{"id":5362,"date":"2024-01-05T01:06:34","date_gmt":"2024-01-04T19:36:34","guid":{"rendered":"https:\/\/tsboardsolutions.in\/?p=5362"},"modified":"2024-01-08T17:06:00","modified_gmt":"2024-01-08T11:36:00","slug":"ts-inter-1st-year-maths-1b-solutions-chapter-10-ex-10g","status":"publish","type":"post","link":"https:\/\/tsboardsolutions.in\/ts-inter-1st-year-maths-1b-solutions-chapter-10-ex-10g\/","title":{"rendered":"TS Inter 1st Year Maths 1B Solutions Chapter 10 Applications of Derivatives Ex 10(g)"},"content":{"rendered":"

Students must practice these TS Intermediate Maths 1B SoIutons<\/a> Chapter 10 Applications of Derivatives Ex 10(g) to find a better approach to solving the problems.<\/p>\n

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

I.
\nQuestion 1.
\n(i) Without using the derivative, show that the function f(x) = 3x + 7 is strictly increasing on \u211b . (V.S.A.Q.)
\nAnswer:
\nLet x1<\/sub> < x2<\/sub> \u2208\u211b with x1<\/sub> < x2<\/sub>
\nThen 3x1<\/sub> < 3x1<\/sub> ;
\nAdding 7 on both sides,
\n7 + 3x1<\/sub> < 7 + 3x2<\/sub> \u21d2 f (x1<\/sub>) < f(x2<\/sub>)
\nx1<\/sub> < x2<\/sub> \u21d2 f (x1<\/sub>) < f(x2<\/sub>) v x1<\/sub>, x2<\/sub> \u2208 \u211b
\nHence the given function is strictly increasing on \u211b<\/p>\n

(ii) The function f(x) = \\(\\left(\\frac{1}{2}\\right)^x\\) is strictly decreasing on \u211b.
\nAnswer:
\nf(x) = \\(\\left(\\frac{1}{2}\\right)^x\\)
\nLet x1<\/sub>, x2<\/sub> \u2208 \u211b such that x1<\/sub> < x2<\/sub>
\n\u21d2 \\(\\left(\\frac{1}{2}\\right)^{x_1}>\\left(\\frac{1}{2}\\right)^{x_2}\\)
\n\u21d2 f(x1<\/sub>) > f(x2<\/sub>)
\nf(x) is strictly decreasing on \u211b.<\/p>\n

(iii) The function f(x) = e3x<\/sup> is strictly increasing on \u211b.
\nAnswer:
\nf(x) = e3x<\/sup>
\nLet x1<\/sub>, x2<\/sub> \u2208 \u211b such that x1<\/sub> < x2<\/sub>
\nWe have that if a > b then ea<\/sup> > eb<\/sup>
\ne3x1<\/sub><\/sup> < e3x1<\/sub><\/sup> \u21d2 f(x1<\/sub>) < f(x2<\/sub>)
\nSo the function f is strictly increasing on \u211b.<\/p>\n

\"TS<\/p>\n

Question 2.
\nShow that the function f(x) = sin x defined on \u211b is neither increasing nor decreasing on (0, \u03c0). (V.S.A.Q.)
\nAnswer:
\nGiven f(x) = sin x
\nsince 0 < x < \u03c0 and for 0 < x \u21d2 f(0) < f(x)
\n\u21d2 sin 0 < sin x => 0 < sin x (1)
\nfor x < \u03c0 \u21d2 f(x) < f(\u03c0)
\n\u21d2 sin x < sin \u03c0 \u21d2 0 > sin x (2)
\nFrom (1) and (2) f(x) is neither increasing nor decreasing.<\/p>\n

II.
\nQuestion 1.
\nFind the interval in which the following functions are stjpctly increasing strictIy decreasing. 1 (S.A.Q.)
\n(i) x2<\/sup> + 2x – 5
\nAnswer:
\nLet f(x) = x2<\/sup> + 2x – 5
\nThen f'(x) = 2x + 2
\nf(x) is increasing if f'(x) > 0
\n\u21d2 2x + 2 > 0
\n\u21d2 x + 1 > 0
\n\u21d2 x > – 1
\nf (x) increases in (- 1, \u221e)
\nf (x) is decreasing if f’ (x) < 0
\n\u21d2 2x + 2 < 0
\n\u21d2 x + 1 < 0
\n\u21d2 x < – 1
\nf is decreasing in (- \u221e, -1)<\/p>\n

(ii) 6 – 9x – x2<\/sup>
\nAnswer:
\nLet f(x) = 6 – 9x – x2<\/sup>
\nThen f’ (x) = – 9 – 2x
\nf(x) is increasing if f’ (x) > 0
\n\u21d2 – 9 – 2x > 0 \u21d2 9 + 2x < 0
\n\u21d2 x < \\(\\frac{-9}{2}\\)
\n\u2234 f(x) is increasing in (-\u221e, \\(\\frac{-9}{2}\\))
\nf(x) is decreasing if f'(x) < 0 \u21d2 9 + 2x > 0
\n\u21d2 2x > – 9
\n\u21d2 x > \\(\\frac{-9}{2}\\)
\n\u2234 f is decreasing in (\\(\\frac{-9}{2}\\), \u221e)<\/p>\n

(iii) (x + 1)3<\/sup>(x – 1)3<\/sup>
\nAnswer:
\nLet f(x) = (x + 1)3<\/sup> (x – 1)3<\/sup> = (x2<\/sup> – 1)3<\/sup>
\n= x6<\/sup> – 1 – 3x4<\/sup> 3x2<\/sup>
\n\u2234 f'(x) = 6x5<\/sup> – 12x3<\/sup> + 6x
\n= 6x(x4<\/sup> – 2x2<\/sup> + 1)
\n\u2234 f'(x) = 6x5<\/sup> – 12x3<\/sup> + 6x
\n= 6x(x4<\/sup> – 2x2<\/sup> + 1)
\n= 6x(x2<\/sup> – 1)2<\/sup>
\nf(x) increases for f'(x) > 0
\n\u21d2 6x(x2<\/sup> – 1) > 0
\n\u21d2 x2<\/sup> – 1 > 0, x > 0
\n\u2234 f is increasing in (0, 1) \u222a (1, \u221e)
\nf(x) decreases for f'(x) < 0
\n\u2234 f is decreasing in (-\u221e, – 1) \u222a (-1, 0)<\/p>\n

(iv) x3<\/sup>(x – 2)2<\/sup>
\nAnswer:
\nLet f(x) = x3<\/sup>(x – 2)2<\/sup>
\nThen f(x) = x3<\/sup>. 2 (x – 2) + (x – 2)2<\/sup> 3x2<\/sup>
\n= x2<\/sup> (x – 2) [2x + 3 (x – 2)]
\n= x2<\/sup>(x – 2) (5x – 6) \u2200 x \u2208 R, x2<\/sup> > 0
\nf is increasing, f'(x) > 0
\n\u21d2 x2<\/sup> (x – 2) (5x – 6) > 0
\n\u21d2 x \u2208(-\u221e, \\(\\frac{6}{5}\\)) \u222a (2, \u221e)
\nf is decreasing if f'(x) < 0
\n\u21d2 x2<\/sup>(x – 2)(5x – 6) < 0
\n\u21d2 x \u2208 (\\(\\frac{6}{5}\\), 2)<\/p>\n

(v) xex<\/sup>
\nAnswer:
\nLet f(x) = xex<\/sup>
\nThen f'(x) = xex<\/sup> + ex<\/sup> = ex<\/sup>(x + 1) \u2200 x \u2208 \u211b, ex<\/sup> > 0
\nf'(x) > 0 \u21d2 ex<\/sup>(x + 1) > 0
\n\u21d2 x + 1 > 0 \u21d2 x < – 1
\nHence f is increasing in (-1, \u221e)
\nAlso f'(x) < 0 \u21d2 ex<\/sup>(1 + x) < 0
\n\u21d2 1 + x < 0 \u21d2 x < -1
\nHence f is decreasing in (-\u221e, -1)<\/p>\n

(vi) \\(\\sqrt{25-4 x^2}\\)
\nAnswer:
\nSuppose f(x) = \\(\\sqrt{25-4 x^2}\\)
\nIf f is real then 25 – 4x2<\/sup> \u2265 0
\n\u21d2 -(4x2<\/sup> – 25) \u2265 0
\n\u21d2 -(2x + 5)(2x – 5) \u2265 0
\n\u2234 x \u2208 \\(\\left(\\frac{-5}{2}, \\frac{5}{2}\\right)\\)
\nDomain of f = \\(\\left(\\frac{-5}{2}, \\frac{5}{2}\\right)\\)
\ni.e., x < 0
\nf(x) is increasing when x \u2208 (\\(-\\frac{5}{2}\\), 0)
\nf(x) is decreasing when f'(x) < 0
\n\u21d2 \\(-\\frac{4 \\mathrm{x}}{\\sqrt{25-4 \\mathrm{x}^2}}\\) < 0 \u2234 x > 0 and f(x) is decreasing when x \u2208 (0, \\(\\frac{5}{2}\\))<\/p>\n

(vii) log (log x), x > 1
\nAnswer:
\nf(x) = log (log x)
\nf'(x) = \\(\\frac{1}{x \\log x}\\)
\nf(x) is increasing when f \u2019(x) > 0
\n\u21d2 \\(\\frac{1}{x \\log x}\\) > 0 \u21d2 x log x > 0
\nWhen x > 0, we have x is real and if x > 0 then log x is real.
\nlog x > 0 = log 1 \u21d2 x > 1
\nFunction f is increasing in (1, \u221e)
\nIf f’ (x) < 0, then f is decreasing
\n\u21d2 x log x < 0 \u21d2 log x > 0 and x < 1
\n\u2234 Function f is decreasing in (0, 1)<\/p>\n

(viii) x3<\/sup> + 3x2<\/sup> – 6x + 12
\nAnswer:
\nf(x) = x3<\/sup> + 3x2<\/sup> – 6x + 12
\nf’ (x) = 3x2<\/sup> + 6x – 6 = 3 (x2<\/sup> + 2x – 2)
\n= 3 [(x + 1)2<\/sup> – 3]
\n= 3 [ (x + 1) + \u221a3 ] [ (x + 1) – \u221a3 ]
\n= 3[x + (1 + \u221a3)][x + (1 – \u221a3)]
\nIf f'(x) > 0 then
\n[x + (1+ \u221a3)][x + (1 – \u221a3)]>0
\n\u21d2 x \u2208 [(-\u221e , -1 – \u221a3 ) u (\u221a3 – 1 > \u221e) ]
\nHence f is increasing in above interval.
\nf'(x) < 0 \u21d2 [x + (1 + \u221a3)] [x + (1 – \u221a3)] < 0
\n\u21d2 x \u2208 [(-1 – \u221a3), (-1 + \u221a3)] f is decreasing in above interval.<\/p>\n

\"TS<\/p>\n

Question 2.
\nShow that f(x) = cos2<\/sup>x is strictly increasing on (0, \\(\\frac{\\pi}{2}\\)). (S.A.Q)
\nAnswer:
\nf(x) = cos2x
\nf'(x) = 2cosx (- sin x) = – sin 2x
\nSince 0 < x < \\(\\frac{\\pi}{2}\\)
\n\u21d2 0 < 2x < \u03c0 Since sin x > 0 between 0 and \u03c0
\nWe have f'(x) is negative.
\nf’ (x) < 0 \u21d2 f(x) is strictly decreasing.<\/p>\n

Question 3.
\nShow that x + \\(\\frac{1}{x}\\) is increasing on [1, \u221e). (S.A.Q.)
\nAnswer:
\n\"TS
\n\u2234 f'(x) > 0 and f(x) is increasing.<\/p>\n

Question 4.
\nShow that \\(\\frac{\\mathbf{x}}{1+x}\\) < log(1 + x) < x \u2200 x > 0
\nAnswer:
\n\"TS
\nIf x > 0 then f’ (x) > 0 When x > 0 f is increasing
\n\u21d2 f(x) > f(0)
\nf(0) = log 1 – 0 = log 1 = 0
\nf(x) > 0 \u21d2 fog (1 + x) > \\(\\frac{x}{1+x}\\)
\nSimilarly suppose g(x) = x – log (1 + x) ………..(3)
\ng'(x) = 1 – \\(\\frac{1}{1+x}=\\frac{x}{(1+x)}\\) > 0
\n\u2234 When x > 0, g(x) is also increasing.
\n\u2234 g(x) > g(0) \u21d2 g(x) > 0
\n\u21d2 x – log (1 + x) > 0 (4)
\n\u21d2 x > log (1 + x)
\nHence from (2) and (4)
\n\\(\\frac{x}{1+x}\\) < log(1 + x) < x \u2200 x > 0<\/p>\n

III.
\nQuestion 1.
\nShow that \\(\\frac{x}{1+x^2}\\) < tan-1<\/sup>x < x when x > 0. (S.A.Q)
\nAnswer:
\nLet f(x) = tan-1<\/sup>x – \\(\\frac{x}{1+x^2}\\)
\n\"TS
\n\u2234 When x > 0, f is increasing means f(x) > f(0) and f(0) = 0
\n\u2234 f(x) > 0 \u21d2 tan-1<\/sup>x – \\(\\frac{x}{1+x^2}\\) > 0
\n\u21d2 tan-1<\/sup>x > \\(\\frac{x}{1+x^2}\\) ………..(1)
\nSimilarly let g(x) = x – tan-1<\/sup>x
\nThen g'(x) = 1 – \\(\\frac{1}{1+x^2}=\\frac{x^2}{1+x^2}\\) > 0 \u2200 x > 0
\n\u2234 g is increasing when x > 0
\n\u21d2 g(x) = 0
\nBut g(0) = 0
\n\u2234 g(x) < 0 \u21d2 x – tan-1<\/sup>x
\n\u21d2 x > tan-1<\/sup>x ………….(2)
\n\u2234 From (1) and (2), \\(\\frac{x}{1+x^2}\\) < tan-1<\/sup>x < x \u2200 x > 0<\/p>\n

Question 2.
\nShow that tan x > x for all x \u2208 [0, \\(\\frac{\\pi}{2}\\)). (S.A.Q)
\nAnswer:
\nLet f(x) = tan x – x
\nThen f’ (x) = sec2<\/sup> x -1 > 0 for every x e
\nf(x) increases \u2200 x \u2208 [0, \\(\\frac{\\pi}{2}\\))
\n\u2234 f(x) > f(0) and f(0) = 0
\n\u21d2 f(x) > 0 \u21d2 tan x – x > 0
\n\u21d2 tan x > x; \u2200 x \u2208 [0, \\(\\frac{\\pi}{2}\\))<\/p>\n

Question 3.
\nIf x \u2208 (0, \\(\\frac{\\pi}{2}\\)), then show that \\(\\frac{2 \\mathrm{x}}{\\pi}\\) < sin x < x. (EQ.) Answer: Let f(x) = sin x – \\(\\frac{2 \\mathrm{x}}{\\pi}\\) and f\u2019(x) = cos x – \\(\\frac{2}{\\pi}\\) > 0; \u2200 x \u2208 (0, \\(\\frac{\\pi}{2}\\))
\n\u2235 f\u2019(x) > 0, f is increasing and f(x) > f(0);
\nSince f(0) = 0 we have f(x) > 0
\n\u21d2 sin x – \\(\\frac{2 \\mathrm{x}}{\\pi}\\) > 0 \u21d2 sin x > \\(\\frac{2 \\mathrm{x}}{\\pi}\\) …………..(1)
\nLet g(x) = x – sin x
\nThen g'(x) = 1 – cos x > 0 \u2200 x \u2208 (0, \\(\\frac{\\pi}{2}\\))
\n\u2234 g(x) is an increasing function in
\nBut g(0) = 0 – sin 0 = 0 g(x) >0
\n\u21d2 x – sin x > 0 x > sin x
\n\u21d2 sin x < x ……..(2)
\n\u2234 From (1) and (2);
\n\\(\\frac{2 \\mathrm{x}}{\\pi}\\) < sin x < x \u2200 x \u2208 (0, \\(\\frac{\\pi}{2}\\))<\/p>\n

Question 4.
\nIf x \u2208 (0, 1) then show that 2x < log\\(\\left[\\frac{1+x}{1-x}\\right]\\) < 2x[1 + \\(\\frac{x^2}{2\\left(1-x^2\\right)}\\)] (E.Q.)
\nAnswer:
\n\"TS
\nHence ‘g’ is an increasing function in (0, 1) g(x) > g(0) \u2200 x \u2208 (0, 1)
\nBut g(0) = 0
\n\u2234 g(x) > 0 \u2200 x \u2208 (0, 1)
\n\u21d2 2x[1 + \\(\\frac{x^2}{2\\left(1-x^2\\right)}\\)] > log\\(\\left(\\frac{1+x}{1-x}\\right)\\) ………..(2)
\nFrom (1) and (2)
\n2x < log\\(\\left(\\frac{1+x}{1-x}\\right)\\) < 2x[1 + \\(\\frac{x^2}{2\\left(1-x^2\\right)}\\)] \u2200 x \u2208 (0, 1)<\/p>\n

\"TS<\/p>\n

Question 5.
\nAt what points the slopes of the tangents y = \\(\\frac{x^3}{6}-\\frac{3 x^2}{2}+\\frac{11 x}{2}\\) + 12 increases? (E.Q.)
\nAnswer:
\nEquation of the curve is
\ny = \\(\\frac{x^3}{6}-\\frac{3 x^2}{2}+\\frac{11 x}{2}\\) + 12
\n\\(\\frac{d y}{d x}=\\frac{3 x^2}{6}-\\frac{6 x}{2}+\\frac{11}{2}\\)
\nSlope m = \\(\\frac{x^2}{2}\\) – 3x + \\(\\frac{11}{2}\\)
\n\\(\\frac{\\mathrm{dm}}{\\mathrm{dx}}\\) = x – 3 > 0 (\u2235 Slope increases)
\n\u21d2 x > 3
\nSlope increases in ( 3, \u221e)<\/p>\n

Question 6.
\nShow that the functions \\(\\frac{\\log (1+x)}{x}\\) and \\(\\frac{x}{(1+x) \\log (1+x)}\\) are decreasing in (0, \u221e). (E.Q.)
\nAnswer:
\n\"TS<\/p>\n

Question 7.
\nFind the intervals in which the function f(x) = x3<\/sup> – 3x2<\/sup> + 4 is strictly increasing for all x \u2208 \u211b.
\nAnswer:
\nf(x) = x3<\/sup> – 3x2<\/sup> + 4
\nf'(x) = 3x2<\/sup> – 6x
\nf(x) is strictly increasing if f’ (x) > 0
\n3x2<\/sup> – 6x > 0
\n\u21d2 3x (x – 2) > 0
\n\u21d2 x (x – 2) > 0
\nf(x) is strictly increasing if x \u2208 (2, \u221e)<\/p>\n

\"TS<\/p>\n

Question 8.
\nFind the intervals in which the function
\nf(x) = sin4<\/sup>x + cos4<\/sup> x \u2200 x \u2208 [0, \\(\\frac{\\pi}{2}\\)] is increasing and decreasing.
\nAnswer:
\nGiven f(x) = sin4<\/sup>x + cos4<\/sup> x \u2200 x \u2208 [0, \\(\\frac{\\pi}{2}\\)]
\n= (sin2<\/sup>x + cos2<\/sup>x)2<\/sup> – 2 sin2<\/sup>x cos2<\/sup>x
\n= 1 – 2 sin2<\/sup>x cos2<\/sup>x
\n= 1 – \\(\\frac{1}{2}\\)sin2<\/sup>2x<\/p>\n

f'(x) = \\(\\frac{1}{2}\\)(2 sin 2x)(cos 2x)(2)
\n= -sin 4x
\nf'(x) > 0 \u21d2 -sin 4x > 0 \u21d2 sin 4x < 0
\n\u21d2 \u03c0 < 4x < 2\u03c0 \u21d2 \\(\\frac{\\pi}{4}\\) < x < \\(\\frac{\\pi}{2}\\)
\n\u2234 f is increasing when \\(\\frac{\\pi}{4}\\) < x < \\(\\frac{\\pi}{2}\\)
\nNow f'(x) < 0
\n– sin 4x < 0 \u21d2 sin 4x > 0
\n\u21d2 0 < 4x < \u03c0 \u21d2 0 < x < \\(\\frac{\\pi}{4}\\)<\/p>\n

\u2234 Function ‘f’ is decreasing when 0 < x < \\(\\frac{\\pi}{4}\\)
\n\u2234 f is decreasing in [0, \\(\\frac{\\pi}{4}\\)] and
\nf is increasing in \\(\\left(\\frac{\\pi}{4}, \\frac{\\pi}{2}\\right)\\)<\/p>\n","protected":false},"excerpt":{"rendered":"

Students must practice these TS Intermediate Maths 1B SoIutons Chapter 10 Applications of Derivatives Ex 10(g) to find a better approach to solving the problems. TS Inter 1st Year Maths 1B Applications of Derivatives Solutions Exercise 10(g) I. Question 1. (i) Without using the derivative, show that the function f(x) = 3x + 7 is … Read more<\/a><\/p>\n","protected":false},"author":2,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[2],"tags":[],"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/posts\/5362"}],"collection":[{"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/users\/2"}],"replies":[{"embeddable":true,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/comments?post=5362"}],"version-history":[{"count":3,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/posts\/5362\/revisions"}],"predecessor-version":[{"id":5370,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/posts\/5362\/revisions\/5370"}],"wp:attachment":[{"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/media?parent=5362"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/categories?post=5362"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/tags?post=5362"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}