Download App

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

CBSE BoardEnglish MediumSTD 12 ScienceMathematicsSTD 12 - 6. Application of derivatives1 Mark
MCQ
The minimum value of the function $y = 2{x^3} - 21{x^2} + 36x - 20$ is
  • ✓
    $-128$
  • B
    $-126$
  • C
    $-120$
  • D
    None of these
✓

Answer

Correct option: A.
$-128$
a
(a) Given, $f(x) = 2{x^3} - 21{x^2} + 36x - 20$

$f'(x) = 6{x^2} - 42x + 36$

Put $f'(x) = 0$==> $6{x^2} - 42x + 36 = 0$==> ${x^2} - 7x + 6 = 0$

==> ${x^2} - 6x - x + 6 = 0$==> $(x - 1)(x - 6) = 0$==> $x = 1,\,6$

Now, $f''\,(x) = 12x - 42$

$f''\,(1) = - 30 = - ve$ and $f''\,(6) = 30 = + ve$

Hence $x = 6$ is the point of minima

$\therefore$ Minimum value = $f(6) = 2{(6)^3} - 21{(6)^2} + 36 \times 6 - 20$

$\therefore$ $f(6) = - 128$.

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Explore more

More "M.C.Q (1 Marks)" from STD 12 - 6. Application of derivatives→STD 12 - 6. Application of derivatives — all question types→Mathematics STD 12 Science question papers→CBSE Board STD 12 Science question papers→CBSE Board question paper generator→

Similar questions

If a matrix has $13$ elements, then the possible dimensions $($orders$)$ of the matrix are:
→
Function  $f(x) = \left\{ {\begin{array}{*{20}{c}}   {sgn \left( {\left[ x \right]} \right)\,\,\,\,;\,\,\,x \ne I} \\   {\left[ {sgn \left( x \right)} \right]\,\,\,;\,\,\,x = I} \end{array}} \right.$ is '( where $sgn ()$ denotes signum function $and$ $[.]$ denotes greatest integer function )
→
Integration of function $\frac{x}{e^{x^2}}$ with respect to $x$ is equal to :
→
Let a function $g:[0,4] \rightarrow R$ be defined as

$g ( x )=\left\{\begin{array}{ll}\max _{0 \leq t \leq x }\left\{ t ^{3}-6 t ^{2}+9 t -3\right\} & , 0 \leq x \leq 3 \\ 4- x & , 3 < x \leq 4\end{array}\right.$ then the number of points in the interval $(0,4)$ where $g(x)$ is NOT differentiable, is $.....$

→
If  $ A$  is a square matrix, then $A + {A^T}$ is
→
If $\text{f(x)}=\begin{vmatrix}0&\text{x}-\text{a}&\text{x}-\text{b}\\\text{x}+\text{a}&0&\text{x}-\text{c}\\\text{x}+\text{b}&\text{x}+\text{c}&0\end{vmatrix},$ then:
→
What is the value of $\int_{-1}^{1}\sin^3\text{x}\cos^2\text{xdx}?$
→
Let $f(x)$ be a polynomial of degree $3$ such that $f(-1)=10, f(1)=-6, f(\mathrm{x})$ has a critical point at $\mathrm{x}=-1$ and $f^{\prime}(\mathrm{x})$ has a critical point at $\mathrm{x}=1$ Then $f(x)$ has a local minima at $x=$
→
$\int_{}^{} {x{{\cos }^2}} xdx = $
→
Function

$f\left( x \right) = \int_1^x {\left\{ {2\left( {t - 1} \right){{\left( {t - 2} \right)}^3} + 3{{\left( {t - 1} \right)}^2}{{\left( {t - 2} \right)}^2}} \right\}} dt$ is maximum when $x$ is equal to

→