IIT-JEE 2008 Paper 1 Offline | Differentiation Question 11 | 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 2008 Paper 1 Offline

MCQ (Single Correct Answer)

-1

Let $$f$$ and $$g$$ be real valued functions defined on interval $$(-1, 1)$$ such that $$g''(x)$$ is continuous, $$g\left( 0 \right) \ne 0.$$ $$g'\left( 0 \right) = 0$$, $$g''\left( 0 \right) \ne 0$$, and $$f\left( x \right) = g\left( x \right)\sin x$$

STATEMENT - 1: $$\mathop {\lim }\limits_{x \to 0} \,\,\left[ {g\left( x \right)\cot x - g\left( 0 \right)\cos ec\,x} \right] = f''\left( 0 \right)$$ and

STATEMENT - 2: $$f'\left( 0 \right) = g\left( 0 \right)$$

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 2007 Paper 2 Offline

MCQ (Single Correct Answer)

-1

$$\frac{d^{2} x}{d y^{2}}$$ equals :

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

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

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

$$-\left(\frac{d^{2} y}{d x^{2}}\right)\left(\frac{d y}{d x}\right)^{-3}$$

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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