) = x3<\/sup> + x2<\/sup> + x + 1
\nSolution:
\n(i) f'(x) = ex<\/sup>;
\nSince f’ (x) \u2260 0 for any value of x.
\nSo f(x) = ex<\/sup> does not have a max. or min.<\/p>\n

(ii) f’ (x) = \\(\\frac { 1 }{ x }\\); Clearly f’ (x) \u2260 0 for any value of x.
\nSo, f’ (x) = log x does not have a maximum or a minimum.<\/p>\n

(iii) We have f(x) = x3<\/sup> + x2<\/sup> + x + 1
\n\u21d2 f’ (x) = 3x2<\/sup> + 2x + 1
\nNow, f’ (x) = 0 = > 3x2<\/sup> + 2x + 1 = 0
\n\\(x=\\frac { -2\\pm \\sqrt { 4-12 } }{ 6 } =\\frac { -1+\\sqrt { -2 } }{ 3 } \\)
\ni.e. f'(x) = 0 at imaginary points
\ni.e. f'(x) \u2260 0 for any real value of x
\nHence, there is neither max. nor min.<\/p>\n

Ex 6.5 Class 12 Maths Ncert Solutions Question 5.<\/strong>
\nFind the absolute maximum value and the absolute minimum value of the following functions in the given intervals:
\n(i) f(x) = x3<\/sup>, x \u2208 [- 2, 2]
\n(ii) f(x) = sin x + cos x, x \u2208 [0, \u03c0]
\n(iii) f(x) = \\(4x-\\frac { 1 }{ 2 } { x }^{ 2 },x\\in \\left[ -2,\\frac { 9 }{ 2 } \\right] \\)
\n(iv) f(x) = \\({ (x-1) }^{ 2 }+3,x\\in \\left[ -3,1 \\right] \\)
\nSolution:
\n(i) We have f'(x) = x3<\/sup> in [ – 2, 2]
\n\u2234 f'(x) = 3x\u00b2; Now, f’ (x) = 0 at x = 0, f(0) = 0
\nNow, f(- 2) = (- 2)3<\/sup> = – 8;
\nf(0) = (0)\u00b3 = 0
\nand f(0) = (2) = 8
\nHence, the absolute maximum value of f (x) is 8 which it attained at x = 2 and absolute minimum value of f(x) = – 8 which is attained at x = -2.<\/p>\n

(ii)
\n\"Class<\/p>\n

(iii) f(x) = \\(4x-\\frac { 1 }{ 2 } { x }^{ 2 },x\\in \\left[ -2,\\frac { 9 }{ 2 } \\right] \\)
\nf'(x) = 4 – x
\nf'(x) = 0 \u21d2 4 – x = 0 \u21d2 x = 4
\nf(- 2) = 4(-2) – \\(\\frac { 1 }{ 2 }\\)(-2)\u00b2 = – 8 – 2 = – 10
\nf(4) = 4(4) – \\(\\frac { 1 }{ 2 }\\)(4)\u00b2 = 16 – 8 = 8
\n\"Exercise<\/p>\n

(iv) f(x) = \\({ (x-1) }^{ 2 }+3,x\\in \\left[ -3,1 \\right] \\)
\nf'(x) = 2(x – 1)
\nf'(x) = 0 \u21d2 2(x – 1) = 0 \u21d2 x = 1
\nf(- 3) = (- 3 – 1)\u00b2 + 3 = 16 + 3 = 19
\nf(1) = (1 – 1)\u00b2 + 3 = 0 + 3 = 3
\n\u2234 Absolute minimum value = Min{19, 3} = 3
\nAbsolute maximum value = Max{19, 3} = 19<\/p>\n

Exercise 6.5 Class 12 Maths Question 6.<\/strong>
\nFind the maximum profit that a company can make, if the profit function is given by p(x) = 41 – 24x – 18x\u00b2
\nSolution:
\nProfit function in p(x) = 41 – 24x – 18x\u00b2
\n\u2234 p'(x) = – 24 – 36x = – 12(2 + 3x)
\nfor maxima and minima, p'(x) = 0
\nNow, p'(x) = 0
\n\u21d2 – 12(2 + 3x) = 0
\n\u21d2 x = \\(-\\frac { 2 }{ 3 }\\),
\np'(x) changes sign from +ve to -ve.
\n\u21d2 p (x) has maximum value at x = \\(-\\frac { 2 }{ 3 }\\)
\nMaximum Profit = 41 + 16 – 8 = \u20b9 49.<\/p>\n

\"NCERT<\/p>\n

Ex 6.5 Class 12 Maths Question 7.<\/strong>
\nFind both the maximum value and the minimum value of 3x4<\/sup> – 8x3<\/sup> + 12x2<\/sup> – 48x + 25 on the interval [0,3].
\nSolution:
\nLet f(x) = 3x4<\/sup> – 8x3<\/sup> + 12x2<\/sup> – 48x + 25
\n\u2234 f'(x) = 12x3<\/sup> – 24x2<\/sup> + 24x – 48
\n= 12(x2<\/sup> + 2)(x – 2)
\nFor maxima and minima, f'(x) = 0
\n\u21d2 12(x2<\/sup> + 2)(x – 2) = 0
\n\u21d2 x = 2
\nNow, we find f (x) at x = 0,2 and 3, f (0) = 25,
\nf (2) = 3 (24<\/sup>) – 8 (23<\/sup>) + 12 (22<\/sup>) – 48 (2) + 25 = – 39
\nand f (3) = (34<\/sup>) – 8 (33<\/sup>) + 12 (32<\/sup>) – 48 (3) + 25
\n= 243 – 216 + 108 – 144 + 25 = 16
\nHence at x = 0, Maximum value = 25
\nat x = 2, Minimum value = – 39.<\/p>\n

Class 12 Ex 6.5 Question 8.<\/strong>
\nAt what points in the interval [0, 2\u03c0], does the function sin 2x attain its maximum value?
\nSolution:
\nWe have f (x) = sin 2x in [0, 2\u03c0], f'(x) = 2 cos 2 x
\nFor maxima and minima f’ (x) = 0 \u21d2 cos 2 x = 0
\n\"Ncert
\nMaximum value of f =
\nMax{1, – 1, 1, – 1, 0} = 1
\n\u2234 sin 2x attains the maximum value at x = \\(\\frac { \u03c0 }{ 4 }\\), x = \\(\\frac { 5\u03c0 }{ 4 }\\)<\/p>\n

6.5 Maths Class 12 Question 9.<\/strong>
\nWhat is the maximum value of the function sin x + cos x?
\nSolution:
\n\"NCERT
\n\u2234 The maximum value of f(x) is \\(\\sqrt{2}\\)<\/p>\n

Class 12 Maths Ex 6.5 Question 10.<\/strong>
\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].
\nSolution:
\n\u2235 f(x) = 2x3<\/sup>\u00a0– 24x + 107
\n\u2234 f(x) = 6x2<\/sup> – 24 ,
\nFor maxima and minima f'(x) = 0;
\n\u21d2 x = \u00b12
\nFor the interval [1, 3], we find the values of f(x)
\nat x = 1,2,3; f(1) = 85, f(2) = 75, f(3) = 89
\nHence, maximum f (x) = 89 at x = 3
\nFor the interval [-3, -1], we find the values of f(x) at x = – 3, – 2, – 1;
\nf(-3) = 125;
\nf(-2) = 139
\nf(-1) = 129
\n\u2234 max.f(x) = 139 at x = – 2.<\/p>\n

6.5 Class 12 Maths Question 11.<\/strong>
\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.
\nSolution:
\n\u2235 f(x) = x4<\/sup> – 62x2<\/sup> + ax + 9, x \u2208 [0, 2]
\n\u2234 f'(x) = 4x3<\/sup> – 124x + a
\nNow f’ (x) = 0 at x = 1
\n\u21d2 4 – 124 + a = 0
\n\u21d2 a = 120
\nNow f” (x) = 12x2<\/sup> – 124:
\nAt x = 1 f” (1) = 12 – 124 = – 112 < 0
\n\u21d2 f(x) has a maximum at x = 1 when a = 120.<\/p>\n

\"NCERT<\/p>\n

Class 12 Maths Ncert Solutions Chapter 6 Exercise 6.5 Question 12.<\/strong>
\nFind the maximum and minimum values of x + sin 2x on [0, 2\u03c0]
\nSolution:
\n\u2234 f(x) = x + sin2x \u2208 [0, 2\u03c0]
\n\u2234 f’ (x) = 1 + 2 cos2x
\nFor maxima and minima f'(x) = 0
\n\"NCERT<\/p>\n

Exercise 6.5 Class 12 Ncert Solutions Question 13.<\/strong>
\nFind two numbers whose sum is 24 and whose product is as large as possible.
\nSolution:
\nLet the required numbers hex and (24 – x)
\n\u2234 Their product, p = x(24 – x) = 24x – x\u00b2
\nNow \\(\\frac { dp }{ dx }\\) = 0 \u21d224 – 2x = 0 \u21d2 x = 12
\nAlso \\(\\frac { { d }^{ 2 }p }{ { dx }^{ 2 } } \\) = – 2 < 0:
\n\u21d2 p is max at x = 12
\nHence, the required numbers are 12 and (24 – 12) i.e. 12.<\/p>\n

Ncert Solutions For Class 12 Maths Chapter 6 Exercise 6.5 Question 14.<\/strong>
\nFind two positive numbers x and y such that x + y = 60 and xy\u00b3 is maximum.
\nSolution:
\ni. x + y = 60
\ny = 60 – x
\nz = xy\u00b3
\n\u2234 z = x(60 – x)\u00b3<\/p>\n

ii. Differentiating z with respect to x, dz
\n\\(\\frac { dz }{ dx }\\) = x(3(60 – x)\u00b2(- 1)) + (60 – x)\u00b3
\n= (60 – x)\u00b2(- 3x + 60 – x)
\n\\(\\frac { dz }{ dx }\\) = (60 – x)\u00b2(60 – 4x)
\n\\(\\frac{d^{2} z}{d x^{2}}\\) = (60 – x)\u00b2 (- 4) + (60 – 4x)(2(60 – x)(-1))
\n= -[4(60 – x)\u00b2 + 2(60 – x)(60 – 4x)]
\nFor maxima, \\(\\frac { dz }{ dx }\\) = 0 dx
\n(60 – x)\u00b2(60 – 4x) = 0
\n\u2234 x = 60 or x = 15
\nWhen x = 60, y = 0
\nWhen x = 15, y = 45
\nAt x = 60,
\n\\(\\frac{d^{2} z}{d x^{2}}\\) = – [4(60 – 60)\u00b2 + 2(60 – 60)(60 – 240)] = 0
\nAt x = 15,
\n\\(\\frac{d^{2} z}{d x^{2}}\\) = – [4(60 – 15)\u00b2 + 2(60 – 15)(60 – 60)]
\n= – [4(45)\u00b2] < 0
\n\u2234 z is maximum at x = 15
\nHence the numbers are 15 and 45.<\/p>\n

\"NCERT<\/p>\n

Ex 6.5 Class 12 Ncert Solutions Question 15.<\/strong>
\nFind two positive numbers x and y such that their sum is 35 and the product x2<\/sup> y5<\/sup> is a maximum.
\nSolution:
\nGiven x + y = 35
\n\u2234 y = 35 – x
\nLet P = x\u00b2y5<\/sup>
\nP = x\u00b2(35 – x)5<\/sup>
\nDifferentiating w.r.t. x,
\n\\(\\frac { dP }{ dx }\\) = x\u00b2(35 – x)4(- 1) + (35 – x)5<\/sup> 2x
\n= x (35 – x)4<\/sup>[- 5x + 70 – 2x]
\n\\(\\frac { dP }{ dx }\\) = x(35 – x)4<\/sup> (70 – 7x)
\n\\(\\frac{d^{2} P}{d x^{2}}\\) = x(35 – x)4<\/sup>(- 7) + x(70 – 7x)4(35 – x)\u00b3(-1) dx + (35 – x)4<\/sup>(70 – 7x)(1)
\nFor maxima, \\(\\frac { dP }{ dx }\\) = 0
\n\u21d2 x(70 – 7x) (35 – x)4<\/sup> = 0
\n\u21d2 x = 0, x = 10, x = 35
\nx = 0 and x = 35 are not possible
\n\u2234 x = 10
\nWhen x = 10, y = 35 – 10 = 25
\nAt x = 10, \\(\\frac{d^{2} P}{d x^{2}}\\) < 0
\n\u2234 P is maximum at x = 10
\nHence the x\u00b2y is maximum when x = 10
\nand y = 25<\/p>\n

Ncert Maths Class 12 Exercise 6.5 Solutions Question 16.<\/strong>
\nFind two positive numbers whose sum is 16 and the sum of whose cubes is minimum.
\nSolution:
\nLet two numbers be x and 16 – x
\n\"NCERT
\nHence, the required numbers are 8 and (16 – 8) i.e. 8 and 8.<\/p>\n

Class 12 Maths Chapter 6 Exercise 6.5 Question 17.<\/strong>
\nA square piece of tin of side 18 cm is 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.
\nSolution:
\nLet each side of the square to be cut off be x cm.
\n\u2234 for the box length = 18 – 2x: breadth = 18 – 2x and height = x
\n\"NCERT
\nx\u00b2 – 12 x + 27 = 0 \u21d2 (x – 9)(x – 3) = 0
\nx = 9 or x = 3,
\nbut x = 9 is not possible.
\n\u2234 x = 3
\nWhen x = 3, \\(\\frac{d^{2} V}{d x^{2}}\\) = 4[6(3) – 36] < 0
\n\u2234 Volume is maximum when x = 3
\n\u2234 Side of the square to be cut off = 3 cm.<\/p>\n

\"NCERT<\/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 comer 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?
\nSolution:
\nLet x be the side of the square cut off and
\nV be the volume of the box.
\n\"NCERT
\nLength of the box = 45 – 2x
\nBreadth of the box = 24 – 2x
\nHeight of the box = x
\nVolume V = (45 – 2x) (24 – 2x)x
\n= (1080 – 90x – 48x + 4x\u00b2)x
\nV = 4x\u00b3 – 138x\u00b2 + 1080x
\n\\(\\frac { dV }{ dx }\\) = 12x\u00b2 – 276x + 1080
\n\\(\\frac{d^{2} V}{d x^{2}}\\) = 24x – 276 dx
\nFor maxima, \\(\\frac { dV }{ dx }\\) = 0
\n\u21d2 12x\u00b2 – 276x + 1080 = 0
\n\u21d2 x\u00b2 – 23x + 90 = 0
\n(x – 18)(x – 5) = 0
\nx = 18 or x = 5 But x = 18 is not possible
\n\u2234 x = 5
\nAt x = 5, \\(\\frac{d^{2} V}{d x^{2}}\\) = 24(5) – 276
\n= 120 – 276 < 0
\nBy second derivative test, V is maximum when x = 5.
\nHence the volume of the box will be maxi\u00acmum when 5 cm is cut off from the side of the square.<\/p>\n

Question 19.
\nShow that of all the rectangles inscribed in a given fixed circle, the square has the maximum area.
\nSolution:
\nLet x and y be the length and breadth of the rectangle inscribed in a circle of radius r.
\n\"NCERT
\n\u2234 x\u00b2 + y\u00b2 = (2a)\u00b2 \u21d2 x\u00b2 + y\u00b2 = 4a\u00b2 … (i)
\n\u2234 Perimeter = 2 (x + y)
\n\"NCERT
\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.
\nSolution:
\nLet x be the radius, y be the height, S be the surface area and V be the volume of the right circular cylinder.
\nS = 2\u03c0xy + 2\u03c0x\u00b2 … (1)
\n\"NCERT
\n\u2234 y = 2x
\ni.e., height = 2x i.e., radius = diameter
\n\u2234 Volume is maximum, when height of the cylinder is equal to the diameter of the cylinder.<\/p>\n

\"NCERT<\/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?
\nSolution:
\nLet r be the radius and h be the height of cylindrical can.
\n\"NCERT<\/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 ?
\nSolution:
\nLet one part be of length x, then the other part = 28 – x
\nLet the part of the length x be converted into a circle of radius r.
\n\"NCERT
\nLet r be the radius of the circle
\nCircumference of the circle is 28 – x
\n\u2234 28 – x = 2\u03c0r,
\n\"NCERT
\nTotal area of the square and circle is given by
\n\"NCERT
\nThus A is minimum
\nLength of the square piece is \\(\\frac { 112 }{ \u03c0+4 }\\) the circular piece = 28 – \\(\\frac { 112 }{ \u03c0+4 }\\) = \\(\\frac { 28 }{ \u03c0+4 }\\)<\/p>\n

Question 23.
\nProve that the volume of the largest cone that can be inscribed in a sphere of radius R is \\(\\frac { 8 }{ 27 }\\) of the volume of the sphere.
\nSolution:
\nLet h, r be the height and base radius of the inscirbed cone with volume V.
\n\"NCERT
\nFrom the figure, (h – R)\u00b2 + r\u00b2 = R\u00b2
\nh\u00b2 – 2RA + R\u00b2 + r\u00b2 = R\u00b2
\nr\u00b2 = 2Rh – h\u00b2 … (i)
\n\"NCERT
\nFor maxima V’ = 0
\nV’ = 0 \u21d2 \\(\\frac { 1 }{ 3 }\\)\u03c0(4Rh – 3h\u00b2) = 0
\ni.e., h (4R – 3h) = 0
\n\u21d2 h = 0 or h = \\(\\frac { 4R }{ 3 }\\)
\nSince h \u2260 0, h = \\(\\frac { 4R }{ 3 }\\)
\nWhen h = \\(\\frac { 4R }{ 3 }\\) , V” = \\(\\frac { -4\u03c0R }{ 3 }\\) < 0
\nV is maximum when h = \\(\\frac { 4R }{ 3 }\\)
\nFrom (ii), Maximum volume
\n\"NCERT
\n\u2234 Volume of the largest cone = \\(\\frac { 8 }{ 27 }\\) of the volume of the sphere.<\/p>\n

\"NCERT<\/p>\n

Question 24.
\nShow that die right circular cone of least curved surface and given volume has an altitude equal to \\(\\sqrt{2}\\) time the radius of the base.
\nSolution:
\nLet r and h be the radius and height of the cone
\n\"NCERT
\nHence h = \\(\\sqrt{2}\\)
\nThe curved surface area of the cone is minimum when altitude equals to \\(\\sqrt{2}\\) times the radius of the base.<\/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<\/sup>\\(\\sqrt{2}\\).
\nSolution:
\nLet v be the volume, l be the slant height and 0 be the semi vertical angle of a cone.
\n\"NCERT
\nLet h, l, r be the height, slant height and radius of the cone and a be the semivertical angle. Given l is a constant.
\nFrom the figure, r\u00b2 = l\u00b2 – h\u00b2
\nVolume, v = \\(\\frac { 1 }{ 3 }\\)\u03c0r\u00b2h
\n\"NCERT<\/p>\n

\"NCERT<\/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) \\)
\nSolution:
\n\"NCERT
\nLet r be the radius, h be the altitude, l be the slant height, V be the volume and S be the surface area and 0 be the semi -vertical angle of a right circular cone.
\nS = \u03c0r\u00b2 + \u03c0rl = constant
\n\"NCERT
\n\u2234 \u03b8 = \\({ sin }^{ -1 }\\left( \\frac { 1 }{ 3 } \\right) \\)
\ni.e., For a given surface area, the volume of a right circular cone is maximum when the semi-vertical angle is \\({ sin }^{ -1 }\\left( \\frac { 1 }{ 3 } \\right) \\)<\/p>\n

\"NCERT<\/p>\n

Question 27.
\nThe point on die curve x\u00b2 = 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)
\nSolution:
\n(a) Let P (x, y) be a point on the curve The other point is A (0,5)
\nZ = PA\u00b2 = x\u00b2 + y\u00b2 + 25 – 10y [\u2235 x\u00b2 = 2y]
\n\"NCERT<\/p>\n

Question 28.
\nFor all real values of x, the minimum value of \\(\\frac { 1-x+{ x }^{ 2 } }{ 1+x+{ x }^{ 2 } } \\)
\n(a) 0
\n(b) 1
\n(c) 3
\n(d) \\(\\frac { 1 }{ 3 }\\)
\nSolution:
\n(d) Let \\(y=\\frac { 1-x+{ x }^{ 2 } }{ 1+x+{ x }^{ 2 } } \\)
\n\"NCERT<\/p>\n

Question 29.
\nThe maximum value of \\({ \\left[ x\\left( x-1 \\right) +1 \\right] }^{ \\frac { 1 }{ 3 } },0\\le x\\le 1\\) is
\n(a) \\({ \\left( \\frac { 1 }{ 3 } \\right) }^{ \\frac { 1 }{ 3 } } \\)
\n(b) \\(\\frac { 1 }{ 2 } \\)
\n(c) 1
\n(d) 0
\nSolution:
\n(c) Let y = \\({ \\left[ x\\left( x-1 \\right) +1 \\right] }^{ \\frac { 1 }{ 3 } },0\\le x\\le 1\\)
\n\"NCERT<\/p>\n","protected":false},"excerpt":{"rendered":"

These NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Ex 6.5 Questions and Answers are prepared by our highly skilled subject experts. https:\/\/mcq-questions.com\/ncert-solutions-for-class-12-maths-chapter-6-ex-6-5\/ NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Exercise 6.5 Ex 6.5 Class 12 Question 1. Find the maximum and minimum values, if any, of the …<\/p>\n

NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Ex 6.5<\/span> Read More »<\/a><\/p>\n","protected":false},"author":9,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"default","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","theme-transparent-header-meta":"default","adv-header-id-meta":"","stick-header-meta":"default","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","spay_email":""},"categories":[3],"tags":[],"yoast_head":"\nNCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Ex 6.5 - MCQ Questions<\/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:\/\/mcq-questions.com\/ncert-solutions-for-class-12-maths-chapter-6-ex-6-5\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Ex 6.5 - MCQ Questions\" \/>\n<meta property=\"og:description\" content=\"These NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Ex 6.5 Questions and Answers are prepared by our highly skilled subject experts. https:\/\/mcq-questions.com\/ncert-solutions-for-class-12-maths-chapter-6-ex-6-5\/ NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Exercise 6.5 Ex 6.5 Class 12 Question 1. 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