{"id":35271,"date":"2022-08-29T12:40:22","date_gmt":"2022-08-29T07:10:22","guid":{"rendered":"https:\/\/apboardsolutions.guru\/?p=35271"},"modified":"2022-08-29T12:40:22","modified_gmt":"2022-08-29T07:10:22","slug":"inter-1st-year-maths-1b-applications-of-derivatives-solutions-ex-10g","status":"publish","type":"post","link":"https:\/\/apboardsolutions.guru\/inter-1st-year-maths-1b-applications-of-derivatives-solutions-ex-10g\/","title":{"rendered":"Inter 1st Year Maths 1B Applications of Derivatives Solutions Ex 10(g)"},"content":{"rendered":"

Practicing the Intermediate 1st Year Maths 1B Textbook Solutions<\/a> Inter 1st Year Maths 1B Applications of Derivatives Solutions Exercise 10(g) will help students to clear their doubts quickly.<\/p>\n

Intermediate 1st Year Maths 1B Applications of Derivatives Solutions Exercise 10(g)<\/h2>\n

I.<\/span><\/p>\n

Question 1.
\nWithout using the derivative, show that
\ni) The function f(x) = 3x + 7 is strictly increasing on R.
\nSolution:
\nLet x1<\/sub>, x2<\/sub> \u2208 R with x1<\/sub> < x2<\/sub>
\nThen 3x1<\/sub> < 3x2<\/sub>
\nAdding 7 on both sides
\n3x1<\/sub> + 7 < 3x2<\/sub> + 7
\n\u21d2 f(x1<\/sub>) < f(x2<\/sub>)
\n\u2234 x1<\/sub> < x2<\/sub> \u21d2 f(x1<\/sub>) < f(x2<\/sub>) \u2200 x1<\/sub> x2<\/sub> \u2208 R
\n\u2234 The given function is strictly increasing on R<\/p>\n

ii) The function f(x) = (\\(\\frac{1}{2}\\))x<\/sup> is strictly decreasing on R.
\nSolution:
\nf(x) = (\\(\\frac{1}{2}\\))x<\/sup>
\nLet x1<\/sub>, x2<\/sub> \u2208 R
\nSuch that x1<\/sub> < x2<\/sub>
\n\u21d2 (\\(\\frac{1}{2}\\))x1<\/sub><\/sup> > (\\(\\frac{1}{2}\\))x2<\/sub><\/sup>
\n\u21d2 f(x1<\/sub>) > f(x2<\/sub>)
\n\u2234 f(x) is strictly decreasing on R.<\/p>\n

iii) The function f(x) = e3x<\/sup> is strictly increasing on R.
\nSolution:
\nf(x) = e3x<\/sup>
\nLet x1<\/sub>, x2<\/sub> \u2208 R such that x1<\/sub> < x2<\/sub>
\nWe know that of a > b then ea<\/sup> > eb<\/sup>
\nThen e3x<\/sup> < e3x2<\/sub><\/sup>
\n\u21d2 f(x1<\/sub>) < f(x2<\/sub>)
\n\u2234 f is strictly increasing on R.<\/p>\n

iv) The function f(x) = 5. – 7x is strictly decreasing on R.
\nSolution:
\nf(x) = 5 – 7x
\nLet x1<\/sub> x2<\/sub> \u2208 R
\nSuch that x1<\/sub> < x2<\/sub>
\nThen 7x1<\/sub> < 7x2<\/sub>
\n-7x1<\/sub> > -7x2<\/sub>
\nAdding 5 on bothsides
\n5 – 7x1<\/sub> > 5 – 7x2<\/sub>
\nf(x1<\/sub>) > (x2<\/sub>)
\n\u2234 x1<\/sub> < x2<\/sub> \u21d2 f(x1<\/sub>) > f(x2<\/sub>) V x1<\/sub> x2<\/sub> \u2208 R.
\nThe given function f is strictly decreasing on R.<\/p>\n

\"Inter<\/p>\n

Question 2.
\nShow that the function f(x) = sin x. Define on R is neither increasing nor decreasing on (0, \u03c0).
\nSolution:
\nf(x) = sin x
\nSince 0 < x < n
\nConsider 0 < x
\nf(0) < f(x)
\nsin 0 < sin x
\n0 < sin x ……….. (1)
\nConsider x < n
\nf(x) < f(\u03c0)
\nsin x < sin \u03c0 0 > sin x ………….. (2)
\nFrom (1) & (2); f(x) is neither increasing nor decreasing.<\/p>\n

II.<\/span><\/p>\n

Question 1.
\nFind the intervals in which the following functions are strictly increasing or strictly decreasing.
\ni) x\u00b2 + 2x – 5
\nSolution:
\nLet f(x) = x\u00b2 + 2x – 5
\nf'(x) = 2x + 2
\nf(x) is increasing if f'(x) > 0
\n\u21d2 2x + 2 < 0 \u21d2 x+ 1 > 0
\nx > -1
\nf(x) is increasing If x \u2208 (-1, \u221e)
\nf(x) is decreasing If f'(x) < 0
\n\u21d2 2x + 2 < 0
\n\u21d2 x + 1 < 0
\n\u21d2 x < -1 f(x) is decreasing if x \u2208 (-\u221e, -1).<\/p>\n

ii) 6 – 9x – x\u00b2.
\nSolution:
\nLet f(x) = 6 – 9x – x\u00b2
\nf'(x) = -9 – 2x
\nf(x) is increasing if f'(x) > 0
\n\u21d2 -9 -2x > 0
\n\u21d2 2x + 9 < 0
\nx < \\(\\frac{-9}{2}\\)
\nf(x) is increasing if x \u2208 (-\u221e, \\(\\frac{-9}{2}\\))
\nf(x) is decreasing if f'(x) < 0
\n\u21d2 2x + 9 > 0
\n\u21d2 x > \\(\\frac{-9}{2}\\)
\nf(x) is decreasing of x \u2208 (\\(\\frac{-9}{2}\\), \u221e)<\/p>\n

iii) (x + 1)\u00b3 (x – 1)\u00b3.
\nSolution:
\nLet f(x) = (x + 1)\u00b3 (x – 1)\u00b3
\n= (x\u00b2 – 1)\u00b3
\nx6<\/sup> – 1 – 3x4<\/sup> + 3x \u00b2
\nf'(x) = 6x5<\/sup> – 12x\u00b3 + 6x
\n= 6(x5<\/sup> – 2x\u00b3 + x)
\n= 6x(x4<\/sup> – 2x\u00b2 +1)
\n= 6x(x\u00b2 – 1)\u00b2
\nf'(x) \u2264 0
\n\u21d2 6x(x\u00b2 – 1)\u00b2 < 0
\nf(x) is decreasing when (-\u221e, -1) \u222a (-1, 0)
\nf'(x) > 0
\nf(x) is increasing when (0, 1) \u222a (1, \u221e)<\/p>\n

iv) x\u00b3(x – 2)\u00b2
\nSolution:
\nf'(x) = x\u00b3. 2(x – 2) + (x – 2)\u00b2.3x\u00b2
\n= x\u00b2 (x – 2) [2x + 3 (x- 2)]
\n= x\u00b2 (x – 2) (2x + 3x – 6)
\n= x\u00b2 (x – 2) (5x – 6) \u2200 x \u2208 R, x\u00b2 \u2265 0
\nFor increasing, f'(x) = 0
\nx\u00b2(x – 2) (5x – 6) > 0
\nx \u2208 (-\u221e, \\(\\frac{6}{5}\\)) \u222a (2, \u221e)
\nFor decreasing, f'(x) < 0
\nx\u00b2(x – 2) (5x – 6) < 0
\nx \u2208 (\\(\\frac{6}{5}\\), 2)<\/p>\n

v) xex<\/sup>
\nSolution:
\nf'(x) = x . ex<\/sup> + ex<\/sup>. 1 = ex<\/sup>(x + 1)
\nex<\/sup> is positive for all real values of x
\nf'(x)>0 \u21d2 x + 1 > 6 \u21d2 x > – 1
\nf(x) is increasing when x > – 1
\nf(x) < 0 \u21d2 x + 1 < 0 \u21d2 x < -1
\nf(x) is decreasing when x < – 1<\/p>\n

vi) \\(\\sqrt{(25-4x^{2})}\\)
\nSolution:
\nf(x) is real only when 25 – 4x\u00b2 > 0
\n-(4x\u00b2 – 25) > 0
\n-(2x + 5) (2x – 5) > 0
\n\u2234 x lies between –\\(\\frac{5}{2}\\) and \\(\\frac{5}{2}\\)
\nDomain of f = (-\\(\\frac{5}{2}\\), \\(\\frac{5}{2}\\))
\nf'(x) = \\(\\frac{1}{2 \\sqrt{25-4 x^{2}}}\\) (-8x)
\n= –\\(\\frac{4x}{\\sqrt{25-4 x^{2}}}\\)
\nf(x) is increasing when f'(x) > 0
\n\u21d2 \\(\\frac{-4x}{\\sqrt{25-4 x^{2}}}\\) > 0
\ni.e., x < o
\nf(x) is increasing when (-\\(\\frac{5}{2}\\), 0)
\nf(x) is decreasing when f'(x) < 0
\n\u21d2 –\\(\\frac{4x}{\\sqrt{25-4 x^{2}}}\\) < 0
\n\u2234 x > 0
\nf(x) is decreasing when (0, \\(\\frac{5}{2}\\)).<\/p>\n

vii) ln (ln(x)); x > 1.
\nSolution:
\nf'(x) = –\\(\\frac{1}{lnx}\u2022\\frac{1}{x}\\)
\nf(x) is decreasing when f'(x) > 0
\n\\(\\frac{1}{x.ln x}\\) >0
\n\u21d2 x. In x > 0
\nln x is real only when x > 0
\n\u2234 ln x < 0 = ln 1
\ni.e., x > 1
\nf(x) is increasing when x > 1 i.e., in (1, \u221e)
\nf(x) is decreasing when f'(x) < 0 \u21d2 ln x > 0 = ln 1
\ni.e., x < 1
\nf(x) is decreasing in (0, 1)<\/p>\n

viii) x\u00b3 + 3x\u00b2 – 6x + 12.
\nSolution:
\nf(x) = x\u00b3 + 3x\u00b2 – 6x + 12
\nf(x) = 3x\u00b2 + 6x – 6
\n= 3(x\u00b2 + 2x – 2)
\n= 3((x + 1)\u00b2 – 3)
\n= 3[(x + 1) + \u221a3] [(x + 1) – \u221a3]
\n= 3(x + (1 + \u221a3) (x + (1 – \u221a3 )
\nf (x) < 0
\n\u21d2 x = -(1+ \u221a3 ) or -(1 – \u221a3 )
\nx = -1 – \u221a3 or \u221a3 – 1
\nf(x) is decreasing in (-1 -\u221a3 , \u221a3 -1)
\nf (x) > 0
\nf(x) increasing when (-\u221e, -1 – \u221a3 ) \u222a (\u221a3 – 1, \u221e)<\/p>\n

\"Inter<\/p>\n

Question 2.
\nShow that f(x) = cos\u00b2x is strictly increasing on (0, \u03c0\/2).
\nSolution:
\nf(x) = cos\u00b2 X
\n\u21d2 f(x) = 2 cos x (-sin x)
\n= -2 sin x cos x
\n= -sin 2x
\nSince 0 < x < \\(\\frac{\\pi}{2}\\)
\n\u21d2 0 < 2x < \u03c0
\nSince ‘sin x’ is +ve between 0 and \u03c0
\n\u2234 f(x) is clearly -ve.
\n\u2234 f'(x) < 0
\n\u2234 f(x) is strictly decreasing.<\/p>\n

Question 3.
\nShow that x + \\(\\frac{1}{x}\\) is increasing on [1, \u221e)
\nSolution:
\nLet f(x) = x + \\(\\frac{1}{x}\\)
\nf'(x) = 1 – \\(\\frac{1}{x^{2}}\\) = \\(\\frac{x^{2}-1}{x^{2}}\\)
\nSince x \u2208 [1, \u221e) = \\(\\frac{x^{2}-1}{x^{2}}\\) > 0
\n\u2234 f'(x) > 0
\n\u2234 f(x) is increasing.<\/p>\n

Question 4.
\nShow that \\(\\frac{x}{1+x}\\) < ln (1 + x) < x \u2200 x > 0
\nSolution:
\nLet f(x) = ln(1 + x)- \\(\\frac{x}{1+x}\\)
\n= ln(1 + x) – \\(\\frac{1+x-1}{1+x}\\)
\n= ln(1 + x) – 1 + \\(\\frac{1}{1+x}\\)
\nf'(x) = \\(\\frac{1}{1+x}\\) – \\(\\frac{1}{(1+x)^{2}}\\)
\n= \\(\\frac{1+x-1}{(1+x)^{2}}\\)
\n= \\(\\frac{x}{(1+x)^{2}}\\) > 0 since x > 0
\nf(x) is increasing when x > 0
\n\u2234 f(x) > f(0)
\nf(0) = ln 1 – \\(\\frac{0}{1+0}\\) = 0 – 0 = 0
\nSince xe [1, \u221e) =
\nln (1 + x) – \\(\\frac{x}{1+x}\\) > 0
\n\u21d2 ln (1 + x) > \\(\\frac{x}{1+x}\\) ……… (1)
\nLet g(x) = x – ln (1 + x)
\ng'(x) = 1 – \\(\\frac{x}{1+x}=\\frac{1+x-1}{1+x}\\)
\n= \\(\\frac{x}{1+x}\\) > 0 since x > 0
\ng(x) is increasing when x > 0
\ni.e., g(x) > g(0)
\ng(0) = 0 – ln (1) = 0 – 0 = 0
\n\u2234 x – ln (1 + x) > 0
\nx > ln(1 + x) ………….. (2)
\nFrom (1), (2) we get
\n\\(\\frac{x}{1+x}\\) < ln (1 + x) < x \u2200 x > 0<\/p>\n

III.<\/span><\/p>\n

Question 1.
\nShow that \\(\\frac{x}{1+x^{2}}\\) < tan-1<\/sup> x < x when x > 0.
\nSolution:
\nLet f(x) = tan-1<\/sup> x – \\(\\frac{x}{1+x^{2}}\\)
\n\"Inter
\nf(x) is increasing when x > 0
\nf(x) > f(0)
\nBut f(0) = tan-1<\/sup> 0 – 0 = 0 – 0 = 0
\ni.e., f(x) > 0
\n\"Inter
\ng(x) is increasing when x > 0
\ng(x) > g(0)
\ng(0) = 0 – tan-1<\/sup> 0 = 0 – 0 = 0
\n\u2234 x – tan-1<\/sup> x > 0
\n\u21d2 x > tan-1<\/sup> x ………. (2)
\nFrom (1), (2) we get
\n\\(\\frac{x}{1+x^{2}}\\) <tan-1<\/sup> x< x for x > 0<\/p>\n

Question 2.
\nShow that tan x > x for all (0, \\(\\frac{\\pi}{2}\\))
\nSolution:
\nLet f(x) = tan x – x
\nf'(x) = sec\u00b2 x – 1 > 0 for every
\nx \u2208 (0, \\(\\frac{\\pi}{2}\\))
\nf(x) is increasing for every x \u2208 (0, \\(\\frac{\\pi}{2}\\))
\ni.e., f(x) > f(0)
\nf(0) = tan 0 – 0 = 0 – 0 = 0
\n\u2234 tan x – x > 0
\n\u21d2 tan x > x for every x \u2208 (0, \\(\\frac{\\pi}{2}\\))<\/p>\n

\"Inter<\/p>\n

Question 3.
\nIf x \u2208 (0, \\(\\frac{\\pi}{2}\\)) then show that \\(\\frac{2x}{\\pi}\\) < sin x < x.
\nSolution:
\nLet f(x) = x – sin x
\nf(x) = 1- cos x > 0 for every x \u2208 (0, \\(\\frac{\\pi}{2}\\))
\nf(x) is increasing for every x \u2208 (0, \\(\\frac{\\pi}{2}\\))
\n\u21d2 f(x) > f(0)
\nf(0) = 0 – sin 0 = 0 – 0 = 0
\n\u2234 x – sin x > 0
\n\u21d2 x > sin x ………….. (1)
\nLet g(x) = sin x – \\(\\frac{2x}{\\pi}\\)
\ng'(x) = cos x – \\(\\frac{2}{\\pi}\\) > 0 for every x \u2208 (0, \\(\\frac{\\pi}{2}\\))
\ng(x) is increasing in (0, \\(\\frac{\\pi}{2}\\))
\ng(x) > g(0)
\ng(0) = sin 0 – 0 = 0 – 0 = 0
\n\u2234 sin x – \\(\\frac{2x}{\\pi}\\) > 0
\n\u21d2 sin x > \\(\\frac{2x}{\\pi}\\) ………… (2)
\nFrom (1), (2) we get
\n\\(\\frac{2x}{\\pi}\\) < sin x < x for every x \u2208 (0, \\(\\frac{\\pi}{2}\\))<\/p>\n

Question 4.
\nIf x e (0,1) then show that 2x < ln [latex]\\frac{(1+x)}{(x-1)}[\/latex] < 2x [1 + \\(\\frac{x^{2}}{2(1+x^{2})}\\)] Solution:
\nLet f(x) = ln \\(\\frac{(1+x)}{1-x}\\) – 2x
\n= ln (1 + x) – ln (1 – x) – 2x
\n\"Inter
\nf(x) is increasing ih (0, 1)
\ni.e., x > 0 \u21d2 f(x) > f(0)
\nf(0) = ln 1 – 0 = 0 – 0 = 0
\n\"Inter
\n\"Inter
\ng(x) is increasing when x > 0
\ng(x) > g(0)
\ng(0) = 0 – ln 1 = 0 – 0 = 0
\n\"Inter
\nfor x \u2208 (0,1)<\/p>\n

Question 5.
\nAt what point the slopes of the tangents y = \\(\\frac{x^{3}}{6}-\\frac{3x^{3}}{2}+\\frac{11x}{2}\\) + 12 increases?
\nSolution:
\nEquation of the curve is
\ny = \\(\\frac{x^{3}}{6}-\\frac{3}{2}x^{2}+\\frac{11x}{2}\\) + 12
\n\"Inter
\nSlope = m = \\(\\frac{x^{2}}{c}-3x+\\frac{11}{2}\\)
\n\\(\\frac{dm}{dx}\\) = \\(\\frac{2x}{2}\\) -3 = x – 3
\nSlope increases \u21d2 m > 0
\nx – 3 > 0
\nx > 3
\nThe slope increases in (3, \u221d)<\/p>\n

Question 6.
\nShow that the functions ln \\(\\frac{(1+x)}{x}\\) and \\(\\frac{x}{(1+x)ln(1+x)}\\) are decrasing on (0, \u221e).
\nSolution:
\ni)
\n\"Inter
\n\"Inter
\n\u2234 f(x) is decreasing for x \u2208 (0, \u221d)<\/p>\n

ii) let f(x) = \\(\\frac{x}{(1+x)ln(1+x)}\\)
\n\"Inter
\n\u2234 f(x) is decreasing for x \u2208 (0, \u221d)<\/p>\n

Question 7.
\nFind the intervals in which the function f (x) = x3 – 3×2 + 4 is strictly increasing all x e R.
\nSolution:
\nf(x) = x\u00b3 – 3x\u00b2 + 4
\nf'(x) = 3x\u00b2 – 6x
\nf(x) is increasing if f'(x) > 0
\n3x\u00b2 – 6x > 0
\n3x(x – 2) > 0
\n(x – 0)(x – 2) > 0
\nf(x) is increasing if x \u00a3 (-\u221e, 0) u (0, \u221e)
\nf(x) is decreasing if f'(x) < 0
\n(x – 0) (x – 2) < 0
\nx \u2208 (0, 2)<\/p>\n

\"Inter<\/p>\n

Question 8.
\nFind the intervals in which the function f(x) = sin4<\/sup>x + cos4<\/sup>x \u2200 x \u2208 [0, \\(\\frac{\\pi}{2}\\)] is increasing and decreasing.
\nSolution:
\nf(x) = sin4<\/sup>x + cos4<\/sup>x
\nf(x) = (sin\u00b2x)\u00b2 + (cos\u00b2x)\u00b2
\n= (sin\u00b2x + cos\u00b2x)\u00b2 – 2sin\u00b2x cos\u00b2x
\n= 1 – \\(\\frac{1}{2}\\) sin\u00b2 2x
\nf'(x) = \\(\\frac{-1}{2}\\) 2sin 2x. cos 2x(2)
\n= -2 sin 2x. cos 2x
\n= -sin 4x
\nLet 0 < x < \\(\\frac{\\pi}{4}\\)
\n\u2234 f(x) is decreasing if f'(x) < 0
\n\u21d2 -sinx < 0 \u21d2 sinx > 0
\n\u2234 x \u2208 (0, \\(\\frac{\\pi}{4}\\))
\nf(x) is increasing if f'(x) > 0
\n\u21d2 – sinx > 0
\n\u21d2 sinx < 0
\n\u2234 x \u2208 (\\(\\frac{\\pi}{4}\\), \\(\\frac{\\pi}{2}\\))<\/p>\n","protected":false},"excerpt":{"rendered":"

Practicing the Intermediate 1st Year Maths 1B Textbook Solutions Inter 1st Year Maths 1B Applications of Derivatives Solutions Exercise 10(g) will help students to clear their doubts quickly. Intermediate 1st Year Maths 1B Applications of Derivatives Solutions Exercise 10(g) I. Question 1. Without using the derivative, show that i) The function f(x) = 3x + …<\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[15],"tags":[],"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts\/35271"}],"collection":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/users\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/comments?post=35271"}],"version-history":[{"count":1,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts\/35271\/revisions"}],"predecessor-version":[{"id":35272,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts\/35271\/revisions\/35272"}],"wp:attachment":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/media?parent=35271"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/categories?post=35271"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/tags?post=35271"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}