1
IIT-JEE 2005 Screening
MCQ (Single Correct Answer)
+2
-0.5
Tangent to the curve $$y = {x^2} + 6$$ at a point $$(1, 7)$$ touches the circle $${x^2} + {y^2} + 16x + 12y + c = 0$$ at a point $$Q$$. Then the coordinates of $$Q$$ are
A
$$(-6, -11)$$
B
$$(-9, -13)$$
C
$$(-10, -15)$$
D
$$(-6, -7)$$
2
IIT-JEE 2005 Screening
MCQ (Single Correct Answer)
+2
-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
A
$$f''\left( x \right) = 2$$ for $$\forall x \in \left( {1,3} \right)$$
B
$$f''\left( x \right) = f'\left( x \right) = 5$$ for some $$x \in \left( {2,3} \right)$$
C
$$f''\left( x \right) = 3$$ for $$\forall x \in \left( {2,3} \right)$$
D
$$f''\left( x \right) = 2$$ for some $$x \in \left( {1,3} \right)$$
3
IIT-JEE 2005 Screening
MCQ (Single Correct Answer)
+2
-0.5
In a triangle $$ABC$$, $$a,b,c$$ are the lengths of its sides and $$A,B,C$$ are the angles of triangle $$ABC$$. The correct relation is given by
A
$$\left( {b - c} \right)\sin \left( {{{B - C} \over 2}} \right) = a\cos {A \over 2}$$
B
$$\left( {b - c} \right)cos\left( {{A \over 2}} \right) = a\,sin{{B - C} \over 2}$$
C
$$\left( {b + c} \right)\sin \left( {{{B + C} \over 2}} \right) = a\cos {A \over 2}$$
D
$$\left( {b - c} \right)cos\left( {{A \over 2}} \right) = 2a\,sin{{B + C} \over 2}$$
4
IIT-JEE 2005 Screening
MCQ (Single Correct Answer)
+2
-0.5
If $$P(x)$$ is a polynomial of degree less than or equal to $$2$$ and $$S$$ is the set of all such polynomials so that $$P(0)=0$$, $$P(1)=1$$ and $$P'\left( x \right) > 0\,\,\forall x \in \left[ {0,1} \right],$$ then
A
$$S = \phi $$
B
$$S = ax + \left( {1 - a} \right){x^2}\,\,\forall \,a \in \left( {0,2} \right)$$
C
$$S = ax + \left( {1 - a} \right){x^2}\,\,\forall \,a \in \left( {0,\infty } \right)$$
D
$$S = ax + \left( {1 - a} \right){x^2}\,\,\forall \,a \in \left( {0,1} \right)$$
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