F(x)=(36^x-9^x-4^x+1)/√2-√1+cos x for x not equal to 0 and K for x=0. Find value of K? Function is continuous at x=0 Thanks

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Answer 1 · 2017-06-28T17:35:08

$K = (4 \sqrt{2}) (ln 4) (ln 9) .$ $K=(4sqrt2)(ln4)(ln9).$

The function $F,$ $F,$ is defined by,

$F(x)=(36^x-9^x-4^x+1)/(sqrt2-sqrt(1+cosx)), if x ne0:$

$F(x)=K, if x=0.$

Knowing that, $1+cos2theta=2cos^2theta, and,$

$1-cos2theta=2sin^2theta," we see that, for "xne0,$

$F(x)=(36^x-9^x-4^x+1)/(sqrt2-sqrt(1+cosx)),$

$={9^x(4^x-1)-1(4^x-1)}/{sqrt2-sqrt(2cos^2(x/2))},$

$={(4^x-1)(9^x-1)}/{(sqrt2)(1-cos(x/2))},$

$={(4^x-1)(9^x-1)}/{(sqrt2)(2sin^2(x/4))},$

$=1/(2sqrt2){(4^x-1)/x}{(9^x-1)/x}(x/sin(x/4))(x/sin(x/4)),$

$=1/(2sqrt2){(4^x-1)/x}{(9^x-1)/x}((x/4*4)/sin(x/4))((x/4*4)/sin(x/4)),$

$=16/(2sqrt2){(4^x-1)/x}{(9^x-1)/x}((x/4)/sin(x/4))((x/4)/sin(x/4)),$

Now, recall that, $lim_(h to 0)(a^h-1)/h=lna, and, lim_(h to 0)sinh/h=1.$

$:. lim_(x to 0)F(x)=(4sqrt2)(ln4)(ln9)(1)(1)=(4sqrt2)(ln4)(ln9).$

$because," F is continuous at "x=0, :., lim_(x to 0)F(x)=F(0)=K,$

$rArr K=(4sqrt2)(ln4)(ln9).$

Enjoy Maths.!