1) \ f(x)\ is\ a\ polynomial\ of\ degree\ 4\ with\ real\ coefficients\ such \that\ f(x)\ =\ 0\ is\ satisfied\ by\ x\ = \1,2,3\ only\ , \ then\ f ' (1). f'(2). f'(3)\ is\ equals\ to \ \\\\\ (a)\ 0\ (b)\ 2\ (c)\ -1\ (d)\ none\ of\ these\
2) \ If\ the\ function\ f(x)\ x^{3} - 9x^{2}+24x+c \ has \ three\ real\ and\ distinct\ roots\ \alpha ,\beta ,\ and\ \gamma ,\ then the value\ of \\\ \left[\alpha \right]+\left[\beta \right]+\left[\gamma \right] is ; \\\\ (a) \ 5,6 \ (b) \ 6,7 \ (c) 7,8 \ (d) \ none \ of\ these
3) \ Let\ f(x)\ = sinx\ + \ ax\ + b\ \,, then\ f(x) \ = 0 \ has \\\ \\\ (a) \ only \ one\ real\ root\ which \ is\ positive\ if\ a >1 , b<0\\ \\\ (b)\ \ only \ one\ real\ root\ which \ is\ negative\ if\ a >1 , b>0\\ \\\ (c) \ \ \ only \ one\ real\ root\ which \ is\ negative\ if\ a <-1 , b<0\\ \\\ (d) \ none\ of\ these
How to solve these type of problems ?
1708(1):\Rightarrow $ Here We will take $f(x)=a.(x-1)^2.(x-2).(x-3)$\\\\ bcz in case of $4^{th}$ degree equation with $3$ real Roots are Given.Then it must\\\\ have Repeated Roots.\\\\ Now Diff. $f(x)$ w. r.to $x$, We get..\\\\ $f^{'}(x)=a.\left \{ (x-1)^2.(x-2+x-3)+(x-2)(x-3).2.(x-1) \right \}$\\\\ Put $x=1$, we get $f^{'}(1)=0$\\\\ So $\boxed{\boxed{f^{'}(1)\times f^{'}(2)\times f^{'}(3) =0}}$
1708(2):\Rightarrow $Here $[\alpha]+[\beta]+[\gamma]=6$ OR $[\alpha]+[\beta]+[\gamma]=7$\\\\ Can anyone conform me........
11Let f(x) = x3 - 9x2 + 24x + c
Let g(x) = x3 - 9x2 + 24x
now g'(x) = 3x2 - 18x + 24
now g'(x) = 0
so ( x - 2 ) (x - 4 ) = 0
so x = 2 , 4
now g''(x) = 6x - 18
now g"(2) = -6 < 0 => max at x = 2
now g"(4) = 6 > 0 => min at x = 4
so max value = g(2) = 20
so min value = g(4) = 16
also from g'(x) we have g(x) dec for 2 > x > 4 and increasing for 2 < x <4.
So we have the graph of g(x) as follows (The black curve).
Now bringing the graph of g(x) down by 16 units we get the blue curve...
Now bringing the graph of g(x) down by 20 units we get the red curve....
Now the value of c in f(x) can vary from ( -16 , -20 )
So the required graph will lie between the red and blue graphs....
Now the blue graph cuts the x axis at x = 1 and the red graph at x = 5.
Thus we have two cases as (beta can lie between 2 & 3 or 3 & 4)
Case 1
(alpha) = ( 1 , 2 )
(beta) = ( 2 , 3 )
(gamma) = ( 4, 5 )
so [(alpha)] + [(beta)] + [(gamma)] = 1 + 2 + 4 = 7
Case 2
(alpha) = ( 1 , 2 )
(beta) = ( 3 , 4 )
(gamma) = ( 4, 5 )
so [(alpha)] + [(beta)] + [(gamma)] = 1 + 3 + 4 = 8
so the ans is 7 , 8
so option c
