NCERT Sol Maths Class XII Ch-6 | Application of Derivatives

 Solution of Important Question of NCERT Book

Class XII Chapter 6 Matrices

Get Free NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivative. Solution of important questions of NCERT book solved by Expert Teachers as per NCERT (CBSE) Book guidelines.

Exercise 6.3 Q 18  For the curve  y = 4x3 – 2x5, find all the points at which the tangent passes through the origin. Solution: Given equation of the curve is  y = 4x3 – 2x5………….(1) Let (h, k) be any point on this curve. Let tangent to the curve at  point  (h, k) passes through the origin (0, 0) Equation of the curve at point (h, k) becomes y = 4x3 – 2x5,   ⇒    k = 4h3 – 2h5  ………..(2) Differentiating eqn. (1)  w.r.t. x we get \[\frac{dy}{dx}=4\times 3x^{2}-2\times 5x^{4}=12x^{2}-10x^{4}\] \[\left (\frac{dy}{dx} \right )_{(h,\;k)}=12h^{2}-10h^{4}\] Equation of the tangent at (h, k) is given by y – k = (4h3 – 2h5)( x – h ) This tangent passes through the origion (0, 0) 0 – k = (12h2 – 10h4)( 0 – h ) k = (12h2 – 10h4)h   ⇒    k = 12h3 – 10h5 Putting the value of k from equation (2) we get 4h3 – 2h5 = 12h3 – 10h5              ⇒  8 h5 - 8 h3 = 0   ⇒   h5 -  h3 = 0                             ⇒  h3(h2 – 1) = 0     ⇒   h3(h + 1)(h – 1) = 0                ⇒  h = 0,  1, -1 Now Putting the values of h in eqn.(2) If h = 0, k = 0,  If h = 1, then k = 4 (1) 3 - 2 (1) 5 = 4 - 2 = 2 If h = -1 then k = 4 (- 1) 3 – 2 (- 1) 5 = - 4 + 2 = -2 Required points on the curve at which the tangent passes through the origin are (0, 0), (1, 2), (- 1, - 2)

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Exercise 6.3  Q 23 Prove that the curves  x = y2 and xy = k cut at right angles if  8k2 = 1 Solution.    The given equations are   x = y2      ……….  (1) xy = k  …………..  (2) First of all we find the point on the curve Putting eqn(1) in eqn(2) we get y2 x y = k        ⇒     y3 = k     ⇒     y = k1/3 Putting the value of y in eqn. (1) we get x = (k1/3)2  =  k2/3 Point on the curve is  (k2/3, k1/3 ) Now differentiating eqn.(1) w. r. t. x we get \[1=2y\frac{dy}{dx}\; \; \Rightarrow\; \; \frac{dy}{dx}=\frac{1}{2y}=m_{1}\] Now differentiating eqn.(2) w. r. t. x we get \[1.y+x.\frac{dy}{dx}= 0\; \; \Rightarrow \; \; \frac{dy}{dx}=\frac{-y}{x}=m_{2}\] Eqn(1) and eqn(2) intersect  Each other at right angles iff \[m_{1}\times m_{2}=-1\] \[\Rightarrow \frac{1}{2y}\times \frac{-y}{x}=-1\; \; \Rightarrow \; \; 2x=1\] Putting x = k2/3  we get \[2(k^{2/3})=1\] Cubing on both sides we get \[\left [2(k^{2/3}) \right ]^{3}=(1)^{3}\; \; \Rightarrow\; 8k^{2}=1\]

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Exercise 6.4 Q 1 (iv)  Using Differential find the approximate value of  (0.009)1/3  up to three decimal places. Solution  (0.009)1/3  = (0.008 + 0.001)1/3   = (x + Δx)1/3    (x + Δx)1/3  = (0.008 + 0.001)1/3   x = 0.008     and    Δx = 0.001 Let y = x1/3 Differentiating it with respect to x we get \[\frac{dy}{dx}=\frac{1}{3}x^{1-\frac{1}{3}}=\frac{1}{3}x^{-2/3}=\frac{1}{3}(0.008)^{-2/3}\] \[\frac{dy}{dx}=\frac{1}{3}\times (0.2)^{-2}=\frac{1}{3\times 0.04}=\frac{1}{0.12}\] \[(x+\Delta x)^{1/3}=x^{1/3}+\frac{dy}{dx}\times \Delta x\] \[\left (0.008+0.001 \right )^{1/3}=(0.008)^{1/3}+\frac{1}{0.12}\times 0.001\]   \[\left (0.009 \right )^{1/3}=0.2+\frac{0.001}{0.120}\] \[\left (0.009 \right )^{1/3}=0.2+\frac{1}{120}=0.2+0.008\] \[\left (0.009 \right )^{1/3}=0.208\: \; \; Ans\]

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