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NCERT Sol Maths Class XII Ch-6 | Application of Derivatives
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Solution of Important Question of NCERT Book
Exercise 6.3 Q 18 For the curve y = 4x^{3} – 2x^{5}, find all the points at which the tangent passes through the origin. Solution: Given equation of the curve is y = 4x^{3} – 2x^{5}………….(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 = 4x^{3} – 2x^{5}, ⇒ k = 4h^{3} – 2h^{5} ………..(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 = (4h^{3} – 2h^{5})( x – h ) This tangent passes through the origion (0, 0) 0 – k = (12h^{2} – 10h^{4})( 0 – h ) k = (12h^{2} – 10h^{4})h ⇒ k = 12h^{3} – 10h^{5} Putting the value of k from equation (2) we get 4h^{3} – 2h^{5} = 12h^{3} – 10h^{5} ⇒ 8 h^{5} - 8 h^{3} = 0 ⇒ h^{5} - h^{3} = 0 ⇒ h^{3}(h^{2} – 1) = 0 ⇒ h^{3}(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) |
Q 23 Prove that the curves x = y^{2} and xy = k cut at right angles if 8k^{2} = 1 Solution. The given equations are x = y^{2} ………. (1) xy = k ………….. (2) First of all we find the point on the curve Putting eqn(1) in eqn(2) we get y^{2} x y = k ⇒ y^{3} = k ⇒ y = k^{1/3} Putting the value of y in eqn. (1) we get x = (k^{1/3})^{2} = k^{2/3} Point on the curve is (k^{2/3}, k^{1/3} ) 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 = k^{2/3} 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 + Î”x)^{1/3} (x + Î”x)^{1/3} = (0.008 + 0.001)^{1/3} x = 0.008 and Î”x = 0.001 Let y = x^{1/3} 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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