$R(x)=\frac{{{x^{16}} – 1}}{{{x^8} + 2{x^7}}} = \frac{{({x^2} – 1)({x^2} + 1)({x^4} + 1)({x^8} + 1)}}{{{x^7}(x + 2)}}$
$\because {x^2} – x – 1 = 0,{x^2} – 1 = x$
$R(x) = \frac{{x({x^2} + 1)({x^4} + 1)({x^8} + 1)}}{{{x^7}(x + 2)}} = \frac{{({x^2} + 1)({x^4} + 1)({x^8} + 1)}}{{{x^6}(x + 2)}}$
$\because {x^2} – x – 1 = 0,{x^2} = x + 1$ Continue reading $x^2-x-1=0$
$R(x)=\frac{{{x^{16}} – 1}}{{{x^8} + 2{x^7}}} = ?$
For $x \in \mathbb{R}$, the number of real roots of the equation $$3{x^2} – 4|{x^2} – 1| + x – 1 = 0$$ is ___ . Continue reading Roots of
$3{x^2} – 4|{x^2} – 1| + x – 1 = 0$
(A) $\{ \mp 4, \pm 2, \pm 1\} $
(B) $\{ \pm 4, \mp 2, \pm 1\} $
(C) $\{ \pm 4, \pm 2, \mp 1\} $
(D) $\{ \pm 4, \pm 2, \pm 1\} $ Continue reading $x^2+4y^2+16z^2=48$
$xy+4yz+2zx=24$
$\{x,y,z\}\equiv$
A car driver going at some speed v engaged in phone call all of a sudden finds a wide wall at a distance d. Are hard brakes applied in panic or car suddenly turned in a circle of radius d more likely to avoid collision with the wall? Assume that the driving balance is not lost.
A capacitor of capacitance ‘C’, is connected across an AC source of voltage V, given by $V=V_0 sin \omega t $. The displacement current between the plates of the capacitor, would then be given by:
(1) $I_d = V_0 \omega C cos \omega t $
(2) $I_d = \frac {V_0}{\omega C} cos \omega t $
(3) $I_d = \frac {V_0}{\omega C} sin \omega t$
(4) $I_d = V_0 \omega C sin \omega t $ Continue reading A capacitor of capacitance ‘C’, is connected ….
We have, $(x-1)^{100}+(x-2)^{200}=(x-1)(x-2)q(x)+r(x)$
Since divisor is quadratic, the remainder must be linear.
So, $r(x)=ax+b$
$(x-1)^{100}+(x-2)^{200}=(x-1)(x-2)q(x)+ax+b$
Putting $x=1$ above yields $a+b=1$
Substituting $x=2$, yields $2a+b=1$
Clearly, a=0 & b=1
So, remainder $=ax+b=1$.
The pressure acting on a submarine is $3 \times 10^5 Pa $ at a certain depth. If the depth is doubled, the percentage increase in the pressure acting on the submarine would be:
(Assume that atmospheric pressure is $1 \times 10^5 Pa$, density of water is $10^3 kgm^{-3}$, $g=10 ms^{-2}$)
(A) $\frac {3}{200} $%
(B) $\frac {5}{200} $%
(C) $\frac {200}{3} $%
(D) $\frac {200}{5} $% Continue reading The pressure acting on a submarine is $3 \times 10^5 Pa $ ….
