Question

If fx=x4-2x^3+3x^2-ax+b is divided by x-1 and x+1,it leaves remainder 5 and 19 respectively.Find the value of a and b.

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Answer 1

Step-by-step explanation:

Answer:

$\begin{gathered}x + 1 = 0 \: \: \: \: , \: \: \: \: x - 1 = 0\\ x = - 1 \: \: \: \: , \: \: \: \: x = 1 \\ f(x) = {x}^{4} - 2 {x}^{3} + 3 {x}^{2} - ax + b \\ f( - 1) = {( - 1)}^{4} - 2 {( - 1)}^{3} + 3 {( - 1)}^{2} - a( - 1) + b=19 \\ = 1 + 2 + 3 + a + b =19\\ = 6 + a + b =19 \\=-13+a+b =0- (i)\\ f( 1) = {( 1)}^{4} - 2 {( 1)}^{3} + 3 {( 1)}^{2} - a( 1) + b=5 \\ = 1 - 2 + 3 - a + b=5 \\ = 2 - a + b =5\\=-3-a+b=0 - (ii)\\ \blue{\mathfrak{\underline{\large{By \: {eq}^{n} \: (i) \: and \: (ii) }}}:}\\ -13 + a + b = -3 - a + b \\ 2a = 10 \\ \boxed{a = 5} \\ \blue{\mathfrak{\underline{\large{ put \: the \: value \: of \: ‘a’ \: in \: (i)}}}:} \\ -13 +5+ b = 0 \\ \boxed{b = 8} \\ \blue{\mathfrak{\underline{\large{Verifying}}}:} \\ f( - 1) = -13 + a + b \\ = -13 +5 +8 = 0\\f( 1) = -3 -a+b \\ = -3-5 +8 = 0 \\ \\ \red{\mathfrak{ \large{\underline{{Hope \: It \: Helps \: You}}}}} \\ \blue{\mathfrak{ \large{\underline{{Mark \: Me \: Brainliest}}}}}\end{gathered}$

Answer 2

Answer:

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happy life

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Answer:

sinα - αcosα

Step-by-step explanation:

Given that,

\lim_{x\to\alpha}\;\;|\frac{x\sin\alpha-\alpha\sin x}{x-\alpha}|lim

x→α

∣

x−α

xsinα−αsinx

∣

Let x - α = h

Thus when x ⇒ α , h ⇒ 0

\lim_{h\to0}\;\;|\frac{(h+\alpha)\sin\alpha-\alpha\sin(h+\alpha)}{h}|lim

h→0

∣

h

(h+α)sinα−αsin(h+α)

∣

Now we will evaluate left hand limit and right and limit separately

For LHL:-

\begin{gathered}\begin{gathered}\lim_{h\to0}\;\;\frac{\alpha\sin\alpha[1-\cos(-h)]}{-h}+\sin\alpha-\frac{\alpha\sin(-h)\cos\alpha}{-h}\\\;\\=\lim_{h\to0}\;\;\frac{\alpha\sin\alpha(1-\cosh)}{-h}+\sin\alpha-\frac{\alpha\sin h\cos\alpha}{h}\\\;\\=\alpha\sin\alpha[\lim_{h\to0}\frac{(1-\cosh)}{-h}]+\sin\alpha-\alpha\cos \alpha[\lim_{h\to0}\frac{\sin h}{h}]\\\;\\=0+\sin\alpha-\alpha\cos\alpha\\\;\\=\sin\alpha-\alpha\cos\alpha\end{gathered} \end{gathered}

h→0

lim

−h

αsinα[1−cos(−h)]

+sinα−

−h

αsin(−h)cosα

=

h→0

lim

−h

αsinα(1−cosh)

+sinα−

h

αsinhcosα

=αsinα[

h→0

lim

−h

(1−cosh)

]+sinα−αcosα[

h→0

lim

h

sinh

]

=0+sinα−αcosα

=sinα−αcosα

For RHL:-

\begin{gathered}\begin{gathered}\lim_{h\to0}\;\;\frac{\alpha\sin\alpha[1-\cosh]}{h}+\sin\alpha-\frac{\alpha\sinh\cos\alpha}{h}\\\;\\=\lim_{h\to0}\;\;\frac{\alpha\sin\alpha(1-\cosh)}{h}+\sin\alpha-\frac{\alpha\sin h\cos\alpha}{h}\\\;\\=\alpha\sin\alpha[\lim_{h\to0}\frac{(1-\cosh)}{h}]+\sin\alpha-\alpha\cos \alpha[\lim_{h\to0}\frac{\sin h}{h}]\\\;\\=0+\sin\alpha-\alpha\cos\alpha\\\;\\=\sin\alpha-\alpha\cos\alpha\end{gathered} \end{gathered}

h→0

lim

h

αsinα[1−cosh]

+sinα−

h

αsinhcosα

=

h→0

lim

h

αsinα(1−cosh)

+sinα−

h

αsinhcosα

=αsinα[

h→0

lim

h

(1−cosh)

]+sinα−αcosα[

h→0

lim

h

sinh

]

=0+sinα−αcosα

=sinα−αcosα

Thus limit is (sinα - αcosα).