{"id":35269,"date":"2022-08-29T12:39:36","date_gmt":"2022-08-29T07:09:36","guid":{"rendered":"https:\/\/apboardsolutions.guru\/?p=35269"},"modified":"2022-08-29T12:39:36","modified_gmt":"2022-08-29T07:09:36","slug":"inter-1st-year-maths-1b-applications-of-derivatives-solutions-ex-10f","status":"publish","type":"post","link":"https:\/\/apboardsolutions.guru\/inter-1st-year-maths-1b-applications-of-derivatives-solutions-ex-10f\/","title":{"rendered":"Inter 1st Year Maths 1B Applications of Derivatives Solutions Ex 10(f)"},"content":{"rendered":"

Practicing the\u00a0Intermediate 1st Year Maths 1B Textbook Solutions<\/a>\u00a0Inter 1st Year Maths 1B Applications of Derivatives Solutions Exercise 10(f) will help students to clear their doubts quickly.<\/p>\n

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

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

Question 1.
\nVerify Rolle’s theorem for the following functions.
\ni) x\u00b2 – 1 on [-1, 1]
\nSolution:
\nLet f(x) = x\u00b2 – 1
\nf is continuous on [-1,1]
\nsince f(-1) = f(1) = 0 and
\nf is differentiable on [-1, 1]
\n\u2234 By Rolle’s theorem \u2203 c \u2208 (-1, 1) such that f'(c) = 0
\nf'(x) = 2x = 0
\n\u2234 = f'(c) = 0
\n2c = 0
\nc = 0
\nThe point c = 0 \u2208 (-1, 1)
\nThen Rolle’s theorem is verified.<\/p>\n

ii) sin x – sin 2x on [0, \u03c0].
\nSolution:
\nLet f(x) = sin x – sin 2x
\nf is continuous on [0, \u03c0]
\nsince f(0) = f(\u03c0)= 0 and
\nf is differentiable on [0, \u03c0]
\nBy Rolle’s theorem \u2203 c \u2208 (0, \u03c0)
\nsuch that f'(c) = 0
\nf'(x) = cos x – 2 cos 2x
\nf'(c) = 0 \u21d2 cosc – 2 cos 2c = 0
\n\u21d2 cosc – 2(2cos\u00b2c – 1):
\ncosc – 4 cos\u00b2c + 2=0 2
\n4 cos\u00b2 c – cosc – 2 = 0
\n\"Inter<\/p>\n

iii) log (x\u00b2 + 2) – log 3 on [-1, 1]
\nSolution:
\nLet f(x) = log (x\u00b2 + 2) – log 3
\nf is continuous on [-1, 1]
\nSince f(-1) = f(1) = 0 and f is
\nDifferentable on [-1, 1]
\nBy Rolle’s theorem \u2203 c \u2208 (-1, 1)
\nSuch that f'(c) = 0
\nf'(x) = \\(\\frac{1}{x^{2}+2}\\) = (2x)
\nf'(x) = \\(\\frac{2c}{c^{2}+2}\\) = 0
\n2c = 0
\nc = 0
\nc = 0 \u2208 (-1, 1).<\/p>\n

\"Inter<\/p>\n

Question 2.
\nIt is given that Rolle’s theorem holds for the function f(x) = x\u00b3 + bx\u00b2 + ax on [1, 3] with c = 2t + \\(\\frac{1}{\\sqrt{3}}\\). Find the values of a and b.
\nSolution:
\nGiven f(x) = x\u00b3 + bx\u00b2 + ax
\nf'(x) = 3x\u00b2 + 2bx + a
\n\u2234 f'(c) = 0 6 \u21d4 3c\u00b2 + 2bc + a = 0
\n\"Inter
\n\u21d4 b = 6 and b\u00b2 – 3a = 3
\n36 – 3 = 3a
\n33 = 3a
\na = 11
\nHence a = 11 and b = -6.<\/p>\n

Question 3.
\nShow that there is no real number k, for which the equation x\u00b2 – 3x + k = 0 has two distinct roots in [0, 1].
\nSolution:
\nClearly f(0) = f(c)
\n0 – 0 + k = 1 – 3 + k
\n0 = -2
\nWhich is not possible
\n\u2234 There is no real number K.<\/p>\n

Question 4.
\nFind a point on the graph of the curve y = (x – 3)\u00b2, where the tangent is parallel to the chord joining (3, 0) and (4, 1).
\nSolution:
\nGiven points (3, 0) and (4, 1)
\nThe slope of chord = \\(\\frac{1-0}{4-3}\\) = 1
\nGiven y = (x- 3)\u00b2
\n\\(\\frac{dy}{dx}\\) = 2(x – 3)
\n\u21d2 Slope = 2(x – 3)
\n1 = 2(x – 3)
\n\\(\\frac{1}{2}\\) = x – 3
\nx = \\(\\frac{1}{2}\\) + 3 = \\(\\frac{7}{2}\\)
\ny = (x – 3)\u00b2 = (\\(\\frac{7}{2}\\) – 3)\u00b2 = \\(\\frac{1}{4}\\)
\n\u2234 The point on the curve is (\\(\\frac{7}{2}\\), \\(\\frac{1}{4}\\))<\/p>\n

Question 5.
\nFind a point on the graph of the curve y = x\u00b3, where the tangent is parallel to the chord joining (1, 1) and (3, 27).
\nSolution:
\nGiven points (1, 1) and (3, 27)
\nSlope of chord = \\(\\frac{27-1}{3-1}\\) = 13
\nGiven y = x\u00b3
\n\\(\\frac{dy}{dx}\\) = 3x\u00b2
\n\u21d2 Slope = 3x\u00b2
\n13 = 3x\u00b2
\n\"Inter
\n\u2234 The point on the curve is (\\(\\frac{\\sqrt{39}}{3}\\), \\(\\frac{13\\sqrt{39}}{9}\\))<\/p>\n

\"Inter<\/p>\n

Question 6.
\nFind ‘c’, so that f'(c) = \\(\\frac{f(b)-f(a)}{b-a}\\) in the following cases.
\ni) f(x) = x\u00b2 – 3x – 1, a = \\(\\frac{-11}{7}\\), b = \\(\\frac{13}{7}\\).
\nSolution:
\n\"Inter
\nf'(x) = 2x – 3
\nf'(c) = 2c – 3
\nGiven f'(c)= \\(\\frac{f(b)-f(a)}{b-a}\\)
\n\"Inter<\/p>\n

ii) f(x) = ex<\/sup> ; a = 0, b = 1
\nSolution:
\nf(b) = f(1) = e’ = e
\nf(a) = f(0) = e\u00b0 = 1
\nGiven f(x) = ex<\/sup>
\nf'(x) = ex<\/sup>
\nGiven condition f'(c) = \\(\\frac{f(b)-f(a)}{b-a}\\)
\n\"Inter<\/p>\n

Question 7.
\nVerify the Rolle’s theorem for the function (x\u00b2 – 1) (x – 2) on [-1, 2]. Find the point in the interval where the derivate vanishes.
\nSolution:
\nLet f(x) = (x\u00b2 – 1) (x- 2) = x\u00b3 – 2x\u00b2 – x + 2
\nf is continous on [-1, 2]
\nsince f(-1) = f(2) = 0 and f is
\nDifferentiable on [-1, 2]
\nBy Rolle’s theorem \u2203 C \u2208 (-1, 2)
\nLet f'(c) = 0
\nf'(x) = 3x\u00b2 – 4x – 1
\n3c\u00b2 – 4c – 1 = 0
\n\"Inter<\/p>\n

Question 8.
\nVerify the conditions of the Lagrange’s mean value theorem for the following functions. In each case find a point ‘c’ in the interval as stated by the theorem
\ni) x\u00b2 -1 on [2, 3]
\nSolution:
\nSolution:
\nLet f(x) = x\u00b2 – 1
\nf is continous on [2, 3]
\nand f is differentiable
\nGiven f(x) = x\u00b2 – 1
\nf'(x) = 2x
\nBy Lagrange’s mean value theorem \u2203 C \u2208 (2, 3) such there
\n\"Inter<\/p>\n

ii) sin x – sin 2x on [0, \u03c0]
\nSolution:
\nLet f(x) = sin x – sin 2x
\nf is continuous on [0, \u03c0] and f is differentiable
\nGiven f(x) = sin x – sin 2x
\nf'(x) = cos x – 2 cos 2x
\nBy Lagrange’s mean value than \u2203 C \u2208 (0, \u03c0) such there
\nf'(c) = \\(\\frac{f(\\pi)-f(0)}{\\pi-0}\\)
\ncosc – 2 cos 2c = 0
\ncosc 2(2cos\u00b2 – 1) = 0
\ncosc – 4 cos\u00b2c + 2 = 0
\n4 cos\u00b2 c – cos c – 2 = 0
\n\"Inter<\/p>\n

\"Inter<\/p>\n

iii) log x on [1, 2].
\nSolution:
\nLet f(x) = log x
\nf is continuous on [1, 2] and f is differentiable
\nGiven f(x) = log x
\nf'(x) = \\(\\frac{1}{x}\\)
\nBy Lagrange’s mean value theorem \u2203 c \u2208 (1, 2) such that
\n\"Inter<\/p>\n","protected":false},"excerpt":{"rendered":"

Practicing the\u00a0Intermediate 1st Year Maths 1B Textbook Solutions\u00a0Inter 1st Year Maths 1B Applications of Derivatives Solutions Exercise 10(f) will help students to clear their doubts quickly. Intermediate 1st Year Maths 1B Applications of Derivatives Solutions Exercise 10(f) I. Question 1. Verify Rolle’s theorem for the following functions. i) x\u00b2 – 1 on [-1, 1] Solution: …<\/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\/35269"}],"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=35269"}],"version-history":[{"count":1,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts\/35269\/revisions"}],"predecessor-version":[{"id":35270,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts\/35269\/revisions\/35270"}],"wp:attachment":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/media?parent=35269"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/categories?post=35269"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/tags?post=35269"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}