IIT-JEE 2008 Paper 1 Offline | Differentiation Question 9 | Mathematics

IIT-JEE 2008 Paper 1 Offline

MCQ (Single Correct Answer)

-1

Consider the functions defined implicitly by the equation $$y^3-3y+x=0$$ on various intervals in the real line. If $$x\in(-\infty,-2)\cup(2,\infty)$$, the equation implicitly defines a unique real valued differentiable function $$y=f(x)$$. If $$x\in(-2,2)$$, the equation implicitly defines a unique real valued differentiable function $$y=g(x)$$ satisfying $$g(0)=0$$

If $$f\left( { - 10\sqrt 2 } \right) = 2\sqrt 2 ,$$ then $$f''\left( { - 10\sqrt 2 } \right) = $$

$${{4\sqrt 2 } \over {{7^3}{3^2}}}$$

$$-{{4\sqrt 2 } \over {{7^3}{3^2}}}$$

$${{4\sqrt 2 } \over {{7^3}3}}$$

$$-{{4\sqrt 2 } \over {{7^3}3}}$$

IIT-JEE 2007

MCQ (Single Correct Answer)

-0.75

$${{{d^2}x} \over {d{y^2}}}$$ equals

$${\left( {{{{d^2}y} \over {d{x^2}}}} \right)^{ - 1}}$$

$$ - {\left( {{{{d^2}y} \over {d{x^2}}}} \right)^{ - 1}}{\left( {{{dy} \over {dx}}} \right)^{ - 3}}$$

$$\left( {{{{d^2}y} \over {d{x^2}}}} \right){\left( {{{dy} \over {dx}}} \right)^{ - 2}}$$

$$ - \left( {{{{d^2}y} \over {d{x^2}}}} \right){\left( {{{dy} \over {dx}}} \right)^{ - 3}}$$

IIT-JEE 2007

MCQ (Single Correct Answer)

-0.75

Let $$\,\,\,$$$$f\left( x \right) = 2 + \cos x$$ for all real $$X$$.

STATEMENT - 1: for eachreal $$t$$, there exists a point $$c$$ in $$\left[ {t,t + \pi } \right]$$ such that $$f'\left( c \right) = 0$$ because
STATEMENT - 2: $$f\left( t \right) = f\left( {t + 2\pi } \right)$$ for each real $$t$$.

Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1

Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1

Statement-1 is True, Statement-2 is False

Statement-1 is False, Statement-2 is True.

IIT-JEE 2005 Screening

MCQ (Single Correct Answer)

-0.5

If $$f(x)$$ is a twice differentiable function and given that $$f\left( 1 \right) = 1;f\left( 2 \right) = 4,f\left( 3 \right) = 9$$, then

$$f''\left( x \right) = 2$$ for $$\forall x \in \left( {1,3} \right)$$

$$f''\left( x \right) = f'\left( x \right) = 5$$ for some $$x \in \left( {2,3} \right)$$

$$f''\left( x \right) = 3$$ for $$\forall x \in \left( {2,3} \right)$$

$$f''\left( x \right) = 2$$ for some $$x \in \left( {1,3} \right)$$

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