For all real values of a0- a1- a2- a3 satisfying a0-a1-2-a2-3-3-4-0- the equation a1x-a2x2-a2x-a3x3-0 has real root in the interval

The correct option is A [0, 1] Let, #160; f(x)=a_3x^4/4+a_2x^3/3+a_1x^2/2+a_0x ∴ f(0)=0, f(1)=a_3/4+a_2/3 #160;+ #160;a_1/2+a_0=0 ⇒ f( ...

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Byju's Answer

Standard XII

Mathematics

Geometrical Applications of Differential Equations

For all real ...

Question

For all real values of $a_{0}, a_{1}, a_{2}, a_{3}$ satisfying $a_{0} + \frac{a_{1}}{2} + \frac{a_{2}}{3} + \frac{α_{3}}{4} = 0$ , the equation $a_{1} x + a_{2} x^{2} + a_{2} x + a_{3} x^{3} = 0$ has real root in the interval

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Solution

The correct option is A $[0, 1]$
Let,
$f (x) = \frac{a_{3} x^{4}}{4} + \frac{a_{2} x^{3}}{3} + \frac{a_{1} x^{2}}{2} + a_{0} x$

$∴ f (0) = 0, f (1) = \frac{a_{3}}{4} + \frac{a_{2}}{3} + \frac{a_{1}}{2} + a_{0} = 0$
$\Rightarrow f (0) = f (1)$

$\Rightarrow f^{'} (x) = 0$ has atleast one real root in $[0, 1]$ . [According to Rolle's theorem]

$∴ f^{'} (x) = a_{3} x^{3} + a_{2} x^{2} + a_{1} x + a_{0}$

Hence, $a_{3} x^{3} + a_{2} x^{2} + a_{1} x + a_{0}$ must has a real root in the interval $[0, 1]$

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Similar questions

Q. Given that

{(1 + x + x^{2})}^{n} = a_{0} + a_{1} x + a_{2} x^{2} + \dots + a_{2 n} x^{2 n},

find the values of
(i)

a_{0} + a_{1} + a_{2} + \dots . + a_{2 n}

(ii)

a_{0} - a_{1} + a_{2} - a_{3} \dots . . + a_{2 n}

$(1 + x + 2 x^{2})^{20}$ = $a_{0} + a_{1} x + a_{2} x^{2}$ ................... $a_{40} x^{40}$ , If $a_{0} + a_{1} + a_{2} + a_{3} . . . . . . . . . . . . . . . . . . a_{40} = 2^{a}$

$a_{0} - a_{1} + a_{2} - a_{3} . . . . . . . . . . . . . . . . . . . . . . a_{40} = 2^{b}$ . Find the value of a + b

___

Q. If

(1 + x + x^{2}) n = a 0 + a^{1} x + a^{2} x + a^{3} x^{3} + . . . . . . . a (2^{n}) x (2^{n})

then prove that

(a^{0} + a^{3} + a^{6} + . . . . . . . . .) = (a^{1} + a^{4} + a^{7}) = (a^{2} + a^{5} + a^{8}) = 3^{(} n - 1),

If $(1 + a x + b x^{2})^{4} = a_{0} + a_{1} x + a_{2} x^{2} + \dots + a_{8} x^{8}; a, b, a_{0}, a_{1} \dots a_{8} ϵ R$ and are such that $a_{0} + a_{1} + a_{2} \neq 0$ and $∣ ∣ ∣ ∣ \begin{matrix} a_{0} & a_{1} & a_{2} a_{1} & a_{2} & a_{0} a_{2} & a_{0} & a_{1} \end{matrix} ∣ ∣ ∣ ∣ = 0$ , then

Q. Let

(1 + x + x^{2})^{2014} = a_{0} + a_{1} x + a_{2} x^{2} + a_{3} x^{3} + . . . . + a_{4028} x^{4028}

and let

A = a_{0} - a_{3} + a_{6} - . . . . . + a_{4026}

B = a_{1} - a_{2} + a_{7} - . . . . . + a_{4027}

C = a_{2} - a_{5} + a_{8} - . . . . . + a_{4028}

Then

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For all real values of a0,a1,a2,a3 satisfying a0+a12+a23+α34=0, the equation a1x+a2x2+a2x+a3x3=0 has real root in the interval

The correct option is A [0,1]Let, f(x)=a3x44+a2x33+a1x22+a0x∴f(0)=0,f(1)=a34+a23+a12+a0=0⇒f(0)=f(1)⇒f′(x)=0 has atleast one real root in [0,1]. [According to Rolle's theorem]∴f′(x)=a3x3+a2x2+a1x+a0Hence, a3x3+a2x2+a1x+a0 must has a real root in the interval [0,1]

For all real values of $a_{0}, a_{1}, a_{2}, a_{3}$ satisfying $a_{0} + \frac{a_{1}}{2} + \frac{a_{2}}{3} + \frac{α_{3}}{4} = 0$ , the equation $a_{1} x + a_{2} x^{2} + a_{2} x + a_{3} x^{3} = 0$ has real root in the interval