IIT-JEE 2008 Paper 1 Offline | JEE Advanced Year Wise Previous Years Questions

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 (More than One Correct Answer)

-2

Let $$f(x)$$ be a non-constant twice differentiable function defined on $$\left( { - \infty ,\infty } \right)$$

such that $$f\left( x \right) = f\left( {1 - x} \right)$$ and $$f'\left( {{1 \over 4}} \right) = 0.$$ Then,

$$f''\left( x \right)$$ vanishes at least twice on $$\left[ {0,1} \right]$$

$$f'\left( {{1 \over 2}} \right) = 0$$

$$\int\limits_{ - 1/2}^{1/2} {f\left( {x + {1 \over 2}} \right)\sin x\,dx} = 0$$

$$\int\limits_0^{1/2} {f\left( t \right){e^{\sin \,\pi t}}dt = } \int\limits_{1/2}^1 {f\left( {1 - t} \right){e^{\sin \,\pi t}}dt} $$

IIT-JEE 2008 Paper 1 Offline

MCQ (Single Correct Answer)

-1

Consider the two curves $${C_1}:{y^2} = 4x,\,{C_2}:{x^2} + {y^2} - 6x + 1 = 0$$. Then,

$${C_1}$$ and $${C_2}$$ touch each other only at one point.

$${C_1}$$ and $${C_2}$$ touch each other exactly at two points

$${C_1}$$ and $${C_2}$$ intersect (but do not touch ) at exactly two points

$${C_1}$$ and $${C_2}$$ neither intersect nor touch each other

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

PREVIOUS NEXT