If the function f(x) = $\left\{\frac{lo{g}_e\left(1-x+x^2\right)+lo{g}_e\left(1+x+x^2\right)}{\begin{array}{cc}\hfill secx-cosx\hfill & \hfill k\hfill \end{array}}\right\},x\in \left(\frac{-\pi }{2},\frac{\pi }{2}\right)-\left\{0\right\}$  is continuous at x = 0, then k is equal to:

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.

LHL = lim (x→2? ) λ|x²-5x+6| / µ (5x-x²-6) = lim (x→2? ) -λ/µ
RHL = lim (x→2? ) e^ (tan (x-2)/ (x- [x]) = e¹
f (2) = µ
For continuity, -λ/µ = e = µ.
⇒ µ=e, -λ=µ²=e², λ=-e²
∴ λ + µ = -e² + e = e (1-e)

$f\left(x\right)=\frac{\left(2^x+2^{-x}\right)tanx\sqrt[]{ta{n}^{-1}\left(2x^2-3x+1\right)}}{{\left(7x^2-3x+1\right)}^3}$

$f\left(x\right)=\left(2^x+2^{-x}\right).tanx.\sqrt[]{ta{n}^{-1}\left(2x^2-3x+1\right)}.{\left(7x^2-3x+1\right)}^{-3}$

$f\text{'}\left(x\right)=\left(2^x+2^{-x}\right).se{c}^{2}x.\sqrt[]{ta{n}^{-1}\left(2x^2-3x+1\right)}.{\left(7x^2-3x+1\right)}^{-3}+tanx.\left(Q\left(x\right)\right)$

$f\text{'}\left(0\right)=2.1\sqrt[]{\frac{\pi }{4}}.1$

$=\sqrt[]{\pi }$

RHL $\underset{x\to 0^{+}}{lim}\frac{co{s}^{-1}\left(1-x^2\right)si{n}^{-1}\left(1-x\right)}{x-x^3}$

$\underset{x\to 0^{+}}{lim}\frac{\pi }{2}\cdot \frac{co{s}^{-1}\left(1-x^2\right)}{x}$

$\frac{\pi }{2}\underset{x\to 0^{+}}{lim}\frac{-1}{\sqrt[]{{\left(1-\left(1-x^2\right)\right)}^2}}\left(-2x\right)$

$=\frac{\pi }{2}\underset{x\to 2^{+}}{lim}\frac{2x}{\sqrt[]{2x^2-x^4}}=\pi \underset{x\to 0^{+}}{lim}\frac{x}{x\sqrt[]{2-x^2}}$

$=\frac{\pi }{\sqrt[]{2}}$

LHL $\underset{x\to 0^{+}}{lim}\frac{co{s}^{-1}\left(1-{\left(1+x\right)}^2\right)si{n}^{-1}\left(1-\left(1+x\right)\right)}{1\cdot \left(1-{\left(1+x\right)}^2\right)}$

$=\underset{x\to 0^{-}}{lim}\frac{co{s}^{-1}\left(-x^2-2x\right),si{n}^{-1}\left(-x\right)}{-x^2-2x}$            

$=\frac{\pi }{2}\underset{x\to 0^{-}}{lim}\frac{-si{n}^{-1}x}{-x\left(x+2\right)}=\frac{\pi }{2}\times \frac{1}{2}=\frac{\pi }{2}$            

Given $\left(x\right)=\{\begin{array}{l}2sin\left(\frac{-\pi x}{2}\right),x<-1\\ \left|a{x}^2+x+b\right|,-1\le x\le 1\\ sin\pi x,x>1\end{array}$

If f (x) is continuous for all $x\in R$ then it should be continuous at x = 1 & x = -1

At x = -1, L.H.L = R.H.L. Þ 2 = |a + b - 1|

->a + b – 3 = 0  OR  a + b + 1 = 0 . (i)

-> a + b + 1 = 0 . (ii)

(i) & (ii), a + b =-1