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Continuity — Maths STD 11 Science — Question

Maharashtra BoardEnglish MediumSTD 11 ScienceMathsContinuity1 Mark
Question
Show that there is a root for the equation $2x^3 – x – 16 = 0$ between $2$ and $3$.
✓

Answer

Let $f(x) = 2x^3 – x – 16$
f(x) is a polynomial function and hence it is continuous for all $x ∈ R$.
A root of f(x) exists, if f(x) = 0 for at least one value of $x$.
$f(2) = 2(2)^3 – 2 – 16 = -2 < 0$
$f(3) = 2(3)^3 – 3 – 16 = 35 > 0$
$\therefore f(2) < 0$ and $f(3) > 0$
$\therefore $ By intermediate value theorem,
there has to be point $‘c’$ between $2$ and $3$ such that $f(c) = 0$.
$\therefore $ There is a root of the given equation between $2$ and $3$.

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