IIT-JEE 2009 Paper 2 Offline | Application of Derivatives Question 36 | Mathematics

IIT-JEE 2009 Paper 2 Offline

MCQ (More than One Correct Answer)

-2

For the function $$$f\left( x \right) = x\cos \,{1 \over x},x \ge 1,$$$

for at least one $$x$$ in the interval $$\left[ {1,\infty } \right)$$, $$f\left( {x + 2} \right) - f\left( x \right) < 2$$

$$\mathop {\lim }\limits_{x \to \infty } f'\left( x \right) = 1$$

for all $$x$$ in the interval $$\left[ {1,\infty } \right)f\left( {x + 2} \right) - f\left( x \right) > 2$$

$$f'(x)$$ is strictly decreasing in the interval $$\left[ {1,\infty } \right)$$

IIT-JEE 2006

MCQ (More than One Correct Answer)

-1

A tangent drawn to the curve $y=f(x)$ at $\mathrm{P}(x, y)$ cuts the X -axis and Y -axis at A and B respectively such that $\mathrm{BP}: \mathrm{AP}=3: 1$, given that $f(1)=1$, then

equation of curve is $x \frac{d y}{d x}-3 y=0$

normal at $(1,1)$ is $x+3 y=4$

curve passes through $(2,1 / 8)$

equation of curve is $x \frac{d y}{d x}+3 y=0$

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

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