IIT-JEE 2006 | JEE Advanced Year Wise Previous Years Questions

IIT-JEE 2006

Numerical

-0

If $a_n=\frac{3}{4}-\left(\frac{3}{4}\right)^2+\left(\frac{3}{4}\right)^3+\cdots \cdots(-1)^{n-1}\left(\frac{3}{4}\right)^n$ and $b_n=1-a_n$, then find the minimum natural number $n_0$ such that $b_n>a_n \forall n>n_0$

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IIT-JEE 2006

Numerical

-0

If $f(x)$ is a twice differentiable function such that $f(A)=0, f(B)=2, f(C)=-1, f(D)=2$, $f(e)=0$, where $a < b < c < d < e$, then the minimum number of zeroes of $g(x)=\left(f^{\prime}(x)\right)^2 +f^{\prime \prime}(x) f(x)$ in the interval $[a, e]$ is :

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IIT-JEE 2006

MCQ (Single Correct Answer)

-0

$$ \text { Normals are drawn at points } \mathrm{P}, \mathrm{Q} \text { and } \mathrm{R} \text { lying on the parabola } y^2=4 x \text { which intersect at }(3,0) \text {. Then } $$

(i)	Area of $\triangle \mathrm{PQR}$	(A)	2
(ii)	Radius of circumcircle of $\triangle \mathrm{PQR}$	(B)	5/2
(iii)	Centroid of $\triangle \mathrm{PQR}$	(C)	(5/2,0)
(iv)	Circumcentre of $\triangle \mathrm{PQR}$	(D)	(2/3,0)

$$ \begin{aligned} & \text { (i)-(A); (ii)-(B); (iii)-(D); } \text { (iv)-(C) } \end{aligned} $$

$$ \begin{aligned} & \text { (i)-(B); (ii)-(A); (iii)-(D); } \text { (iv)-(C) } \end{aligned} $$

$$ \begin{aligned} & \text { (i)-(A); (ii)-(B); (iii)-(C); } \text { (iv)-(D) } \end{aligned} $$

$$ \begin{aligned} & \text { (i)-(A); (ii)-(D); (iii)-(B); } \text { (iv)-(C) } \end{aligned} $$

IIT-JEE 2006

MCQ (Single Correct Answer)

-0

$$ \text { Match the following : } $$

(i)	$$ \int_0^{\pi / 2}(\sin x)^{\cos x}\left(\cos x \cot x-\log \left(\sin ^x\right)^{\sin } x\right) \mathrm{d} x $$	(A)	1
(ii)	$$ \text { Area bounded by }-4 y^2=x \text { and } x-1=-5 y^2 $$	(B)	0
(iii)	Cosine of the angle of intersection of $y=3^{x-1} \log x$ and $y=x^{x-1}$ is	(C)	6 In 2
(iv)	$$ \frac{d y}{d x}=\frac{2}{(x+y)} ; y\left(-\frac{2}{3}\right)=0 \text {, then value of constant }(\mathrm{k})= $$	(D)	4/3

$$ \begin{aligned} & \text { (i)-(A); (ii)-(D); (iii)-(B); }\text { (iv)-(D) } \end{aligned} $$

$$ \begin{aligned} & \text { (i)-(A); (ii)-(C); (iii)-(B); }\text { (iv)-(D) } \end{aligned} $$

$$ \begin{aligned} & \text { (i)-(A); (ii)-(D); (iii)-(A); }\text { (iv)-(D) } \end{aligned} $$

$$ \begin{aligned} & \text { (i)-(A); (ii)-(B); (iii)-(C); }\text { (iv)-(D) } \end{aligned} $$

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