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

Le tf(x) be a polynomial of degree 6 in x, in which the coefficient of x⁶ is unity and it has extrema at x = -1 and x = 1. If $\underset{x \to 0}{l i m} \frac{f (x)}{x^{3}} = 1,$ then 5 f(2) is equal to……………….

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alok kumar singh | Contributor-Level 10

a month ago

Let f(x) = x⁶ + ax⁵ + bx⁴ + ax³ + dx² + ex + f

$\underset{x \to 0}{l i m} \frac{f (x)}{x^{3}} = 1$ Non zero finite

So, d = e = f = 0

f(x) = x⁶ + ax⁵ + bx⁴ + cx³

$\underset{x \to 0}{l i m} \frac{f (x)}{x^{3}} = 1$ Non zero finite

f’(x) = $6 x^{5} + 5 a x^{4} + 4 b x^{3} + 3 x^{2}$

f’(1) = 0

6 + 5a + 4b + 3 = 0

5a + 4b = - 9 .(i)

f’(-1) = 0

-6 + 5a – 4b + 3 = 0 .(ii)

Solving (i) and (ii)

a -3/5, b = -3/2

$∴ f (x) = x^{6} + (\frac{- 3}{5}) x^{5} + (\frac{- 3}{2}) x^{4} + x^{3}$

$∴ 5 . f (2) = 144$

Let f(x) = x6 + ax5 + bx4 + ax3 + dx2 + ex + f <math> <mrow> <munder> <mrow> <mi>l</mi> <mi>i</mi> <mi>m</mi> </mrow> <mrow> <mi>x</mi> <mo>→</mo> <mn>0</mn> </mrow> </munder> <mfrac> <mrow> <mi>f</mi> <mrow> <mo>(</mo> <mrow> <mi>x</mi> </mrow> <mo>)</mo> </mrow> </mrow> <mrow> <msup> <mrow> <mi>x</mi> </mrow> <mrow> <mn>3</mn> </mrow> </msup> </mrow> </mfrac> <mo>=</mo> <mn>1</mn> </mrow> </math>                   Non zero finiteSo, d = e = f = 0f(x)  = x6 + ax5 + bx4 + cx3 <math> <mrow> <munder> <mrow> <mi>l</mi> <mi>i</mi> <mi>m</mi> </mrow> <mrow> <mi>x</mi> <mo>→</mo> <mn>0</mn> </mrow> </munder> <mfrac> <mrow> <mi>f</mi> <mrow> <mo>(</mo> <mrow> <mi>x</mi> </mrow> <mo>)</mo> </mrow> </mrow> <mrow> <msup> <mrow> <mi>x</mi> </mrow> <mrow> <mn>3</mn> </mrow> </msup> </mrow> </mfrac> <mo>=</mo> <mn>1</mn> </mrow> </math>          Non zero finitef’(x) = <math> <mrow> <mn>6</mn> <msup> <mrow> <mi>x</mi> </mrow> <mrow> <mn>5</mn> </mrow> </msup> <mo>+</mo> <mn>5</mn> <mi>a</mi> <msup> <mrow> <mi>x</mi> </mrow> <mrow> <mn>4</mn> </mrow> </msup> <mo>+</mo> <mn>4</mn> <mi>b</mi> <msup> <mrow> <mi>x</mi> </mrow> <mrow> <mn>3</mn> </mrow> </msup> <mo>+</mo> <mn>3</mn> <msup> <mrow> <mi>x</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msup> </mrow> </math>  f’(1) = 06 + 5a + 4b + 3 = 05a + 4b = - 9 .(i)f’(-1) = 0-6 + 5a – 4b + 3 = 0 .(ii)Solving (i) and (ii)a  -3/5, b = -3/2 <math> <mrow> <mo>∴</mo> <mi>f</mi> <mrow> <mo>(</mo> <mrow> <mi>x</mi> </mrow> <mo>)</mo> </mrow> <mo>=</mo> <msup> <mrow> <mi>x</mi> </mrow> <mrow> <mn>6</mn> </mrow> </msup> <mo>+</mo> <mrow> <mo>(</mo> <mrow> <mfrac> <mrow> <mo>−</mo> <mn>3</mn> </mrow> <mrow> <mn>5</mn> </mrow> </mfrac> </mrow> <mo>)</mo> </mrow> <msup> <mrow> <mi>x</mi> </mrow> <mrow> <mn>5</mn> </mrow> </msup> <mo>+</mo> <mrow> <mo>(</mo> <mrow> <mfrac> <mrow> <mo>−</mo> <mn>3</mn> </mrow> <mrow> <mn>2</mn> </mrow> </mfrac> </mrow> <mo>)</mo> </mrow> <msup> <mrow> <mi>x</mi> </mrow> <mrow> <mn>4</mn> </mrow> </msup> <mo>+</mo> <msup> <mrow> <mi>x</mi> </mrow> <mrow> <mn>3</mn> </mrow> </msup> </mrow> </math> <math> <mrow> <mo>∴</mo> <mn>5</mn> <mo>.</mo> <mi>f</mi> <mrow> <mo>(</mo> <mrow> <mn>2</mn> </mrow> <mo>)</mo> </mrow> <mo>=</mo> <mn>1</mn> <mn>4</mn> <mn>4</mn> </mrow> </math>

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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
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b = ∫? (2t² - 15t + 10)dt = [2t³/3 - 15t²/2 + 10t] from 0 to 7.
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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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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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Hence $Δ$ be equilateral having each side of length $\sqrt[]{3} r .$

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Le tf(x) be a polynomial of degree 6 in x, in which the coefficient of x⁶ is unity and it has extrema at x = -1 and x = 1. If $\underset{x \to 0}{l i m} \frac{f (x)}{x^{3}} = 1,$ then 5 f(2) is equal to……………….

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Le tf(x) be a polynomial of degree 6 in x, in which the coefficient of x6 is unity and it has extrema at x = -1 and x = 1. If l i m x → 0 f ( x ) x 3 = 1 , then 5 f(2) is equal to……………….

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Le tf(x) be a polynomial of degree 6 in x, in which the coefficient of x⁶ is unity and it has extrema at x = -1 and x = 1. If $\underset{x \to 0}{l i m} \frac{f (x)}{x^{3}} = 1,$ then 5 f(2) is equal to……………….