1
IIT-JEE 2008 Paper 1 Offline
MCQ (Single Correct Answer)
+3
-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: $$f'\left( 0 \right) = g\left( 0 \right)$$

A
Statement - 1 is True, Statement - 2 is True; Statement - 2 is a correct explanation for Statement - 1
B
Statement - 1 is True, Statement - 2 is True; Statement - 2 is NOT a correct explanation for Statement - 1
C
Statement - 1 is True, Statement -2 is False
D
Statement - 1 is False, Statement -2 is True
2
IIT-JEE 2008 Paper 1 Offline
MCQ (Single Correct Answer)
+3
-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 $$f''\left( { - 10\sqrt 2 } \right) = $$

A
$${{4\sqrt 2 } \over {{7^3}{3^2}}}$$
B
$$-{{4\sqrt 2 } \over {{7^3}{3^2}}}$$
C
$${{4\sqrt 2 } \over {{7^3}3}}$$
D
$$-{{4\sqrt 2 } \over {{7^3}3}}$$
3
IIT-JEE 2007
MCQ (Single Correct Answer)
+3
-0.75
$${{{d^2}x} \over {d{y^2}}}$$ equals
A
$${\left( {{{{d^2}y} \over {d{x^2}}}} \right)^{ - 1}}$$
B
$$ - {\left( {{{{d^2}y} \over {d{x^2}}}} \right)^{ - 1}}{\left( {{{dy} \over {dx}}} \right)^{ - 3}}$$
C
$$\left( {{{{d^2}y} \over {d{x^2}}}} \right){\left( {{{dy} \over {dx}}} \right)^{ - 2}}$$
D
$$ - \left( {{{{d^2}y} \over {d{x^2}}}} \right){\left( {{{dy} \over {dx}}} \right)^{ - 3}}$$
4
IIT-JEE 2007
MCQ (Single Correct Answer)
+3
-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 $$t$$.

A
Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1
B
Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1
C
Statement-1 is True, Statement-2 is False
D
Statement-1 is False, Statement-2 is True.
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