If f, g, h are real functions given by $f (x) = x^{2}, g (x) = t a n x$ and $h (x) = l o g_{e} x$ , then write the value of $h o g o f (\sqrt{\frac{π}{4}})$

Q. Let

f (x) = x^{2} - \frac{1}{x^{2}}

and

g (x) = x - \frac{1}{x}, x \in R - - 1, 0, 1

. If

h (x) = \frac{f (x)}{g (x)}

then the local minimum value of

h (x)

is:

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If f : R → R, g : R → R and h : R → R is such that f(x)=x2,g(x)=tanx and h(x)=logx, then the value of [ho(gof)], if x=π2 will be

The correct option is A 0[ho(gof)](x)=(hof)(f(x))=(hog)(x2)=(hof)(π4)=h(g(π4))=h(tanπ4)=h(1)=log1=0

If f : R $\to$ R, g : R $\to$ R and h : R $\to$ R is such that $f (x) = x^{2}, g (x) = t a n x$ and $h (x) = l o g x$ , then the value of [ho(gof)], if $x = \frac{π}{2}$ will be

The correct option is A 0
$[h o (g o f)] (x) = (h o f) (f (x))$
$= (h o g) (x^{2}) = (h o f) (\frac{π}{4}) = h (g (\frac{π}{4}))$
$= h (t a n \frac{π}{4}) = h (1) = l o g 1 = 0$