46. Find the point on the curve y = x3 – 11x + 5 at which the tangent is y = x – 11.

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

46. Find the point on the curve y = x³ – 11x + 5 at which the tangent is y = x – 11.

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Vishal Baghel | Contributor-Level 10

4 months ago

The given eqⁿ of the curve is $y = x^{3} - 11 x + 5$

slope of tangent to the curve $\frac{d y}{d x} = 3 x^{2} - 11$

Then eqⁿ of tangent is $y = x - 11$ $\Rightarrow x - y - 11 = 0$ which gives us slope $= \frac{- 1}{- 1} = 1$

So, $3 x^{2} - 11 = 1$

$\Rightarrow 3 x^{2} = 1 + 11 = 12$

$\Rightarrow x^{2} = 4$

$\Rightarrow x = \pm 2$

When x = 2, $y = 2^{3} - 11 (2) + 5 = 8 - 22 + 5 = - 9 .$

And when x = 2, $y = {(- 2)}^{3} - 11 (- 2) + 5 = - 8 + 22 + 5 = 19 .$

The point $(2, - 9)$ when put into $y = x - 11 .$ we get

$- 9 = 2 - 11$

$\Rightarrow - 9 = - 9$ which is true.

and the point $(- 2, 19)$ when put into $y = x - 11$ gives,

$19 = - 2 - 11$

$\Rightarrow 19 = - 13$ which is not true.

Hence, the required point is $(2, - 9)$

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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..........

Raj Pandey

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)
c = e^ (1/ (e-1)

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:

Vishal Baghel

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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46. Find the point on the curve y = x³ – 11x + 5 at which the tangent is y = x – 11.

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