{"id":19661,"date":"2020-01-13T17:11:19","date_gmt":"2020-01-13T11:41:19","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=19661"},"modified":"2021-07-02T15:29:39","modified_gmt":"2021-07-02T09:59:39","slug":"2nd-puc-maths-question-bank-chapter-6-ex-6-3","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/2nd-puc-maths-question-bank-chapter-6-ex-6-3\/","title":{"rendered":"2nd PUC Maths Question Bank Chapter 6 Application of Derivatives Ex 6.3"},"content":{"rendered":"

Students can Download Maths Chapter 6 Application of Derivatives Ex 6.3 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 6.3<\/h2>\n

2nd PUC Maths Application of Derivatives NCERT Text Book Questions and Answers Ex 6.3<\/h3>\n

Question 1.
\nFind the slope of the tangent to the curve y = 3x4<\/sup> – 4x at x = 4.
\nAnswer:
\n\"2nd<\/p>\n

Question 2.
\nFind the slope of the tangent to the curve
\n\\(y=\\frac{x-1}{x-2}, x \\neq 2 \\text { at } x=10 \\)
\nAnswer:
\n\"2nd<\/p>\n

Question 3.
\nFind the slope of the tangent to curve y = x3<\/sup> – x + 1 at the point whose x- coordinate is 2.
\nAnswer:
\n\"2nd<\/p>\n

Question 4.
\nFind the slope of the tangent to the curve y = x3<\/sup> – 3x + 2 at the point whose x-coordinate is 3.
\nAnswer:
\n\\(\\frac{d y}{d x}\\) = 3x2<\/sup> – 3 dx
\nslope at x = 3 is 3 (9) – 3 = 24.<\/p>\n

\"KSEEB<\/p>\n

Question 5.
\nFind the slope of the normal to the curve
\n\\(x=a \\cos ^{3} \\theta, y=a \\sin ^{3} \\theta \\text { at } \\theta=\\frac{\\pi}{4}\\)
\nAnswer:
\n\"2nd
\n\"2nd<\/p>\n

Question 6.
\nFind the slope of the normal to the curve
\n\\(x=1-a \\sin \\theta, y=b \\cos ^{2} \\theta \\text { at } \\theta=\\frac{\\pi}{2}\\)
\nAnswer:
\n\"2nd<\/p>\n

Question 7.
\nFind points at which the tangent to the curve y = x3<\/sup> – 3x2<\/sup> – 9x + 7 is parallel to
\nthe x-axis.
\nAnswer:
\n\\(\\frac{d y}{d x}\\) = 3x2<\/sup> – 6x – 9, slope of the tangent
\nsince the tangent is parallel to x-axis \\(\\frac{d y}{d x}\\) = 0
\n3x2<\/sup> – 6x – 9 = 0
\n3 (x + 1) (x – 3) = 0, x = 3, x = -1
\nwhen x = 3, y = 27 – 27 – 27 4- 7 = -20
\nwhen x = -1, y = -1 -3 + 9 + 7 = 12
\nThe points at which the tangent parallel to x – axis are (3, -20) and (-1, 12).<\/p>\n

Question 8.
\nFind a point on the curve y = (x – 2)2<\/sup> at which the tangent is parallel to the chord joining the points (2, 0) and (4, 4).
\nAnswer:
\n\"2nd
\n\"2nd<\/p>\n

Question 9.
\nFind the point on the curve y = x3<\/sup> – 11x + 5 at which the tangent is y = x – 11.
\nAnswer:
\nSlope of the line y = x – 11 is 1
\nslope of the curve \\(\\frac{d y}{d x}\\) = 3x2<\/sup> – 11
\n\u2234 3x2<\/sup> – 11 = 1\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 ‘
\n3x2<\/sup> = 12 \u21d2 x2<\/sup> = 4, x = +2
\nwhen x = 2, y = (2)3<\/sup> – 11 (2) + 5 = -9
\nwhen x = -2, y = -8 + 22 + 5 = 19
\npoints are (2, -9) and (-2, 19)
\nequation of tangent at (2, -9) and slope is 1
\ny + 9 = 1 (x – 2)
\ny = x- 11 equation of tangent at (-2, 19)
\ny – 19 = 1 (x + 2) \u21d2 y = x + 211
\n\u2234 (2, -9) is the only point at which the tangent is y = x – 11.10.<\/p>\n

\"KSEEB<\/p>\n

Question 10.
\nFind the equation of all lines having slope – 1 that are tangents to the curve
\n\\(y=\\frac{1}{x-1}, x \\neq 3\\)
\nAnswer:
\n\"2nd
\nx = 2, x = 0
\nwhen x = 2, y = 1
\nwhen x = 0, y = -1
\nThe points are (2, 1) and (0,-1)
\nequation of line though (2,1) having slope = -1
\nis y – 1 = -1 x (x – 2) \u21d2 y + x = 3
\nor x + y – 3 = 0 and equation of line through (0,- 1)y + 1 = -1 (x – 0)
\n\u2234\u00a0 y + x + 1 = 0.<\/p>\n

Question 11.
\nFind the equation of all lines having slope 2 which are tangents to the curve
\n\\(y=\\frac{1}{x-3}, x \\neq 3\\)
\nAnswer:
\nslope of the line = 2
\nslope of the curve = \\(\\frac{-1}{(x-3)^{2}}\\)
\n\\(\\frac{-1}{(x-3)^{2}}\\) =2
\n\u21d2 2 (x -3)2<\/sup> = -1
\n\u21d2 2 (x2<\/sup> – 6x + 9) = -1
\n\u21d2 2x2<\/sup> – 12x + 19 = 0
\nwhich has no real roots b2<\/sup> – 4ac < 0
\nhence there is no point on the curve.<\/p>\n

Question 12.
\nFind the equations of all lines having slope 0 which are tangent to the curve
\n\\(y=\\frac{1}{x^{2}-2 x+3}\\)
\nAnswer:
\n\"2nd<\/p>\n

Question 13.
\nFind the points of curve \\(\\frac{x^{2}}{9}+\\frac{y^{2}}{16}=1\\) which the tangents are
\n(i) parallel to x-axis
\n(ii) parallel to y-axis.
\nAnswer:
\n16x2<\/sup> + 9y2<\/sup> = 144
\n16 x 2 x + 9 x 2y x \\(\\frac{d y}{d x}\\) = 0
\n\"2nd
\n(0,4) and (0, -4) are the points on the curve at which tangent is parallel to x – axis.<\/p>\n

\"KSEEB<\/p>\n

(ii) For tangents parallel to y – axis \\(\\frac{d y}{d x}\\) = \\(\\frac{1}{0}\\)
\n\\(\\frac{-16 x}{9 y}=\\frac{1}{0} \\Rightarrow y=0\\)
\nwhen y = 0, x2<\/sup> = 9x = \u00b1 3 point is (3,0) and (-3,0)
\n(3, 0) and (-3, 0) are the points at which the curve is parallel to y – axis.<\/p>\n

Question 14.
\nFind the equations of the tangent and normal to the given curves at the indicated points:
\n(i) y = x4<\/sup> – 6x3<\/sup> + 13x2<\/sup> – 10x + 5 at (0, 5)
\nAnswer:
\ny = x4<\/sup> – 6x3<\/sup> + 13x2<\/sup> – 10x + 5 at (0, 5) dy
\n\\(\\frac{d y}{d x}\\) = 4x3<\/sup> – 18x2<\/sup> + 26x – 10 dx
\nslope at (0,5) = -10
\n\u2234 Equation of tangent at (0, 5) is
\ny – 5 = -10 (x – 0)
\ny – 5 = -10 x 10 x + y – 5 = 0
\nslope of the normal at (0,5)
\n\\((0,5)=\\frac{-1}{-10}=\\frac{1}{10}\\)
\n\u2234 equation of normal is
\ny – 5 =\\(\\frac{1}{10}\\) (x – 0) = 10y – 50 = x
\nx – 10y + 50 = 0<\/p>\n

(ii) y = x4<\/sup> – 6x3<\/sup> + 13x2<\/sup> – 10x + 5 at (1, 3)
\nAnswer:
\n\\(\\frac{d y}{d x}\\) = 4x3<\/sup> – 18x2<\/sup> + 26x – 10 dx
\nslope of the tangent at x = 1<\/p>\n

(iii) y = x3<\/sup> at (1,1)
\nAnswer:
\n\\(\\frac{d y}{d x}=3 x^{2}\\)\u00a0
\n<\/sup>\u2234 slope of the tangent at x = 1 = 3
\n\u2234 Equation of tangent is y – 1 = 3(x – 1)
\n3x – y – 2 = 0
\n\u2234 slope of normal = \\(\\frac{-1}{3} \\)
\n\u2234 equation of normal is y – 1 = \\(\\frac{-1}{3} \\)(x – 1)
\n3y – 3 = – (x – 1)
\nx + 3y – 4 = 0.<\/p>\n

\"KSEEB<\/p>\n

(iv) y = x2<\/sup> at (0, 0)
\nAnswer:
\n\\(\\frac{d y}{d x}=2 x\\)
\n\u2234 slope at x = 0 =0
\n\u2234 Equation of tangent is y – 0 = 0 (x – 0) \u21d2 y = 0
\nslope of the normal = -1\/(0)
\n\u2234 equation of normal \\(y – 0=\\frac{-1}{0}(x-0) \\Rightarrow x=0\\)<\/p>\n

(v) x = cos t, y = sin t at \\(\\frac{\\pi}{4}\\)
\nAnswer:
\n\"2nd
\n\"2nd<\/p>\n

Question 15.
\nFind the equation of the tangent line to the curve y = x2<\/sup> – 2x +7 which is
\n(a) parallel to the line 2x – y + 9 = 0
\nAnswer:
\n\\(\\frac{d y}{d x}\\)\u00a0= 2x – 2 also slope = 2
\n2 x – 2 = 2 \u21d22x = 4 \u21d2 x = 2
\nwhen x = 2, y = (2)2<\/sup> – 2 (2) + 7 = 7
\n\u2234 equation of tangent is y – 7 = 2 (x – 2)
\n2x – y + 3 = 0 \u21d2 2x – y + 3 = 0.<\/p>\n

\n

(b) perpendicular to the line 5y – 15x = 13.
\nAnswer:
\n\"2nd
\n12x + 36y – 227 = 0<\/p>\n

Question 16.
\nShow that the tangents to the curve y = 7x3<\/sup> + 11 at the points where x = 2 and x = – 2 are parallel.
\nAnswer:
\n\"2nd
\n\"KSEEB<\/p>\n

Question 17.
\nFind the points on the curve y = x3<\/sup> at which the slope of the tangent is equal to the y-coordinate of the point.
\nAnswer:
\ny = x3
\n<\/sup>\\(\\frac{d y}{d x}\\) = 3x2,
\n<\/sup>Given that \\(\\frac{d y}{d x}\\) = y = x3<\/sup>
\n\u2234 3x2<\/sup> = x3<\/sup> \u21d2 x2<\/sup> (3 – x) = 0 \u21d2 x = 0 or x = 3
\nwhen x = 3, y = 33<\/sup> = 27, when x = 0, y = 0
\n\u2234 The required points are (0, 0), (3, 27).<\/p>\n

Question 18.
\nFor the curve y = 4x3<\/sup> – 2x5<\/sup>, find all the points at which the tangent passes through the origin.
\nAnswer:
\ny = 4x3<\/sup> – 2x5 <\/sup>dy
\n\\(\\frac{d y}{d x}\\) = 12x2<\/sup> – 10x4 <\/sup>dx
\nLet (a, b) be the point on the curve at which the tangent passes through the origin.
\n\u2234 Equation of tangent is
\ny – b = (12a2<\/sup> – 10a2<\/sup>) (x – a)
\nbut this passes through the origin
\n\u2234 0 – b = (12a2<\/sup> – 10a4<\/sup>) (-a) b = 12a3<\/sup> – 10a5<\/sup> ….(1)
\nAlso from the equation b = 4a3<\/sup> – 2a5<\/sup> …. (2)
\nfrom (1) and (2) 12a3<\/sup> – 10a5<\/sup> = 4a3<\/sup> – 2a5
\n<\/sup>8a3<\/sup> = 8a5
\n<\/sup>a3<\/sup> (1 – a2<\/sup>) = 0 \u21d2 a = 0, a = + 1
\nwhen a = 0, b = 0, (0, 0)
\na = 1,b= 12(1)- 10(1) = 2, (1,2)
\na = -1, b = 12 (-1) -10 (-1) = -2, (-1, -2)
\nHence the required points are (0,0), (1,2), (-1,-2).<\/p>\n

Question 19.
\nFind the points on the curve x2 <\/sup>+ y2 <\/sup>– 2x – 3 = 0 at which the tangents are parallel to the x-axis.
\nAnswer:
\n\"2nd<\/p>\n

\"KSEEB<\/p>\n

Question 20.
\nFind the equation of the normal at the point (am2<\/sup>,am3<\/sup>) for the curve ay2<\/sup> = x3<\/sup>.
\nAnswer:
\n\"2nd<\/p>\n

Question 21.
\nFind the equation of the normal to the curve y = x3<\/sup> + 2x + 6 which are parallel to the line x + 14y + 4 = 0.
\nAnswer:
\ny = x3<\/sup> + 2x + 6
\n\\(\\frac{d y}{d x}\\) = 3x2<\/sup>+ 2
\n\"2nd<\/p>\n

Question 22.
\nFind the equations of the tangent and normal to the parabola y2<\/sup> = 4ax at the
\npoint (at2<\/sup>, 2at).
\nAnswer:
\n\"2nd
\n\"2nd<\/p>\n

Question 23.
\nProve that the curves x = y2<\/sup> and xy = k cut at right angles* if 8k2<\/sup> = 1.
\nAnswer:
\n\"2nd<\/p>\n

Question 24.
\nFind the equations of the tangent and normal to the hyperbola
\n\\(\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1\\) at the point (x0<\/sub>, y0<\/sub>)
\nAnswer:
\n\"2nd
\n\"2nd<\/p>\n

Question 25.
\nFind the equation of the tangent to the curve
\n\\(y=\\sqrt{3 x-2}\\) which is parallel to the line 4x – 2y +5 = 0
\nAnswer:
\n\"2nd<\/p>\n

\"2nd<\/p>\n

\"KSEEB<\/p>\n

Choose the correct answer in Exercises 26 and 27.<\/p>\n

Question 26.
\nThe slope of the normal to the curve y = 2x2<\/sup> + 3 sin x at x = 0 is
\n(A) 3
\n(B) 1\/3
\n<\/sub>(C) -3
\n(D) -1\/3
\nAnswer:
\n\"2nd<\/p>\n

Question 27.
\nThe line y = x + 1 is a tangent to the curve y2<\/sup> = 4x at the point
\n(A) (1, 2)
\n(B) (2,1)
\n(C) (1, – 2)
\n(D) (- 1, 2)
\nAnswer:
\n\"2nd
\n\"2nd<\/p>\n<\/div>\n","protected":false},"excerpt":{"rendered":"

Students can Download Maths Chapter 6 Application of Derivatives Ex 6.3 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 6.3 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\/19661"}],"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=19661"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/19661\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=19661"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=19661"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=19661"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}