Maths Continuity and Differentiability

f (x) is an even function

$f\left(-\frac{1}{4}\right)=f\left(-\frac{1}{2}\right)=f\left(\frac{1}{2}\right)=f\left(\frac{1}{4}\right)=0$  

So, f (x) has at least four roots in (-2, 2)

$g\left(\frac{-3}{4}\right)=g\left(\frac{3}{4}\right)=0$  

So, g (x) has at least two roots in (-2, 2)

now number of roots of f (x) $\cdot g"\left(x\right)=f\text{'}\left(x\right)\cdot g\text{'}\left(x\right)=0$  

It is same as number of roots of  $\frac{d}{dx}\left(f\left(x\right)\cdot g\text{'}\left(x\right)\right)=0$ will have atleast 4 roots in (-2, 2)

$f\left(x\right)=\{\begin{array}{cc}\hfill x+a,\hfill & \hfill x?0\hfill \\ \hfill \left|x?4\right|,\hfill & \hfill x>0\hfill \end{array}\text{?}\text{?}\text{?}\text{?}and\text{?}\text{?}\text{?}g\left(x\right)=\{\begin{array}{cc}\hfill x+1,\hfill & \hfill x<0\hfill \\ \hfill {\left(x?4\right)}^2+b,\hfill & \hfill x?0\hfill \end{array}$

$?$    f (x) and g (x) are continuous on R $?$  a = 4 and b = 1 – 16 = 15

then (gof) (2) + (fog) (-2) = g (2) + f (-1) = -11 + 3 = -8

$f\left(x\right)=\{\begin{array}{l}\frac{lo{g}_e\left(1-x+x^2\right)+lo{g}_e\left(1+x+x^2\right)}{secx-cosx},x\in \left(\frac{-\pi }{2},\frac{\pi }{2}\right)-\left\{0\right\}\\ k,x=0\end{array}$ for continuity at x = 0

      $\underset{x\to 0}{lim}f\left(x\right)=k\therefore k=\underset{x\to 0}{lim}\frac{lo{g}_e\left(1+x^2+x^4\right)}{secx-cosx}\left(\frac{0}{0}form\right)=\underset{x\to 0}{lim}\frac{cosxlo{g}_e\left(1+x^2+x^4\right)}{si{n}^{2}x}=1$  

f (x) = f (6 – x) Þ f' (x) = -f' (6 – x) …. (1)

put x = 0, 2, 5

f' (0) = f' (6) = f' (2) = f' (4) = f' (5) = f' (1) = 0

and from equation (1) we get f' (3) = -f' (3)

$?f\text{'}(3)=0$

So f' (x) = 0 has minimum 7 roots in $x?\left[0,6\right]?f\text{'}\text{'}\left(x\right)$  has min 6 roots in   $x?\left[0,6\right]$

h (x) = f' (x) . f' (x)

h' (x) = (f' (x)² + f' (x) f' (x)

h (x) = 0 has 13 roots in x $?$   [0, 6]

h' (x) = 0 has 12 roots in x $?$ [0, 6]

$f\left(x\right)=[\begin{array}{l}-\frac{1}{x}x\in \left(-\infty ,-1\right)\\ a{x}^2+bx\in \left(-1,1\right)\\ \frac{1}{x}x\in \left[1,\infty \right)\end{array}$

cont at x = 1, a + b = 1

diff at x = 1, 2a = 1 $\Rightarrow a=-\frac{1}{2}\Rightarrow b=\frac{3}{2}$

$Inx\in \left(-2,2\right)$  

 |x|- 1| is not differentiable at x = -1, 0, 1

|cospx| is not differentiable at x =  $\frac{3}{2},-\frac{1}{2},\frac{1}{2},\frac{3}{2}$ -  

RHL&LHL lim (x→0) (sin (2x²/a) + cos (3x/b)^ (ab/x²)
= e^ (lim (x→0) (sin (2x²/a) + cos (3x/b) - 1) (ab/x²) = e^ (4b²-9a)/2b)
f (0) = e³
For continuity at x = 0
Limit = f (0)
(4b² - 9a)/2b = 3 ⇒ 4b² – 6b – 9a = 0∀b ∈ R
⇒ D ≥ 0 ⇒ a ≥ -1/4
a? = -1/4
⇒ |1/a? | = 4.

  $\underset{x\to 1^{-}}{lim}f\left(x\right)=\underset{x\to 1^{+}}{lim}f\left(x\right)\Rightarrow a+b=2$   

  $\therefore \underset{x\to 3^{-}}{lim}f\left(x\right)=\underset{x\to 3^{+}}{lim}f\left(x\right)\Rightarrow 3a+b=6tan\frac{3\pi }{12}$          

$\therefore$ a = 2, b = 0

1 + x? - x? = a? (1+x)? + a? (1+x) + a? (1+x)² . + a? (1+x)?
Differentiate
4x³ - 5x? = a? + 2a? (1+x) + 3a? (1+x)².
12x² - 20x³ = 2a? + 6a? (1+x).
Put x = -1
12 + 20 = 2a? ⇒ a? = 16

Differentiable ⇒ continuous
Continuous at x = 2 ⇒ 2a + b = 0
Continuous at x = 3
0 = 9p + 3q + 1
Differentiable at x=2
a = 2*2 - 5 ⇒ a = -1
Differentiable at x=3
2*3 - 5 = 2p*3 + q ⇒ 6p + q = 1
a = -1, b = 2, p = 4/9, q = -5/3