Let fx be a polynomial of degree 3 such that fk-2k for k - 2- 3- 4- 5- Then the value of 52 - 10 f10 is equal to

Let P(k) = kf(k) + 2 ⇒ kf(k) + 2 = a(x 2)(x 3)(x 4)(x 5) If k = 0 ⇒ 2 = a( 2)( 3)( 4)( 5) ∴ a=160 ⇒ kf(k)+2=160(x 2)(x 3)(x 4)(x 5) Putting k = 10 ⇒ 10f ...

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Byju's Answer

Standard XII

Mathematics

Descarte's Rule for Positive Roots

Let fx be a p...

Question

Let $f (x)$ be a polynomial of degree $3$ such that $f (k) = - \frac{2}{k}$ for $k = 2, 3, 4, 5$ . Then the value of $52 - 10 f (10)$ is equal to

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Solution

Let $P (k) = k f (k) + 2$
$\Rightarrow k f (k) + 2 = a (x - 2) (x - 3) (x - 4) (x - 5)$
If $k = 0$
$\Rightarrow 2 = a (- 2) (- 3) (- 4) (- 5)$
$∴ a = \frac{1}{60}$
$\Rightarrow k f (k) + 2 = \frac{1}{60} (x - 2) (x - 3) (x - 4) (x - 5)$
Putting $k = 10$
$\Rightarrow 10 f (10) + 2 = \frac{1}{60} \cdot 8 \cdot 7 \cdot 6 \cdot 5 = 28$
$∴ 10 f (10) = 26$
$\Rightarrow 52 - 10 f (10) = 26$

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Let f(x) be a polynomial of degree 3 such that f(k)=−2k for k=2,3,4,5. Then the value of 52−10f(10) is equal to

Let P(k)=kf(k)+2 ⇒kf(k)+2=a(x−2)(x−3)(x−4)(x−5) If k=0 ⇒2=a(−2)(−3)(−4)(−5) ∴a=160 ⇒kf(k)+2=160(x−2)(x−3)(x−4)(x−5) Putting k=10 ⇒10f(10)+2=160⋅8⋅7⋅6⋅5=28 ∴10f(10)=26 ⇒52−10f(10)=26

Let $f (x)$ be a polynomial of degree $3$ such that $f (k) = - \frac{2}{k}$ for $k = 2, 3, 4, 5$ . Then the value of $52 - 10 f (10)$ is equal to