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 $$g(0)=0$$

$$\int\limits_{ - 1}^1 {g'\left( x \right)dx = } $$

$$2g(-1)$$

$$0$$

$$-2g(1)$$

$$2g(1)$$

IIT-JEE 2008 Paper 1 Offline

MCQ (Single Correct Answer)

-1

The area of the region bounded by the curve $$y=f(x),$$ the
$$x$$-axis, and the lines $$x=a$$ and $$x=b$$, where $$ - \infty < a < b < - 2,$$ is :

$$\int\limits_a^b {{x \over {3\left( {{{(f(x))}^2} - 1} \right)}}} dx + bf\left( b \right) - af\left( a \right)$$

$$ - \int\limits_a^b {{x \over {3\left( {{{(f(x))}^2} - 1} \right)}}} dx + bf\left( b \right) - af\left( a \right)$$

$$\int\limits_a^b {{x \over {3\left( {{{(f(x))}^2} - 1} \right)}}} dx - bf\left( b \right) + af\left( a \right)$$

$$ - \int\limits_a^b {{x \over {3\left( {{{(f(x))}^2} - 1} \right)}}} dx - bf\left( b \right) + af\left( a \right)$$

IIT-JEE 2008 Paper 1 Offline

MCQ (Single Correct Answer)

-1

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 $$\left[ {0,1} \right]$$

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

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

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

PREVIOUS NEXT