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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Vishal Baghel | Contributor-Level 10

2 weeks 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> <!- [endif]-><!- [if gte mso 9]><xml> <o:OLEObject Type="Embed" ProgID="Equation.DSMT4" ShapeID="_x0000_i1027" DrawAspect="Content" ObjectID="_1819630341"> </o:OLEObject></xml><! [endif]->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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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……………….