A cubic $f\left( x \right)$ vanishes at $x = - 2$ and has relative minimum/maximum at $x = - 1\,and\,x = \frac{1}{3}$ such that $\mathop \smallint \limits_{ - 1}^1 f\left( x \right)dx = \frac{{14}}{3}$ Then, $f\left( x \right)$ is

JEE MATHEMATICS MCQ: MAXIMA & MINIMA

1 Hard

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Application +4 | -1

A cubic

f\left( x \right)

vanishes at

x = - 2

and has relative minimum/maximum at

x = - 1\,and\,x = \frac{1}{3}

such that

\int \limits_{ - 1}^1 f\left( x \right)dx = \frac{{14}}{3}

Then,

f\left( x \right)

View answer and explanation

Correct option: C

{x^3} + {x^2} - x + 2

Option D (

{x^3} + {x^2} - x + 2

) is correct because it satisfies the given conditions: vanishes at x = -2, has relative extrema at x = -1 and x = 1/3, and the integral from -1 to 1 is 14/3.

More Questions from MAXIMA & MINIMA

Practice All Chapter MCQs →

Let

f(x)=\left(1+b^2\right) x^2+2 b x+1

. Let

m(b)

be the minimum value of

f(x)

(that is least value). As

b

varies, the range of

m(b)

is Let

f: R \rightarrow R

be a function defined by

f(x)=(x-3)^{n_1}(x-5)^{n_2}, n_1, n_2 \in N

. Then, which of the following is NOT true? The equation

|2 a x-3|+|a x+1|+|5-a x|=\frac{1}{2}

possesses Let the function

f(x)=2 x^{3}+(2 p-7) x^{2}+3(2 p-9) x

-6 have a maxima for some value of

x<0

and a minima for some value of

x>0

. Then, the set of all values of

p

is The least value of the

f\left( x \right)

given by

f\left( x \right) = {\tan ^{ - 1}}x - \frac{1}{2}{\log _e}x

in the interval

\left[ {1/\sqrt 3 ,\;\sqrt 3 } \right]

, is If the total maximum value of the function

f(x)=\left(\frac{\sqrt{3 e}}{2 \sin x}\right)^{\sin ^{2} x}, x \in\left(0, \frac{\pi}{2}\right),

\frac{k}{e}

, then

\left(\frac{k}{e}\right)^{8}+\frac{k^{8}}{e^{5}}+k^{8}

is equal to