Concept Problems

1. If x4−14x2+24x−k=0$x^4-14x^2+24x-k=0$ has four real and unequal roots, find the range where k lies. 
If this equation has has to be 4 real roots, it must have 2 minima, both must happen below the x - axis and and 1 maxima, to be happen above x - axis.  Then only the graph cuts the x-axis at 4 points. 
Now, To find where this local maxima and minimas happen, we differentiate f(x) and equate to zero.
f1(x)=4x3−28x+24$f^1(x)=4x^3-28x+24$ = 4(x-1)(x-2)(x+3)

So f1(x)$f^1(x)$ has the roots -3, 1, 2.
Now f(2) = 16 - 56 + 48 - k = 8 - k
and f(1) =  1 -14 + 24 - k = 11 - k
and f(-3) = 81 - 126 -72 -k = -117 - k

We know that f(x) is an increasing function. So it attains minima firstly at 2.  So 8 - k < 0.  and attains maxima at 1. So 11 - k > 0. and again attains minima at - 3. so -117 - k < 0.  Solving above equations we get k > 8, k > -117, and k < 11

We have to get the region which satisfies all the three equations. So k must lie in between 8 and 11.  

2.  A ≡ (–1, 2), B ≡ (1, 4). Point K lies on the x-axis. What is the area of ΔAKB when ∠AKB is maximum, with K on the positive x-axis?
 1)16 sq. units   2) 6 sq. units  3) 4 sq. units  4) Cannot be determined
Sol:

Slope of the line AK = y2−y1x2−x1${\displaystyle \frac{y_2-y_1}{x_2-x_1}}$ = 2−1−a${\displaystyle \frac{2}{-1-a}}$
Slope of the line BK = 41−a${\displaystyle \frac{4}{1-a}}$

Let us find the angle between the lines using slopes. We know that Tanθ=m1−m21+m1m2$Tan\theta ={\displaystyle \frac{m_1-m_2}{1+m_1{m}_2}}$
⇒Tanθ=−21+a−41−a1+−81−a2$\Rightarrow Tan\theta ={\displaystyle \frac{{\displaystyle \frac{-2}{1+a}-{\displaystyle \frac{4}{1-a}}}}{1+{\displaystyle \frac{-8}{1-a^2}}}}$
⇒−2+2a−4−4a1−a21−a2−81−a2=−2a−6−a2−7=2a+6a2+7$\Rightarrow {\displaystyle \frac{{\displaystyle \frac{-2+2a-4-4a}{1-a^2}}}{{\displaystyle \frac{1-a^2-8}{1-a^2}}}={\displaystyle \frac{-2a-6}{-a^2-7}={\displaystyle \frac{2a+6}{a^2+7}}}}$

We need to maximum θ$\theta$ so we have to differentiate the given function w.r.t to independent variable 'a'

Sec2θ.dθda=(a2+7)2−2a(2a+6)(a2+7)2$Se{c}^2\theta .{\displaystyle \frac{d\theta }{da}={\displaystyle \frac{(a^2+7)2-2a(2a+6)}{{\left(a^2+7\right)}^2}}}$

To find the maxima, we should equate this function to '0'
So we get 2a2+14−4a2−12a$2a^2+14-4a^2-12a$ = 0
Or, a2+6a−7$a^2+6a-7$
Or a = 1 or - 7. 

Given K is on the x - axis. So a = 1
We have to now find Area of the triangle with co-ordinates (-1, 2), (1, 4), and (1,0)
Add (-1, 0) to all coordinates = (-2, 2), (0, 4), (0,0)

The area of the triangle whose coordinates are (0,0), (x1,y1)$\left(x_{1,}{y}_1\right)$, (x2,y2)$\left(x_{2,}{y}_2\right)$ is 12|[x1.y2−x2.y1]|${\displaystyle \frac{1}{2}\left|\left[x_1.y_2-x_2.y_1\right]\right|}$

So By applying above formula we get area of the triangle = 4 units.