JEE Main 2022 — Mathematics Calculus

Given,

$$f(x)={3}^{{({x}^{2}-2)}^{3}+4}$$f(x)=3(x2−2)3+4

Now on rearranging we get,

$$f(x)=81.{3}^{{({x}^{2}-2)}^{3}}$$f(x)=81.3(x2−2)3

Now differentiating both side we get,

$${f}^{'}(x)=81.{3}^{{({x}^{2}-2)}^{3}}\cdot \mathrm{ln}3\cdot 3{({x}^{2}-2)}^{2}\cdot 2x$$f′(x)=81.3(x2−2)3⋅ln3⋅3(x2−2)2⋅2x

$$=(81\times 6){3}^{{({x}^{2}-2)}^{3}}\times {({x}^{2}-2)}^{2}\mathrm{ln}3$$=(81×6)3(x2−2)3×(x2−2)2ln3

So by using first derivate test we get $x=0$x=0 is point of local minima

Now differentiating the function ${f}^{'}(x)=\underset{k}{\underset{⏟}{(486\cdot \mathrm{ln}3)}}\underset{g(x)}{\underset{⏟}{{3}^{{({x}^{2}-2)}^{3}}x{({x}^{2}-2)}^{2}}}$f′(x)=k⏟(486⋅ln3)​​g(x)⏟3(x2−2)3x(x2−2)2​​

We get,${g}^{'}(x)={3}^{{({x}^{2}-2)}^{3}}{({x}^{2}-2)}^{2}+x\cdot {3}^{{({x}^{2}-2)}^{3}}\cdot 4x\cdot ({x}^{2}-2)$g′(x)=3(x2−2)3(x2−2)2+x⋅3(x2−2)3⋅4x⋅(x2−2)

$$+x\cdot {({x}^{2}-2)}^{2}\cdot {3}^{{({x}^{2}-2)}^{3}}\mathrm{ln}3\cdot 3{({x}^{2}-2)}^{2}\cdot 2x$$+x⋅(x2−2)2⋅3(x2−2)3ln3⋅3(x2−2)2⋅2x

$$={3}^{{({x}^{2}-2)}^{3}}({x}^{2}-2)[{x}^{2}-2+4{x}^{2}+6{x}^{2}\mathrm{ln}3{({x}^{2}-2)}^{3}]$$=3(x2−2)3(x2−2)[x2−2+4x2+6x2ln3(x2−2)3]

$${g}^{'}(x)={3}^{{({x}^{2}-2)}^{3}}({x}^{2}-2)[5{x}^{2}-2+6{x}^{2}\mathrm{ln}3{({x}^{2}-2)}^{3}]$$g′(x)=3(x2−2)3(x2−2)[5x2−2+6x2ln3(x2−2)3]

$${f}^{''}(x)=k\cdot {g}^{'}(x)$$f′′(x)=k⋅g′(x)

$${f}^{''}(\sqrt{2})=0,{f}^{''}({\sqrt{2}}^{+})>0,{f}^{''}({\sqrt{2}}^{-})<0$$f′′(2​)=0,f′′(2​+)>0,f′′(2​−)<0

Since double derivate is changing the sign at given point so we can say that $x=\sqrt{2}$x=2​ is point of inflection

Also, ${f}^{''}(x)>0$f′′(x)>0 for $x>\sqrt{2}$x>2​ so ${f}^{'}(x)$f′(x) is increasing