লিমিট

Let ff be differentiable (xϵR)(x\epsilon R), if f(2)=2f(2) = -2 and f(x)2f'(x) \geq 2 for xϵ[1,6]x\epsilon [1, 6], then-

হানি নাটস

By LMVT there is c(1,6) c \in(1,6)

Such that f(6)f(1)61=f(3) \frac{f(6)-f(1)}{6-1}=f^{\prime}(3)

f(6)+25=f(3)f(6)=5f(3)25.22f(6)8=>f(6)6 \begin{array}{l} \frac{f(6)+2}{5}=f^{\prime}(3) \\ f(6)=5 f^{\prime}(3)-2 \geq 5.2-2 \\ f(6) \geqslant 8=>f(6) \geq 6 \end{array}

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