If the function f(x)= begincases ( log e(1-x+x2)+ loge(1+x+x2)/ sec x- cos x), x ∈((-π/2), (π/2))- 0 k x=0 endcases is continuous at x=0, then k is equal to :

Question

If the function f(x)= begincases ( log e(1-x+x2)+ loge(1+x+x2)/ sec x- cos x), x ∈((-π/2), (π/2))- 0 k x=0 endcases is continuous at x=0, then k is equal to : (A) 1 (B) -1 (C) e (D) 0

Tardigrade Admin · Accepted Answer

displaystyle lim x arrow 0 (( ln (1+x2+x4)) cos x/1- cos 2 x) displaystyle lim x arrow 0 ((( ln (1+x2+x4)/x2+x4)) x2(1+x2) cos x/(( sin 2 x)x2) x2)=1 ∴ k=1

Q. If the function
$f (x) = {\frac{l o g _{e} ( 1 - x + x ^{2} ) + l o g _{e} ( 1 + x + x ^{2} )}{s e c x - c o s x}, x \in (\frac{- π}{2}, \frac{π}{2}) - {0} k x = 0$
is continuous at $x = 0$ , then $k$ is equal to :

1

-1

e

0

$x \to 0 lim \frac{( ln ( 1 + x ^{2} + x ^{4} ) ) cos x}{1 - cos ^{2} x}$
$x \to 0 lim \frac{( \frac{l n ( 1 + x ^{2} + x ^{4} )}{x ^{2} + x ^{4}} ) x ^{2} ( 1 + x ^{2} ) cos x}{( \frac{s i n ^{2} x}{x ^{2}} ) x ^{2}} = 1$
$∴ k = 1$

Q. If the function f(x)={secx−cosxloge​(1−x+x2)+loge​(1+x+x2)​,x∈(2−π​,2π​)−{0}kx=0​ is continuous at x=0, then k is equal to :

1

-1

e

0

x→0lim​1−cos2x(ln(1+x2+x4))cosx​ x→0lim​(x2sin2x​)x2(x2+x4ln(1+x2+x4)​)x2(1+x2)cosx​=1 ∴k=1

Q. If the function
$f (x) = {\frac{l o g _{e} ( 1 - x + x ^{2} ) + l o g _{e} ( 1 + x + x ^{2} )}{s e c x - c o s x}, x \in (\frac{- π}{2}, \frac{π}{2}) - {0} k x = 0$
is continuous at $x = 0$ , then $k$ is equal to :

$x \to 0 lim \frac{( ln ( 1 + x ^{2} + x ^{4} ) ) cos x}{1 - cos ^{2} x}$
$x \to 0 lim \frac{( \frac{l n ( 1 + x ^{2} + x ^{4} )}{x ^{2} + x ^{4}} ) x ^{2} ( 1 + x ^{2} ) cos x}{( \frac{s i n ^{2} x}{x ^{2}} ) x ^{2}} = 1$
$∴ k = 1$