NCERT Solutions for Class 12 Maths<\/a> Chapter 6 Application of Derivatives Ex 6.3 Questions and Answers are prepared by our highly skilled subject experts. https:\/\/mcq-questions.com\/ncert-solutions-for-class-12-maths-chapter-6-ex-6-3\/<\/p>\nNCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Exercise 6.3<\/h2>\n <\/p>\n
Ex 6.3 Class 12 Question 1.<\/strong> \nFind the slope of the tangent to the curve y = 3x4<\/sup> – 4x at x = 4. \nSolution: \ny = 3x4<\/sup> – 4x \nDifferentiating w.r.t. x, \n\u2234 \\(\\frac { dy }{ dx }\\) = 12x3<\/sup> – 4 \n\u2234 Req. slope = \\({ \\left( \\frac { dy }{ dx } \\right) }_{ x=4 }\\) \n= 12(43<\/sup>) – 4 = 764.<\/p>\nExercise 6.3 Class 12 Question 2.<\/strong> \nFind the slope of the tangent to the curve y = \\(\\frac { x-1 }{ x-2 }\\), x \u2260 2 at x = 10. \nSolution: \ny = \\(\\frac { x-1 }{ x-2 }\\), x \u2260 2 \nDifferentiating w.r.t. x, \n <\/p>\n6.3 Class 12 Question 3.<\/strong> \nFind the slope of the tangent to curve y = x3<\/sup> – x + 1 at the point whose x-coordinate is 2. \nSolution: \nThe curve is y = x3<\/sup> – x + 1 \nDifferentiating w.r.t. x, \n\\(\\frac { dy }{ dx }\\) = 3x\u00b2 – 1 \n\u2234 slope of tangent = \\(\\frac { dy }{ dx }\\) \n= 3x\u00b2 – 1 \n= 3 x 2\u00b2 – 1 \n= 11<\/p>\nEx6.3 Class 12 Question 4.<\/strong> \nFind the slope of the tangent to the curve y = x3<\/sup> – 3x + 2 at the point whose x-coordinate is 3. \nSolution: \nThe curve is y = x3<\/sup> – 3x + 2 \n\\(\\frac { dy }{ dx }\\) = 3x\u00b2 – 3 \n\u2234 slope of tangent = \\(\\frac { dy }{ dx }\\) \n= 3x\u00b2 – 3 \n= 3 x 3\u00b2 – 3 \n= 24<\/p>\n <\/p>\n
Ex 6.3 Class 12 Maths Ncert Solutions Question 5.<\/strong> \nFind the slope of the normal to the curve x = a cos3<\/sup> \u03b8, y = a sin3<\/sup> \u03b8 at \u03b8 = \\(\\frac { \\pi }{ 4 } \\). \nSolution: \ny = a sin3<\/sup>\u03b8 and x = a cos3<\/sup>\u03b8 \nDifferentiating w.r.t. x, \n <\/p>\nExercise 6.3 Class 12 Maths Question 6.<\/strong> \nFind the slope of the normal to the curve x = 1 – a sin \u03b8, y = b cos\u00b2 \u03b8 at \u03b8 = \\(\\frac { \\pi }{ 2 } \\) \nSolution: \nx = 1 – a sin \u03b8 and y = b cos\u00b2 \u03b8 \nDifferentiating w.r.t. x, \n <\/p>\n6.3 Maths Class 12 Question 7.<\/strong> \nFind points at which the tangent to the curve y = x3<\/sup> – 3x2<\/sup> – 9x + 7 is parallel to the x-axis. \nSolution: \ny = x3<\/sup> – 3x2<\/sup> – 9x + 7 \nDifferentiating w.r.t. x, \n\\(\\frac { dy }{ dx }\\) = 3(x – 3)(x + 1) \nTangent is parallel to x-axis if the slope of tangent = 0 \nor \\(\\frac { dy }{ dx }=0\\) \n\u21d2 3(x + 3)(x + 1) = 0 \n\u21d2 x = – 1, 3 \nwhen x = – 1, y = 12 & When x = 3, y = – 20 \nHence the tangent to the given curve are parallel to x-axis at the points (- 1, – 12), (3, – 20)<\/p>\n <\/p>\n
Exercise 6.3 Class 12 Maths Ncert Solutions Question 8.<\/strong> \nFind a point on the curve y = (x – 2)\u00b2 at which the tangent is parallel to the chord joining the points (2,0) and (4,4). \nSolution: \nThe equation of the curve is y = (x – 2)\u00b2 \nDifferentiating w.r.t x \n\\(\\frac { dy }{ dx }\\) = 2(x – 2 ) \nThe point A and B are (2,0) and (4,4) respectively. \n \nSlope of AB = \\(\\frac { { y }_{ 2 }-{ y }_{ 1 } }{ { x }_{ 2 }-{ x }_{ 1 } }\\) \n= \\(\\frac { 4-0 }{ 4-2 } =\\frac { 4 }{ 2 } \\) = 2 … (i) \nSlope of the tangent = 2 (x – 2) ….(ii) \nfrom (i) & (ii) 2 (x – 2) = 2 \n\u2234 x – 2 = 1 or x = 3 \nwhen x = 3,y = (3 – 2)\u00b2 = 1 \n\u2234 The tangent is parallel to the chord AB at (3, 1)<\/p>\n6.3 Class 12 Maths Question 9.<\/strong> \nFind the point on the curve y = x3<\/sup> – 11x + 5 at which the tangent is y = x – 11. \nSolution: \nHere, y = x3<\/sup> – 11x + 5 \n\u21d2 \\(\\frac { dy }{ dx }\\) = 3x\u00b2 – 11 \nThe slope of tangent line y = x – 11 is 1 \n\u2234 3x\u00b2 – 11 = 1 \n\u21d2 3x\u00b2 = 12 \n\u21d2 x\u00b2 = 4, x = \u00b12 \nWhen x = 2, y = – 9 & when x = -2,y = -13 \nBut (-2, -13) does not lie on the curve \n\u2234 y = x – 11 is the tangent at (2, -9)<\/p>\n <\/p>\n
Ncert Class 12 Maths Exercise 6.3 Solutions Question 10.<\/strong> \nFind the equation of all lines having slope -1 that are tangents to the curve y = \\(\\frac { 1 }{ x-1 }\\), x \u2260 1 \nSolution: \ny = \\(\\frac { 1 }{ x-1 }\\), x \u2260 1 \nDifferentiating w.r.t x \n\\(\\frac { dy }{ dx }\\) = \\(\\frac{-1}{(x-1)^{2}}\\) \nSince the tangent have slope – 1, \n\\(\\frac { dy }{ dx }\\) = – 1 \u21d2 \\(\\frac{-1}{(x-1)^{2}}\\) = – 1 \n\u21d2 (x – 1)\u00b2 = 1 \u21d2 x – 1 = \u00b1 1 \n\u21d2 x = 2 or x = 0 \nWhen x = 0, y = \\(\\frac { 1 }{ 0 – 1 }\\) = – 1 \nWhen x = 2, y = \\(\\frac { 1 }{ 2 – 1 }\\) = 1 \n\u2234 The required points are (0, – 1) and (2, 1) \nEquation of the tangent at (0, – 1) is \ny – 1 = – 1 (x – 0) \u21d2 x + y + 1 = 0 \nEquation of the tangent at (2, 1) is \ny – 1 = – 1(x – 2) \u21d2 x + y – 3 = 0<\/p>\nClass 12 Ex 6.3 Question 11.<\/strong> \nFind the equation of ail lines having slope 2 which are tangents to the curve y = \\(\\frac { 1 }{ x-3 }\\), x \u2260 3. \nSolution: \nHere y = \\(\\frac { 1 }{ x-3 }\\) \n\\(\\frac { dy }{ dx } ={ (-1)(x-3) }^{ -2 }=\\frac { -1 }{ { (x-3) }^{ 2 } } \\) \n\u2235 slope of tangent = 2 \n\\(\\frac { -1 }{ { (x-3) }^{ 2 } } =2\\Rightarrow { (x-3) }^{ 2 }=-\\frac { 1 }{ 2 } \\) \nWhich is not possible as (x – 3)\u00b2 > 0 \nThus, no tangent to y = \\(\\frac { 1 }{ x-3 }\\) \nHence there is no tangent to the curve with slope 2.<\/p>\nClass 12 Maths Ex 6.3 Question 12.<\/strong> \nFind the equations of all lines having slope 0 which are tangent to the curve y = \\(\\frac { 1 }{ { x }^{ 2 }-2x+3 }\\) \nSolution: \ny = \\(\\frac { 1 }{ { x }^{ 2 }-2x+3 }\\) \nDifferentiating w.r.t x \n\u2234 \\(\\frac { dy }{ dx }\\) = \\(\\frac{-(2 x-2)}{\\left(x^{2}-2 x+3\\right)^{2}}\\) \nSince the slope of the tangent is zero, \n\\(\\frac { dy }{ dx }\\) = 0 \n <\/p>\n <\/p>\n
Class 12 Maths Chapter 6 Exercise 6.3 Question 13.<\/strong> \nFind points on the curve \\(\\frac { { x }^{ 2 } }{ 9 } + \\frac { { y }^{ 2 } }{ 16 } \\) = 1 at which the tangents are \ni. parallel to x-axis \nii. parallel to y-axis \nSolution: \nThe equation of the curve is \\(\\frac { { x }^{ 2 } }{ 9 } +\\frac { { y }^{ 2 } }{ 16 }\\) = 1 … (i) \nDifferentiating both sides w.r.t x \n\\(\\frac { 2 }{ 9 }\\)x + \\(\\frac { 2y }{ 16 }\\)\\(\\frac { dy }{ dx }\\) = 0 \n\u2234 \\(\\frac { dy }{ dx }\\) = \\(\\frac { -16x }{ 9y }\\)<\/p>\ni. Since the tangent is parallel to x axis, \n\\(\\frac { dy }{ dx }\\) = 0 \u21d2 \\(\\frac{-16x}{9y}\\) \nHence x = 0 \nWhen x = 0, we get \\(\\frac{0}{9}\\) + \\(\\frac { { y }^{ 2 } }{ 16 }\\) = 1 \n\u21d2 y\u00b2 = 16 \u21d2 y = \u00b14 \n\u2234 (0,4) and (0, – 4) are the points at which the tangents are parallel to the x axis<\/p>\n
ii. Since the tangent is parallel to y axis, its slope is not defined, then the normal is parallel to x-axis whose slope is zero. \n \n\u2234 (3, 0) and (- 3, 0) are the points at which the tangents are parallel to they axis.<\/p>\n
Class 12 Ch 6 Ex 6.3 Question 14.<\/strong> \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) \n(ii) y = x4<\/sup> – 6x3<\/sup> + 13x2<\/sup> – 10x + 5 at (1,3) \n(iii) y = x3<\/sup> at (1, 1) \n(iv) y = x2<\/sup> at (0,0) \n(v) x = cos t, y = sin t at t = \\(\\frac { \\pi }{ 4 } \\) \nSolution: \n(i) y = x4<\/sup> – 6x3<\/sup> + 13x2<\/sup> – 10x + 5 \nDifferentiating w.r.t x \n \nEquation of the tangent at (0, 5) is \n(y- 5) = – 10(x- 0) \ni.e., 10x+y-5 = 0 \nEquation of the normal at (0, 5) is \n(y – 5) = – 10(x – 0) \ni.e., x – 10y + 50 = 0<\/p>\n(ii) y = x4<\/sup> – 6x3<\/sup> + 13x2<\/sup> – 10x + 5 \nDifferentiating w.r.t. x, \n\\(\\frac { dy }{ dx }\\) = 4x3<\/sup> – 18x2<\/sup> + 26x – 10 \n\\(\\frac { dy }{ dx }\\)(1, 3)<\/sub> = 4(1) – 18(1) + 26(1) – 10 = 2 \nAt (1, 3), the slope of the tangent = 2 \nAt (1, 3), the slope of the normal = \\(\\frac { – 1 }{ 2 }\\) \nEquation of the tangent at (1, 3) is y – 3 = 2( x – 1) \ni.e., y = 2x + 1 \nEquation of the normal at (1, 3) is \ny – 3 = y(x – 1) \ni.e., x + 2y – 7 = 0<\/p>\n(iii) y = x3<\/sup> \nDifferentiating w.r.t. x, \n\\(\\frac { dy }{ dx }\\) = 3x2<\/sup> \nAt(1, 1), \\(\\frac { dy }{ dx }\\) = 3 x 1\u00b2 = 3 \nAt (1, 1), the slope of the tangent = 3 \nEquation of the tangent at(1, 1) is \ny – 1 = 3(x – 1) \ni.e., y = 3x – 2 \nAt (1, 1), the slope o fthe normal = \\(\\frac { -1 }{ 3 }\\) \nEquation of the normal at (1, 1) is \ny – 1 = \\(\\frac { -1 }{ 3 }\\) (x – 1) \ni.e., x + 3y – 4 = 0<\/p>\n(iv) y = x2<\/sup> \nDifferentiating w.r.t. x, \n\\(\\frac { dy }{ dx }\\) = 2x \n[ \\(\\frac { dy }{ dx }\\) ](0, 0)<\/sub> = 0 \nAt (0, 0), the slope of the tangent = 0 \nEquation of the tangent at (0, 0) is \ny – 0 = 0(x – 0) \ni.e., y = 0 or the x axis \nSince the tangent is the x axis, the normal is parallel to y axis. \n\u2234 The equation of the normal at (0, 0) is x = 0, since the normal passes through (0,0)<\/p>\n(v) y = sin t and x = cos t \nDifferentiating w.r.t. t, we get \\(\\frac { dy }{ dt }\\) = cos t \n <\/p>\n
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Ncert Solutions For Class 12 Maths Chapter 6 Exercise 6.3 Question 15.<\/strong> \nFind the equation of the tangent line to the curve y = x2<\/sup> – 2x + 7 which is \n(i) parallel to the line 2x – y + 9 = 0 \n(ii) perpendicular to the line 5y – 15x = 13. \nSolution: \ny = x2<\/sup> – 2x + 7 \nDifferentiating w.r.t. x, \n\\(\\frac { dy }{ dx }\\) = 2x – 2 \n\\(\\frac { dy }{ dx }\\) = 2(x – 1) … (1) \n(i) Slope of the line 2x – y + 9 = 0 is 2 \nSince the tangent to the curve is parallel to the line 2x – y + 9 = 0, we get \n\\(\\frac { dy }{ dx }\\) = 2 \nFrom (1), we get 2(x – 1) = 2 \u21d2 x = 2 \nWhen x = 2, y = 2\u00b2 – 2 x 2 + 7 = 7 \n\u2234 At (2, 7) the tangent is parallel to 2x – y + 9 = 0 \nThe equation of the tangent to the given curve at (2, 7) is y – 7 = 2(x – 2) \ni.e., 2x – y + 3 = 0 or y – 2x – 3 = 0 \nSlope of the line – 15x = 13 is 3<\/p>\n(ii) Since the tangent to the curve is perpendicular to the line 5y – 15x = 13, \n\\(\\frac { dy }{ dx }\\) = \\(\\frac { -1 }{ 3 }\\) \nFrom (1), we get 2(x – 1) = \\(\\frac { -1 }{ 3 }\\) \n \nThe tangent is perpendicular to the line 5y – 15x = 13 \nThe equation of the tangent to the given \n <\/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. \nSolution: \nHere, y = 7x3<\/sup> + 11 \n\u21d2 x \\(\\frac { dy }{ dx }\\) = 21 x\u00b2 \nNow m1<\/sub> = slope at x = 2 is \n\\({ \\left( \\frac { dy }{ dx } \\right) }_{ x=2 }\\) \n= 21 x 2\u00b2 = 84 \nand m2<\/sub> = slope at x = -2 is \n\\({ \\left( \\frac { dy }{ dx } \\right) }_{ x=-2 }\\) \n= 21 x (- 2)\u00b2 = 84 \nHence, m1<\/sub> = m2<\/sub> Thus, the tangents to the given curve at the points where x = 2 and x = – 2 are parallel<\/p>\n <\/p>\n
Question 17. \nFind the points on the curve y = x\u00b3 at which the slope of the tangent is equal to the y-coordinate of the point \nSolution: \ny = x\u00b3 \nDifferentiating w.r.t. x, \n\\(\\frac { dy }{ dx }\\) = 3x\u00b2 \nThe slope of the tangent = 3x\u00b2 \nSince slope of the tangent = y coordinate, \n3x\u00b2 = y or 3x\u00b2 = x\u00b3 \n= 3x\u00b2 – x\u00b3 = 0 = x\u00b2(3 – x) = 0 \n\u21d2 x = 3 or x = 0 \nWhen x = 0, y = (0)\u00b3 = O \nWhen x = 3, y= (3)\u00b3 = 27 \n\u2234 (0, 0) and (3, 27) are the points at which the slope of the tangent is equal to the y coordinate of the point.<\/p>\n
Question 18. \nFor the curve y = 4x3<\/sup> – 2x5<\/sup>, find all the points at which the tangent passes through the origin. \nSolution: \nLet, (h, k) be the point at which the tangent passes through the origin. \ny = 4x\u00b3 – 2x5<\/sup> \nDifferentiating w.r.t. x, \n\\(\\frac { dy }{ dx }\\) = 12x\u00b2 – 10x4<\/sup> \nAt (h, k) the slope of the tangent = \\(\\frac { dy }{ dx }\\)(h,k)<\/sub> = 12h\u00b2 – 10h4<\/sup> \nEquation of the tangent at (h, k) is \ny – k = (12h\u00b2 – 10h4<\/sup>) (x – h) \nSince the tangent passes through the origin, \n(0 – k) = (12h\u00b2 – 10h4<\/sup>) (0 – h) \n\u21d2 k = 12h\u00b3 – 10h5<\/sup> … (1) \n(h, k) is a point on the curve y = 4x\u00b3 – 2x5<\/sup> \n\u2234 k = 4h\u00b3 – 2h5<\/sup> … (2) \nFrom (1) & (2) we get \n12h\u00b3 – 10h5<\/sup> = 4h\u00b3 – 2h5<\/sup> \n\u21d2 8h\u00b3 – 8h5<\/sup> = 0 \n\u21d2 8h\u00b3 (1 – h\u00b2) = 0 \n\u21d2 8h\u00b3 (1 – h) (1 +h) = 0 \n\u21d2 h = 0 or h= 1 or h = – 1 \nWhen h = 0, k = 0 \nWhen h = 1, k = 2 \nWhen h = – 1, k = – 2 \n\u2234 (0, 0), (1, 2) and (- 1, – 2) are the points at which the tangent passes through the origin.<\/p>\n <\/p>\n
Question 19. \nFind the points on the curve x2<\/sup>\u00a0+ y2<\/sup> – 2x – 3 = 0 at which the tangents are parallel to the x-axis. \nSolution: \nHere, x2<\/sup> + y2<\/sup> – 2x – 3 = 0 \n\u21d2 \\(\\frac { dy }{ dx } =\\frac { 1-x }{ y }\\) \nTangent is parallel to x-axis, if \\(\\frac { dy }{ dx }\\) = 0 i.e. \nif 1 – x = 0 \n\u21d2 x = 1 \nPutting x = 1 in (i) \n\u21d2 y = \u00b12 \nHence, the required points are (1,2), (1, -2) i.e. (1, \u00b12).<\/p>\nQuestion 20. \nFind the equation of the normal at the point (am\u00b2, am\u00b3 ) for the curve ay\u00b2 = x\u00b3 . \nSolution: \nay\u00b2 = x\u00b3 \nDifferentiating w.r.t. x, \n <\/p>\n
Question 21. \nFind the equation of the normal’s to the curve y = x3<\/sup> + 2x + 6 which are parallel to the line x + 14y + 4 = 0. \nSolution: \ny = x3<\/sup> + 2x + 6 \nDifferentiating w.r.t. x, \n\\(\\frac { dy }{ dx }\\) = 3x\u00b2 + 2 \nSlope of the normal = \\(\\frac{-1}{3 x^{2}+2}\\) \nSlope of the line x + 14y + 4 = 0 is \\(\\frac { -1 }{ 14 }\\) \nSince the normal is parallel to x + 14y + 4 = 0 \n\\(\\frac{-1}{3 x^{2}+2}\\) = \\(\\frac { -1 }{ 14 }\\) \u21d2 3x\u00b2 + 2 = 14 \n\u21d2 3x\u00b2 = 12 \u21d2 x\u00b2 = 4 \u21d2 x = \u00b1 2 \nWhen x = 2, y = 18 and when x = – 2, y = – 6 \n\u2234 At (2, 18) and (- 2, – 6), the normals are parallel to x + 14y + 4 = 0. \nSlope of the normal = \\(\\frac { -1 }{ 14 }\\) \nEquation of the normal at (2, 18) is \ny – 8 = \\(\\frac { -1 }{ 14 }\\)(x – 2) \ni.e., x + 14y – 254 = 0 \nEquation of the normal at ( -2, -6) is \ny + 6 = \\(\\frac { -1 }{ 14 }\\) (x + 2) i.e., x + 14y + 86 = 0<\/p>\n <\/p>\n
Question 22. \nFind the equations of the tangent and normal to the parabola y\u00b2 = 4ax at the point (at\u00b2,2at). \nSolution: \ny\u00b2 = 4ax \nDifferentiating w.r.t. x, \n2y\\(\\frac { dy }{ dx }\\) = 4a \n \nEquation of the tangent at (at\u00b2, 2at) is \ny – 2at = \\(\\frac { 1 }{ t }\\)(x – at\u00b2) \nyt – 2at\u00b2 = x – at\u00b2 \nx – yt + at\u00b2 = 0 or ty = x + at\u00b2 \nSlope of the normal at (at\u00b2, 2at) is – t. \nEquation of the normal at (at\u00b2, 2at) is \ny – 2at = – t(x – at\u00b2) \nxt + y = at\u00b3 + 2at \ny = – tx + 2at + at\u00b3<\/p>\n
Question 23. \nProve that the curves x = y\u00b2 and xy = k cut at right angles if 8k\u00b2 = 1. \nSolution: \nThe curve are \nx = y\u00b2 … (1) \nxy = k … (2) \nLet P be the point of intersection \nSubstituting x = y\u00b2 in (2), \n \nSince the two curves cut at right angles at P, the product of their slopes at P is – 1. \n \nTaking the cubes we get k\u00b2 = \\(\\frac { 1 }{ 8 }\\) or 8k\u00b2 = I<\/p>\n
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Question 24. \nFind the equations of the tangent and normal to the hyperbola \\(\\frac { { x }^{ 2 } }{ { a }^{ 2 } } -\\frac { { y }^{ 2 } }{ { b }^{ 2 } }\\) = 1 at the point (x0<\/sub>, y0<\/sub>). \nSolution: \n <\/p>\nQuestion 25. \nFind the equation of the tangent to the curve y = \\(\\sqrt { 3x – 2 } \\) which is parallel to the line 4x – 2y + 5 = 0. \nSolution: \ny = \\(\\sqrt { 3x – 2 } \\) \nDifferentiating w.r.t. x, \n\\(\\frac { dy }{ dx }\\) = \\(\\frac{3}{2 \\sqrt{3 x-2}}\\) \nSlope of the tangent = \\(\\frac{3}{2 \\sqrt{3 x-2}}\\) \nSlope of the line 4x – 2y + 5 = 0 is 2 \nSince the tangent is parallel to the line dy \n4x – 2y + 5 = 0, we get \\(\\frac { dy }{ dx }\\) = 2 \n \n\u2234 At(\\(\\frac { 41 }{ 48 }\\), \\(\\frac { 3 }{ 4 }\\)), the tangent is parallel to the line 4x – 2y + 5 = 0 \nEquation of the tangent at (\\(\\frac { 41 }{ 48 }\\), \\(\\frac { 3 }{ 4 }\\)) is \ny – \\(\\frac { 3 }{ 4 }\\) = 2(x – \\(\\frac { 41 }{ 48 }\\) \n24y – 18 = 48x – 41 \ni.e., 48x – 24y = 23<\/p>\n
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Question 26. \nThe slope of the normal to the curve y = 2x\u00b2 + 3 sin x at x = 0 is \n(a) 3 \n(b) \\(\\frac { 1 }{ 3 }\\) \n(c) -3 \n(d) \\(-\\frac { 1 }{ 3 }\\) \nSolution: \n(d) \u2235 y = 2x\u00b2 + 3sinx \n\u2234 \\(\\frac { dy }{ dx }\\) = 4x+3cosx \nat \nx = 0, \\(\\frac { dy }{ dx }\\) = 3 \n\u2234 slope = 3 \n\u21d2 slope of normal is = \\(\\frac { 1 }{ 3 }\\)<\/p>\n
Question 27. \nThe line y = x + 1 is a tangent to the curve y\u00b2 = 4x at the point \n(a) (1,2) \n(b) (2,1) \n(c) (1,-2) \n(d) (-1,2) \nSolution: \n(a) The curve is y\u00b2 = 4x, \n\u2234 \\(\\frac { dy }{ dx } =\\frac { 4 }{ 2y } =\\frac { 2 }{ y } \\) \nSlope of the given line y = x + 1 is 1 \n\u2234 \\(\\frac { 2 }{ y }\\) = 1 \ny = 2 Putting y = 2 in \n\u21d2 y\u00b2 = 4x \n\u21d2 2\u00b2 = 4x \n\u21d2 x = 1 \n\u2234 The Point of contact is (1, 2)<\/p>\n","protected":false},"excerpt":{"rendered":"
These NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Ex 6.3 Questions and Answers are prepared by our highly skilled subject experts. https:\/\/mcq-questions.com\/ncert-solutions-for-class-12-maths-chapter-6-ex-6-3\/ NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Exercise 6.3 Ex 6.3 Class 12 Question 1. Find the slope of the tangent to the curve y …<\/p>\n
NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Ex 6.3<\/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 6.3 - MCQ Questions<\/title>\n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n\t \n\t \n\t \n