IIT-JEE 2008 Paper 2 Offline | Differentiation Question 7 | Mathematics

IIT-JEE 2008 Paper 2 Offline

MCQ (Single Correct Answer)

-1

Consider the function $$f:\left( { - \infty ,\infty } \right) \to \left( { - \infty ,\infty } \right)$$ defined by

$$f\left( x \right) = {{{x^2} - ax + 1} \over {{x^2} + ax + 1}},0 < a < 2.$$

Which of the following is true?

$${\left( {2 + a} \right)^2}f''\left( 1 \right) + {\left( {2 - a} \right)^2}f''\left( { - 1} \right) = 0$$

$${\left( {2 - a} \right)^2}f''\left( 1 \right) - {\left( {2 + a} \right)^2}f''\left( { - 1} \right) = 0$$

$$f'\left( 1 \right)f'\left( { - 1} \right) = {\left( {2 - a} \right)^2}$$

$$f'\left( 1 \right)f'\left( { - 1} \right) = -{\left( {2 + a} \right)^2}$$

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 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

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

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

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

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

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