{"id":31614,"date":"2022-03-29T15:00:53","date_gmt":"2022-03-29T09:30:53","guid":{"rendered":"https:\/\/mcq-questions.com\/?p=31614"},"modified":"2022-03-29T15:33:14","modified_gmt":"2022-03-29T10:03:14","slug":"ncert-solutions-for-class-12-maths-chapter-6-miscellaneous-exercise","status":"publish","type":"post","link":"https:\/\/mcq-questions.com\/ncert-solutions-for-class-12-maths-chapter-6-miscellaneous-exercise\/","title":{"rendered":"NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Miscellaneous Exercise"},"content":{"rendered":"

These NCERT Solutions for Class 12 Maths<\/a> Chapter 6 Application of Derivatives Miscellaneous Exercise Questions and Answers are prepared by our highly skilled subject experts. https:\/\/mcq-questions.com\/ncert-solutions-for-class-12-maths-chapter-6-miscellaneous-exercise\/<\/p>\n

NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Miscellaneous Exercise<\/h2>\n

\"NCERT<\/p>\n

Question 1.
\nUsing differentials, find the approximate value of each of the following.
\na. \\(\\left(\\frac{17}{81}\\right)^{\\frac{1}{4}}\\)
\nb. \\(\\text { (33) }^{-\\frac{1}{5}}\\)
\nSolution:
\n\"NCERT
\n\u2206y is approximately equal to dy
\n\u2234 (1) \u21d2 (33)\\(\\frac { -1 }{ 5 }\\)<\/sup> \u2248 2-1<\/sup> – 0.003
\n= 0.5 – 0.003 = 0.497
\nHence (33)\\(\\frac { -1 }{ 5 }\\)<\/sup> \u2248 0.497<\/p>\n

Question 2.
\nShow that the function given by f(x) = \\(\\frac { log x }{ x }\\) has maximum at x = e.
\nSolution:
\n\"NCERT
\nFor maxima, f'(x) = 0
\ni.e., \\(\\frac{1-\\log x}{x^{2}}\\) = 0 \u21d2 1 – log x = 0
\n\u21d2 log x = 1 \u21d2 x = e
\nAt x = e, f”(e) = \\(\\frac{-3+2 \\log e}{e^{3}}\\) < 0,
\n\u2234 f has a maximum at x = e<\/p>\n

Question 3.
\nThe two equal sides of an isosceles triangle with fixed base b are decreasing at the rate of 3 cm per second. How fast is the area decreasing when the two equal sides are equal to the base?
\nSolution:
\n\"NCERT
\n\u2206 ABC be an isosceles triangle with AB = AC
\nLet AB = x, BC = b, \\(\\frac { dx }{ dt }\\) = – 3cm\/sec at
\nDraw AD \u22a5 BC. Then D is the midpoint of BC.
\ni.e., BD = \\(\\frac { b }{ 2 }\\)
\n\"NCERT
\nHence area is decreasing at the rate of \\(\\sqrt{3}\\) bcm\u00b2\/sec.<\/p>\n

\"NCERT<\/p>\n

Question 4.
\nFind the equation of the normal to curve y\u00b2 = 4x at the point (1,2).
\nSolution:
\ny\u00b2 = 4x
\nDifferentiating w.r.t. x,
\n2y\\(\\frac { dy }{ dx }\\) = 4
\n\\(\\frac { dy }{ dx }\\) = \\(\\frac { 2 }{ y }\\)
\nAt(1, 2), \\(\\frac { dy }{ dx }\\) = \\(\\frac { 2 }{ 2 }\\) = 1
\nSlope of the tangent at (1, 2) = \\(\\frac { -1 }{ 1 }\\) = – 1
\nSlope of the normal at(1, 2)
\nEquation of the tangent at (1, 2) is
\ny – 2 = 1(x – 1)
\ny – x – 1 = 0
\nEquation of the normal at (1, 2) is
\ny – 2 = 1(x – 1)
\nx + y – 3 = 0<\/p>\n

Question 5.
\nShow that the normal at any point \u03b8 to the curve x = a cos\u03b8 + a\u03b8 sin\u03b8, y = a sin\u03b8 – a\u03b8 cos\u03b8 is at a constant distance from the origin.
\nSolution:
\nx = acos\u03b8 + a0sin\u03b8
\ny = asin\u03b8 – a\u03b8 cos\u03b8
\nDifferentiating w.r.t \u03b8, dx
\n\"NCERT
\nSlope of the tangent at the point 0 is \\(\\frac { sin\u03b8 }{ cos\u03b8 }\\)
\n\u2234 Slope of the normal at the point \u03b8 is \\(\\frac { – cos\u03b8 }{ sin\u03b8 }\\)
\n\u2234 Equation of the normal at the point \u03b8 is
\n\"NCERT
\ni.e., xcos\u03b8 + ysin \u03b8 = a, which is the equation of the straight line in the normal form. Hence \u2018a\u2019 is the distance of the normal line from the origin. Thus the normal at any point 0 to the curve is at a constant distance from the origin.<\/p>\n

Question 6.
\nFind the intervals in which the function f given by
\nf(x) = \\(\\frac{4 \\sin x-2 x-x \\cos x}{2+\\cos x}, 0 \\leq x \\leq 2 \\pi\\) (i) increasing (ii) decreasing.
\nSolution:
\n\"NCERT
\nSince f is continuous
\ni. f is increasing: 0 \u2264 x \u2264 \\(\\frac { \u03c0 }{ 2 }\\) and \\(\\frac { 3\u03c0 }{ 2 }\\) \u2264 2\u03c0
\nii. is decreasing: \\(\\frac { \u03c0 }{ 2 }\\) \u2264 x \u2264 \\(\\frac { 3\u03c0 }{ 2 }\\)<\/p>\n

\"NCERT<\/p>\n

Question 7.
\nFind the intervals in which the function f given by f(x) = x\u00b3 + \\(\\frac{1}{x^{3}}\\), x \u2260 0 is
\ni. increasing
\nii. decreasing
\nSolution:
\n\"NCERT
\nThe critical points are x = – 1, x = 0 and x =1. x = – 1, x = 0, x = 1 divide Rinto disjoint open intervals as
\n\"NCERT
\n(-\u221e, -1), (-1, 0),(0, 1) and (1, \u221e)
\ni. f is increasing: x < – 1 and x > 1
\nii. f is decreasing: – 1 < x < 0 and 0 < x < 1
\n\"NCERT<\/p>\n

Question 8.
\nFind the maximum area of an isosceles tri-angle inscribed in the ellipse \\(\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1\\) = 1 with its vertex at one end of the major axis.
\nSolution:
\n\"NCERT
\nConsider the ellipse \\(\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1\\) = 1 , a > b
\nLet A be one end point of the major axis and P(x, y) be a point on the ellipse. Draw PM perpendicular to x axis and extend it to meet the ellipse at Q.
\nThen the coordinate of Q is (x, – y)
\n\u2206APQ is an isosceles triangle
\nLet S be the area of \u2206APQ
\nS = \\(\\frac { 1 }{ 2 }\\)PQ.AM = \\(\\frac { 1 }{ 2 }\\)(2y)(x + a)
\nS = y(x + a) = \\(\\frac { b }{ 2 }\\)\\(\\sqrt{a^{2}-x^{2}}\\)(x + a)
\nSince S is positive, S\u00b2 is maximum when S is maximum
\n\"NCERT
\nWhen x = – a, the point P coincides with A which is not possible.
\n\u2234 x = – \\(\\frac { a }{ 2 }\\)
\nWhen x = \\(\\frac { a }{ 2 }\\), \\(\\frac{d^{2}}{d x^{2}}\\left(\\mathrm{~S}^{2}\\right)\\)
\n= \\(\\frac{-12 b^{2}}{a^{2}} \\frac{a}{2}\\left(\\frac{a}{2}+a\\right)\\) < 0
\n\u2234 S\u00b2 is maximum when x = \\(\\frac { a }{ 2 }\\)
\ni.e., S is maximum when x = \\(\\frac { a }{ 2 }\\)
\nThe maximum value of S
\n\"NCERT<\/p>\n

Question 9.
\nA tank with rectangular base and rectangular sides, open at the top is to be constructed so that its depth is 2 m and volume is 8m\u00b3. If building of tank costs \u20b9 70 per sq.metre for the base and \u20b9 45 per square metre for sides, what is the cost of least expensive tank?
\nSolution:
\nLet x and y be the length and breadth of the base of the tank (in metre). Let C be the cost of building the tank.
\n\"NCERT
\nVolume of the tank = 8m\u00b3 (x)(y).(2) = 8 \u21d2 xy = 4
\n\u2234 y = \\(\\frac { 4 }{ x }\\)
\nArea of base = xy
\nArea of 4 sides = 2(x + y)(2) = 4(x + y)
\nCost of construction
\n\"NCERT
\nBut x = – 2 is not possible, since length of the base cannot be negative
\n\u2234 x = 2 When x = 2,
\n\\(\\frac{d^{2} \\mathrm{C}}{d x^{2}}=\\frac{180 \\times 8}{8}\\) > 0
\nThe minimum cost of construction,
\nC = 280 + 180(2 + \\(\\frac { 4 }{ 2 }\\)) = \u20b9 1000<\/p>\n

\"NCERT<\/p>\n

Question 10.
\nThe sum of the perimeter of a circle and square is k, where k is some constant. Prove that the sum of their areas is least when the side of square is double the radius of the circle.
\nSolution:
\nLet r be the radius and x be the side of the square.
\nThen 2\u03c0r + 4x = k
\n\u2234 r = \\(\\frac{k-4 x}{2 \\pi}\\) … (1)
\nLet S be the sum of areas of circle and square
\n\"NCERT<\/p>\n

Question 11.
\nA window is in the form of a rectangle sur-mounted by a semicircular opening. The total perimeter of the window is 10 m. Find the dimensions of the window to admit maximum light through the whole opening.
\nSolution:
\nLength of the window = 2x,
\nHeight of the rectangular portion = y
\ni. Perimeter of the window = 10 m
\n2x + 2y + \u03c0x = 10
\n(\u03c0 + 2)x + 2y = 10<\/p>\n

ii. Let A be the area of the window
\n\"NCERT<\/p>\n

iii. Differentiating A w.r.t. x,
\n\u2234 A’ = 10 – 2 \u03c0x – 4x + \u03c0 x
\nA’ = 10 – \u03c0 x – 4x
\nA” = – \u03c0 – 4 = – (\u03c0 + 4)
\nFor maxima, A’ = 0
\n10 – \u03c0x – 4x = 0
\nx(\u03c0 + 4) = 10
\nx = \\(\\frac { 10 }{ \u03c0 + 4 }\\)
\nWhen x = \\(\\frac { 10 }{ \u03c0 + 4 }\\), A” = – (\u03c0 + 4) < 0
\n\u2234 A is maximum when x = \\(\\frac { 10 }{ \u03c0 + 4 }\\)
\n2y = 10 – (\u03c0 + 2)x
\n= 10 – (\u03c0 + 2)\\(\\frac { 10 }{ \u03c0 + 4 }\\)
\n= \\(\\frac{10 \\pi+40-10 \\pi-20}{\\pi+4}\\) = \\(\\frac { 20 }{ \u03c0 + 4 }\\)
\n\u2234 y = \\(\\frac { 10 }{ \u03c0 + 4 }\\)
\n\u2234 Length of the window = 2x = \\(\\frac { 20 }{ \u03c0 + 4 }\\)
\nLength of larger side of window = y = \\(\\frac { 10 }{ \u03c0 + 4 }\\).<\/p>\n

Question 12.
\nA point on the hypotenuse of a triangle is at distance a and b from the side of the tri\u00acangle. Show that the minimum length of the hypotenuse is \\(\\left(a^{\\frac{2}{3}}+b^{\\frac{2}{3}}\\right)^{\\frac{3}{2}}\\).
\nSolution:
\n\"NCERT
\nLet \u2206ABC be right angled at B and P be a point on the hypotenuse AC at a distance of a from BC and b from AB.
\n\u2234 PQ = a and PR = b
\nLet \u2220A = \u03b8 then \u2220C = 90 – \u03b8
\nFrom right triangles ARP and PQC, we get
\nAP = \\(\\frac { b }{ sin \u03b8 }\\) = b cosec \u03b8
\nPC = \\(\\frac { a }{ sin(90 – \u03b8) }\\) = \\(\\frac { a }{ cos \u03b8 }\\) = asec \u03b8
\nLet l be the length of the hypotenuse
\n\u2234 l = AP + PC = asec\u03b8 + bcosec\u03b8,
\n0 < \u03b8 < \\(\\frac { \u03c0 }{ 2 }\\)
\n\\(\\frac { dl }{ d\u03b8 }\\) = asec\u03b8 tan\u03b8 – bcosec\u03b8 cot\u03b8
\n\\(\\frac{d^{2} l}{d \\theta^{2}}\\) = a[sec\u03b8.sec\u00b2\u03b8 + tan\u03b8.sec\u03b8 tan\u03b8] – b[cosec\u03b8(-cosec\u00b2\u03b8) + cot \u03b8(-cosec\u03b8 cot\u03b8)] = a[sec\u00b3\u03b8 + sec\u03b8tan\u00b2\u03b8] + b[cosec\u00b3\u03b8 + cosec\u03b8 cot\u00b2\u03b8]
\nFor minima
\n\\(\\frac { dl }{ d\u03b8 }\\) = 0 \u21d2 asec\u03b8 tan\u03b8 – bcosec\u03b8 cot\u03b8 = 0
\n\u21d2 a sec\u03b8 tan\u03b8 = b cosec\u03b8 cot\u03b8
\n\"NCERT
\nMinimum length of the hypotenuse = \\(\\left(a^{\\frac{2}{3}}+b^{\\frac{2}{3}}\\right)^{\\frac{3}{2}}\\).<\/p>\n

\"NCERT<\/p>\n

Question 13.
\nFind the points at which the function given by f(x) = (x – 2)4<\/sup> (x + 1)\u00b3 has
\ni. local maxima
\nii. local minima
\niii. point of in flexion
\nSolution:
\nf(x) = (x – 2)4<\/sup> (x + 1)\u00b3
\nf\u2019(x) = (x – 2)4<\/sup>[3(x + 1)\u00b2] + (x+ 1)\u00b3[4(x – 2)\u00b3]
\n= (x – 2)\u00b3(x + 1)\u00b2[3x – 6 + 4x + 4]
\n= (x – 2)\u00b3(x + 1)\u00b2(7x – 2)
\nf\u2019(x) = 0 \u21d2 (x – 2)\u00b3(x + 1)\u00b2(7x – 2) = 0
\n\u21d2 x = – 1, and x = \\(\\frac { 2 }{ 7 }\\) x = 2
\nLet us apply first derivative test.
\n\"NCERT
\ni. local maxima is at x = \\(\\frac { 2 }{ 7 }\\)
\nii local minima \u00a1s at x = 2
\niii. point of inflexion is at x = – 1<\/p>\n

Question 14.
\nFind the absolute maximum and minimum values of the function f given by
\nf(x) = cos\u00b2 x + sin x, x \u2208 [0, \u03c0]
\nSolution:
\nf(x) = cos\u00b2 x + sin x, x \u2208 [0, \u03c0]
\nf'(x) = – 2cosx sinx + cosx = cosx (1 – 2sinx)
\nf'(x) = 0 \u21d2 cosx (1 – 2sinx) = 0
\n\u21d2 cosx = 0 or sin x = \\(\\frac { 1 }{ 2 }\\)
\n\u21d2 x = \\(\\frac { \u03c0 }{ 2 }\\) or x = \\(\\frac { \u03c0 }{ 6 }\\) and x = \\(\\frac { 5\u03c0 }{ 6 }\\)
\n\u2234 x = \\(\\frac { \u03c0 }{ 6 }\\), \\(\\frac { \u03c0 }{ 2 }\\), \\(\\frac { 5\u03c0 }{ 6 }\\)
\nf(0) = cos\u00b20 + sin0 = 1\u00b2 + 0 = 1
\n\"NCERT<\/p>\n

Question 15.
\nShow that the altitude of the right circular cone of maximum volume that can be in-scribed in a sphere of radius r is \\(\\frac { 4r }{ 3 }\\)
\nSolution:
\nLet R be the radius and A be the height of the right circular cone inscribed in a sphere of radius r.
\n\"NCERT
\ni.e., AB = h, OA = OC = r and BC = R
\nIn right triangle OBC, OB = h – r
\nand OC\u00b2 = OB\u00b2 + BC\u00b2
\ni.e., r\u00b2 = (h – r)\u00b2 + R\u00b2
\nr\u00b2 = h\u00b2 – 2 hr + r\u00b2 + R\u00b2
\n\u2234 R\u00b2 = 2hr – h\u00b2
\nVolume of the cone V = \\(\\frac { 1 }{ 3 }\\) \u03c0R\u00b2h
\n\"NCERT
\n\u2234 When h = \\(\\frac { 4r }{ 3 }\\), the volume is maximum.<\/p>\n

\"NCERT<\/p>\n

Question 16.
\nLet f be a function defined on [a, b] such that f'(x) > 0, for all x \u2208 (a, b). Then prove that f is an increasing function on (a, b)
\nSolution:
\nLet c1<\/sub>, c2<\/sub> \u2208 [a , b\\ such that c1<\/sub>, c2<\/sub>.
\nSince f'(x) > 0 in (a, b), f is continuous in [a, b] and differentiable in (a, b).
\n\u2234 By mean value theorem, there exists c \u2208 (c1<\/sub>, c2<\/sub>) such that
\n\"NCERT
\nSince c1<\/sub> and c2<\/sub> are arbitrary, we get f(x) is increasing on (a, b).<\/p>\n

Question 17.
\nShow that the height of the cylinder of maxi-mum volume that can be inscribed in a sphere of radius R is \\(\\frac{2 \\mathrm{R}}{\\sqrt{3}}[\/latex . Also find the maximum volume.
\nSolution:
\n\"NCERT
\nLet y be the radius and x be the height of the cylinder.
\n\"NCERT
\n\u2234 V is maximum when x = [latex]\\frac{2 \\mathrm{R}}{\\sqrt{3}}\\)
\nThe height of cylinder with maximum volume is \\(\\frac{2 \\mathrm{R}}{\\sqrt{3}}\\)
\n\u2234 (1) \u2192 Maximum volume,
\nV = \\(\\frac{\\pi}{4}\\left(4 \\mathrm{R}^{2} \\cdot \\frac{2 \\mathrm{R}}{\\sqrt{3}}-\\frac{8 \\mathrm{R}^{3}}{3 \\sqrt{3}}\\right)=\\frac{4 \\pi \\mathrm{R}^{3}}{3 \\sqrt{3}}\\)<\/p>\n

Question 18.
\nShow that the height of the cylinder of greatest volume which can be inscribed in a right circular cone of height h and semi vertical angle a is one-third that of the cone and the greatest volume of cylinder is \\(\\frac { 4 }{ 27 }\\)\u03c0h\u00b3 tan\u00b2\u03b1.
\nSolution:
\n\"NCERT
\nLet y be the radius and x be the height of the right circular cylinder inscribed in a cone of height h and semivertical angle \u03b1.
\nThen AB h, CD = y, BC = x
\n\u2234 AC = h – x
\nFrom right triangle ACD, tan \u03b1 = \\(\\frac { CD }{ AC }\\)
\n= \\(\\frac { y }{ h – x }\\)
\n\u21d2 y = (h – x)tan \u03b1
\nVolume of the cylinder V = \u03c0y\u00b2x
\n= \u03c0(h – x)\u00b2 tan\u00b2 \u03b1.x
\n\"NCERT
\n\u2234 V is maximum when the height of the cylinder is one third of the height of the cone.
\n(1) \u2192 Maximum volume,
\n\"NCERT<\/p>\n

Question 19.
\nA cylindrical tank of radius 10 m is being filled with wheat at the rate of 314 cubic metre per hour. Then the depth of the wheat is increasing at the rate of
\na. 1 m\u00b3\/h
\nb. 0.1 m\u00b3\/h
\nc. 1.1 m\u00b3\/h
\nd. 0.5 m\u00b3\/h
\nSolution:
\na. 1 m\u00b3\/h
\nLet h be the height of the wheat at time t and V be the volume at that instant
\n\"NCERT<\/p>\n

Question 20.
\nThe slope of the tangent to the curve x = t\u00b2 + 3t – 8, y = 2t\u00b2 – 2t – 5 at the point (2, – 1) is
\na. \\(\\frac { 22 }{ 7 }\\)
\nb. \\(\\frac { 6 }{ 7 }\\)
\nc. \\(\\frac { 7 }{ 6 }\\)
\nd. \\(\\frac { – 6 }{ 7 }\\)
\nSolution:
\nb. \\(\\frac { 6 }{ 7 }\\)
\nx = t\u00b2 + 3t – 8 and y = 2t\u00b2 – 2t – 5
\nPut x = 2 and y = – 1, we get
\nt\u00b2 + 3t – 8 = 2 and 2t\u00b2 – 2t – 5 = – 1
\ni.e., t\u00b2 + 3t – 10 = 0 and 2t\u00b2 – 2t – 4 = 0
\ni.e, (t + 5) (t – 2) = 0 and 2(t – 2) (t+ 1) = 0
\ni.e., t = -5 or t = 2 and t = 2 or t = – 1
\nHence t = 2
\n\"NCERT<\/p>\n

\"NCERT<\/p>\n

Question 21.
\nThe line y = mx + 1 is a tangent to the curve y\u00b2 = 4x if the value of m is
\na. 1
\nb. 2
\nc. 3
\nd. \\(\\frac { 1 }{ 2 }\\)
\nSolution:
\ny\u00b2 = 4x
\nDifferentiating w.r.t x,
\n2y\\(\\frac { dy }{ dx }\\) = 4 \u21d2 \\(\\frac { dy }{ dx }\\) = \\(\\frac { 2 }{ y }\\)
\n\u2234 Slope of the tangent, m = \\(\\frac { 2 }{ y }\\)
\n\u2234 y = mx + 1 becomes y = \\(\\frac { 2 }{ y }\\) x + 1
\ny\u00b2 = 2x + y
\ni.e., y\u00b2 = \\(\\frac { y\u00b2 }{ 2 }\\) + y since y\u00b2 = 4x
\n\\(\\frac { y\u00b2 }{ 2 }\\) = y \u21d2 y\u00b2 = 2y
\n\u21d2 y(y – 2) = 0
\ny \u2260 0 since slope = \\(\\frac { 2 }{ y }\\)
\n\u2234 y = 2
\n\u2234 slope m = \\(\\frac { 2 }{ y }\\) = \\(\\frac { 2 }{ 2 }\\) = 1<\/p>\n

Question 22.
\nThe normal at the point (1,1) on the curve 2y + x\u00b2 – 3 is
\na. x + y = 0
\nb. x – y = 0
\nc. x + y + 1 = 0
\nd. x – y = 0
\nSolution:
\nb. x – y = 0
\n2y + x\u00b2 = 3
\nDifferentiating w.r.t x,
\n2\\(\\frac { dy }{ dx }\\) + 2x = 0
\n\\(\\frac { dy }{ dx }\\) = – x
\nSlope of the tangent at (1, 1)
\n= \\(\\frac { dy }{ dx }\\) at (1, 1) = – 1
\n\u2234 Slope of the normal at (1, 1) = \\(\\frac { – 1 }{ – 1 }\\) = 1
\nEquation of the normal at (1, 1) is
\ny – 1 = 1(x – 1)
\n\u21d2 x – y = 0<\/p>\n

\"NCERT<\/p>\n

Question 23.
\nThe normal to the curve x\u00b2 = 4y passing through (1, 2) is
\na. x + y = 3
\nb. x – y = 3
\nc. x + y = 1
\nd. x – y = 1
\nSolution:
\na. x + y = 3
\nx\u00b2 = Ay
\nDifferentiating w.r.t. x,
\n2x = 4\\(\\frac { dy }{ dx }\\)
\n\u2234\\(\\frac { dy }{ dx }\\) = \\(\\frac { x }{ 2 }\\)
\nSlope of the tangent = \\(\\frac { x }{ 2 }\\)
\n\u2234 Slope of the normal = \\(\\frac { – 2 }{ x }\\)
\nLet (x1<\/sub>, y1<\/sub>) be a point on the curve
\n\u2234 Slope of normal at (x1<\/sub>, y1<\/sub>) = \\(\\frac{-2}{x_{1}}\\)
\nEquation of the normal at (x1<\/sub>, y1<\/sub>) is
\ny – y1<\/sub> = \\(\\frac{-2}{x_{1}}\\)(x – x1<\/sub>)
\nThe normal passes through the point (1, 2)
\n\u2234 (2 – y1<\/sub>) = \\(\\frac{-2}{x_{1}}\\)(1 – x1<\/sub>)
\nEquation of the normal at(x1<\/sub>, y1<\/sub>) is
\ny – y1<\/sub> = \\(\\frac{-2}{x_{1}}\\)(1 – x1<\/sub>)
\n2 – y1<\/sub> = \\(\\frac{2}{x_{1}}\\) + 2
\n\u2234 y1<\/sub> = \\(\\frac{2}{x_{1}}\\)
\nSince (x1<\/sub>, y1<\/sub>) is a point on x\u00b2 = 4y, we get
\nx1<\/sub>\u00b2 = 4y1<\/sub>
\ni.e., x1<\/sub>\u00b3 = \\(\\frac{4 \\times 2}{x_{1}}\\), sjnce y1<\/sub> = \\(\\frac{2}{x_{1}}\\)
\nx1<\/sub>\u00b2 = 8 \u21d2 x, = 2
\nWhen x1<\/sub> = 2, y1<\/sub> = \\(\\frac { 2 }{ 2 }\\) = 1
\n\u2234 Equation of the normal at (2, 1) is
\ny – 1 = \\(\\frac { – 2 }{ 2 }\\)(x – 2)
\ny – 1 = – (x – 2)
\ni.e., x + y = 3<\/p>\n

Question 24
\nThe points on the curve 9y\u00b2 = x\u00b3, where the normal to the curve makes equal intercepts with the axes are
\na. \\(\\left(4, \\pm \\frac{8}{3}\\right)\\)
\nb. \\(\\left(4, \\pm \\frac{-8}{3}\\right)\\)
\nc. \\(\\left(4, \\pm \\frac{3}{8}\\right)\\)
\nd. \\(\\left(\\pm 4, \\frac{3}{8}\\right)\\)
\nSolution:
\n9y\u00b2 = x\u00b3
\nLet (x1<\/sub>, y1<\/sub>) be the point at which the nor\u00acmal make equal intercepts on the coordinate axes.
\nDifferentiating 9y\u00b3 = x\u00b3 w.r.t. x, we get
\n18y\\(\\frac { dy }{ dx }\\) = 3x\u00b2
\n\\(\\frac { dy }{ dx }\\) = \\(\\frac{x^{2}}{6 y}\\)
\nSlope of the tangent at (x1<\/sub>, y1<\/sub>) = \\(\\frac{x_{1}^{2}}{6 y_{1}}\\)
\n\u2234 Slope of normal at (x1<\/sub>, y1<\/sub>) = \\(\\frac{-6 y_{1}}{x_{1}^{2}}\\)
\nEquation of the normal at (x1<\/sub>, y1<\/sub>) is
\n\"NCERT
\nwhich is in the intercept form.
\nSince the intercepts of the normal are equal,we get
\n\"NCERT
\nBut x1<\/sub> \u2260 0 since x1<\/sub> = 0 makes the intercepts meaningless.
\n\u2234 x1<\/sub> = 4
\nWhen x1<\/sub>, 9y\u00b21<\/sub> = x\u00b31<\/sub> becomes
\n9y\u00b21<\/sub> = 4\u00b3 = 64
\ny\u00b21<\/sub> = \\(\\frac { 64 }{ 9 }\\) \u21d2 y1<\/sub> = \u2260 \\(\\frac { 8 }{ 3 }\\)
\n\u2234 The required points are (4, \\(\\frac { 8 }{ 3 }\\))<\/p>\n","protected":false},"excerpt":{"rendered":"

These NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Miscellaneous Exercise Questions and Answers are prepared by our highly skilled subject experts. https:\/\/mcq-questions.com\/ncert-solutions-for-class-12-maths-chapter-6-miscellaneous-exercise\/ NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Miscellaneous Exercise Question 1. Using differentials, find the approximate value of each of the following. a. b. Solution: …<\/p>\n

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