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Ncert Solutions Maths class 12th Maths Class 12th Application of Derivatives

51. Find the equations of the tangent and normal to the given curves at the indicated points:

$\begin{array}{l} (i) y = x^{4} - 6 x^{3} + 13 x^{2} - 10 x + 5, a t, (0, 5) \\ (i i) y = x^{4} - 6 x^{3} + 13 x^{2} - 10 x + 5, a t, (1, 3) \\ (i i i) y = x^{3}, a t, (1, 1) \\ (i v) y = x^{2}, a t, (0, 0) \\ (v) x = c o s y, y = s i n t, a t, t = \frac{π}{4} \end{array}$

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Vishal Baghel | Contributor-Level 10

4 months ago

(i) we have, $y = x^{4} - 6 x^{3} + 13 x^{2} - 10 x + 5$

slope of tangent, $\frac{d y}{d x} = 4 x^{3} - 18 x^{2} + 26 x - 10$

${\frac{d y}{d x} |}_{(x, y) = (0, 5)} = - 10 .$

slope of normal $= \frac{- 1}{- 10} = \frac{1}{10}$

Hence eqⁿ of tangent at (0, 5) is

$y - 5 = - 10 (x - 0) \to 10 x + y - 5 = 0$

And eqⁿ of normal at (0, 5) is

$y - 5 = \frac{1}{10} (x - 0)$

$\Rightarrow 10 y - 50 = x$

$\Rightarrow x - 10 y + 50 = 0$

(ii) We have, $y = x^{4} - 6 x^{3} + 13 x^{2} - 10 x + 5$

Slope of tangent, $\frac{d y}{d x} = 4 x^{3} - 18 x^{2} + 26 x - 10 .$

${\frac{d y}{d x} |}_{(x, y) = (1, 3)} = 4 {(1)}^{3} - 18 {(1)}^{2} + 26 (1) - 10$

$= 4 - 18 + 26 - 10$

= 30 28

= 2

Slope of normal $= - \frac{1}{2}$

Hence eqⁿ of tangent at (1, 3) is

$y - 3 = 2 (x - 1)$

$\Rightarrow y - 3 = 2 x - 2$

$\to 2 x - y + 1 = 0$

And eqⁿ of normal at (1,3) is

$(y - 3) = - \frac{1}{2} (x - 1)$

$\Rightarrow 2 y - 6 = - x + 1$

$\to x + 2 y - 7 = 0$

(iii) We have, $y = x^{3}$

Slope of tangent, $\frac{d y}{d x} = 3 x^{2}$

${\frac{d y}{d x} |}_{(1, 1)} = 3 {(1)}^{2} = 3 .$

And slope of normal $= - \frac{1}{3}$

Hence, eqn of tangent at (1, 1) is

$y - 1 = 3 (x - 1)$

$\Rightarrow y - 1 = 3 x - 3$

$\to 3 x - y - 2 = 0$

And eqⁿ of normal at (1,1) is

$y - 1 = - \frac{1}{3} (x - 1)$

$\to 3 y - 3 = - x + 1$

$\Rightarrow x + 3 y - 4 = 0 .$

(iv) We have, $y = x^{2}$

Slope of tangent $\frac{d y}{d x} = 2 x$

${\frac{d y}{d x} |}_{(0, 0)} = 0 .$

So, eqⁿ of the tangent at (0,0) is

$(y - 0) = 0 (x - 0)$

$\Rightarrow y = 0 .$

ie, x- axis

Hence, the eqⁿ of normal is x = 0 ie, y-axis

(v) We have, $x = c o s t y = s i n t$

$\frac{d x}{d t} = - s i n t \cdot \frac{d y}{d t} = c o s t$

So, slope of tangent $\frac{d y}{d x} = \frac{d y / d t}{d x / d t} = \frac{c o s t}{- s i n t} = - c o t t$

${\frac{d y}{d x} |}_{t = \frac{π}{4}} = - c o t \frac{π}{4} = - 1 .$

And slope of normal $= \frac{- 1}{- 1} = 1$

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If the normal to the curve y(x) = ∫(from 2 to x) (2t² - 15t + 10)dt at a point (a, b) is parallel to the line x + 3y = -5, a > 1, then the value of |a + 6b| is equal to..........

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y (x) = ∫? (2t² - 15t + 10)dt
dy/dx = 2x² - 15x + 10.
For tangent at (a, b), slope is m = dx/dy = 1 / (dy/dx) = 1 / (2a² - 15a + 10).
Given slope is -1/3.
2a² - 15a + 10 = -3
2a² - 15a + 13 = 0 (The provided solution has 2a²-15a+7=0, suggesting a different problem or a typo)
Following the image: 2a² - 15a + 7 = 0
(2a - 1) (a - 7) = 0
a = 1/2 or a = 7.
a = 1/2 Rejected as a > 1. So a = 7.
b = ∫? (2t² - 15t + 10)dt = [2t³/3 - 15t²/2 + 10t] from 0 to 7.
6b = [4t³ - 45t² + 60t] from 0 to 7 = 4 (7)³ - 45 (7)² + 60 (7) = 1372 - 2205 + 420 = -413.
|a + 6b| = |7 - 413| = |-406|

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If the tangent to the curve, y = f(x) = xlog? x, (x > 0) at a point (c, f(c)) is parallel to the line - segment joining the points (1,0) and (e, e) then c is equal to:

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f' (c) = 1 + lnc = e/ (e-1)
lnc = e/ (e-1) - 1 = (e - (e-1)/ (e-1) = 1/ (e-1)
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The lengths of the sides of a triangle are 10 + x2, 10 + x2 and 20 – 2x2. If for x = k, the area of the triangle is maximum, then 3k2 is equal to:

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$C D = \sqrt[]{{(10 + x^{2})}^{2} - {(10 - x^{2})}^{2}} = 2 \sqrt[]{10} | x |$

Area

$= \frac{1}{2} \times C D \times A B = \frac{1}{2} \times 2 \sqrt[]{10} | x | (20 - 2 x^{2})$

$\Rightarrow 10 - x^{2} = 2 x$

3x² = 10

x = k

3k² = 10

Let F₁(A,B,C) = $(A \land \sim B) \lor$ $[\sim C \land (A \lor B)] \lor \sim A a n d F_{2} (A, B) = (A \lor B) \lor (B \to \sim A)$ be two logical expressions. Then:

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By truth table

So F₁ (A, B, C) is not a tautology

Now again by truth table

So F₂ (A, B) be a tautology.

The triangle of maximum area that can be inscribed in a given circle of radius ‘r’ is:

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From option let it be isosceles where AB = AC then

$x = \sqrt[]{r^{2} - {(h - r)}^{2}}$

= $\sqrt[]{r^{2} - h^{2} - r^{2} + 2 r h}$

$x = \sqrt[]{2 h r - h^{2}} . . . . . . . . (i)$

Now ar $(Δ A B C) = Δ = \frac{1}{2} B C \times A L$

$Δ = \frac{1}{2} \times 2 \sqrt[]{2 h r - a^{2}} \times h$

then $x = \sqrt[]{2 \times \frac{3 r}{2} \times r - \frac{9 r^{2}}{4}} = \frac{\sqrt[]{3}}{2} r f r o m (i)$

$\Rightarrow B C = \sqrt[]{3} r$

So $A B = \sqrt[]{h^{2} + x^{2}} = \sqrt[]{\frac{9 r^{2}}{4} + \frac{3 r^{2}}{4}} = \sqrt[]{3} r$ .

Hence $Δ$ be equilateral having each side of length $\sqrt[]{3} r .$

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51. Find the equations of the tangent and normal to the given curves at the indicated points: (i)y=x4−6x3+13x2−10x+5,at,(0,5)(ii)y=x4−6x3+13x2−10x+5,at,(1,3)(iii)y=x3,at,(1,1)(iv)y=x2,at,(0,0)(v)x=cosy,y=sint,at,t=π4

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