IIT-JEE 2006 | Application of Derivatives Question 41 | Mathematics

IIT-JEE 2006

MCQ (More than One Correct Answer)

-1

$f(x)$ is cubic polynomial which has local maximum at $x=-1$. If $f(2)=18, f(1)=-1$ and $f(x)$ has local minima at $x=0$, then

the distance between $(-1,2)$ and $(a, f(A)$, where $x=a$ is the point of local minima is $2 \sqrt{5}$

$f(x)$ is increasing for $x \in[1,2 \sqrt{5}]$

$f(x)$ has local minima at $x=1$

the value of $f(0)=5$

IIT-JEE 2006

MCQ (More than One Correct Answer)

-1

$$ \begin{aligned} & f(x)=\left\{\begin{array}{cc} e^x, & 0 \leq x \leq 1 \\ 2-e^{x-1}, & 1 < x \leq 2 \\ x-e, & 2 < x \leq 3 \end{array} \quad\right. \text { and } \\ & g(x)=\int_0^x f(t) d t, x \in[1,3] \text { then } g(x) \text { has } \end{aligned} $$

local maxima at $x=1+\ln 2$ and local $\operatorname{minima}$ at $x=e$

local maxima at $x=1$ and local minima at $x=2$

no local maxima

no local minima

IIT-JEE 1999

MCQ (More than One Correct Answer)

-0.75

The function $$f\left( x \right) = \int\limits_{ - 1}^x {t\left( {{e^t} - 1} \right)\left( {t - 1} \right){{\left( {t - 2} \right)}^3}\,\,\,{{\left( {t - 3} \right)}^5}} $$ $$dt$$ has a local minimum at $$x=$$

$$0$$

$$1$$

$$2$$

$$3$$

IIT-JEE 1998

MCQ (More than One Correct Answer)

-0.5

Let $$h\left( x \right) = f\left( x \right) - {\left( {f\left( x \right)} \right)^2} + {\left( {f\left( x \right)} \right)^3}$$ for every real number $$x$$. Then

$$h$$ is increasing whenever $$f$$ is increasing

$$h$$ is increasing whenever $$f$$ is decreasing

$$h$$ is decreasing whenever $$f$$ is decreasing

nothing can be said in general.

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