STD 12 - 6. Application of derivatives — Mathematics STD 12 Science — Question

$\therefore $ $f'(x) = 8{e^{2x}} - 18{e^{ - 2x}}$

Put $f'(x) = 0 \Rightarrow 8{e^{2x}} - 18{e^{ - 2x}} = 0$

${e^{2x}} = 3/2 \Rightarrow x = \log {(3/2)^{1/2}}$

Again $f''(x) = 16{e^{2x}} + 36{e^{ - 2x}} > 0$

Now $f(\log {(3/2)^{1/2}}) = 4{e^{2.(\log {{(3/2)}^{1/2}})}} + 9{e^{ - 2(\log {{(3/2)}^{1/2}})}}$

$=  4 \times \frac{3}{2} + 9 \times \frac{2}{3}  =  6 + 6 = 12$

Hence minimum value $=  12.$