JEE Advance - Mathematics (2004)
- 2$${P_1}$$ and $${P_2}$$ are planes passing through origin. $${L_1}$$ and $${L_2}$$ are two line on $${P_1}$$ and $${P_2}$$ respectively such that their intersection is origin. Show that there exists points $$A, B, C,$$ whose permutation $$A',B',C'$$ can be chosen such that (i) $$A$$ is on $${L_1},$$ $$B$$ on $${P_1}$$ but not on $${L_1}$$ and $$C$$ not on $${P_1}$$ (ii) $$A'$$ is on $${L_2},$$ $$B'$$ on $${P_2}$$ but not on $${L_2}$$ and $$C'$$ not on $${P_2}$$回答(A)The statement is always true and can be proved using linear independence and suitable vector choices.
- 3If $$\overrightarrow a ,\overrightarrow b ,\overrightarrow c $$ and $$\overrightarrow d $$ are distinct vectors such that
$$\,\overrightarrow a \times \overrightarrow c = \overrightarrow b \times \overrightarrow d $$ and $$\overrightarrow a \times \overrightarrow b = \overrightarrow c \times \overrightarrow d \,.$$ Prove that
$$\left( {\overrightarrow a - \overrightarrow d } \right).\left( {\overrightarrow b - \overrightarrow c } \right) \ne 0\,\,i.e.\,\,\,\overrightarrow a .\overrightarrow b + \overrightarrow d .\overrightarrow c \ne \overrightarrow d .\overrightarrow b + \overrightarrow a .\overrightarrow c $$回答BE - 5A parallelopiped $$'S'$$ has base points $$A, B, C$$ and $$D$$ and upper face points $$A',$$ $$B',$$ $$C'$$ and $$D'.$$ This parallelopiped is compressed by upper face $$A'B'C'D'$$ to form a new parallelopiped $$'T'$$ having upper face points $$A'',B'',C''$$ and $$D''.$$ Volume of parallelopiped $$T$$ is $$90$$ percent of the volume of parallelopiped $$S.$$ Prove that the locus of $$'A''',$$ is a plane.回答(B)The locus of A'' is a plane parallel to the base ABCD.
- 6A box contains $$12$$ red and $$6$$ white balls. Balls are drawn from the box one at a time without replacement. If in $$6$$ draws there are at least $$4$$ white balls, find the probability that exactly one white is drawn in the next two draws. (binomial coefficients can be left as such)回答(E)$$(\frac{\binom{6}{4}\binom{12}{2} + \binom{6}{5}\binom{12}{1} + \binom{6}{6}\binom{12}{0}}{\binom{18}{6}}) \times (\frac{\binom{2}{1} \binom{10}{1}}{\binom{12}{2}})$$
- 7The unit vector which is orthogonal to the vector $$3\overrightarrow i + 2\overrightarrow j + 6\overrightarrow k $$ and is coplanar with the vectors $$\,2\widehat i + \widehat j + \widehat k$$ and $$\,\widehat i - \widehat j + \widehat k$$$$\,\,\,$$ is回答(C)$${{3\widehat i - \widehat k} \over {\sqrt {10} }}$$
- 8A curve $$'C''$$ passes through $$(2,0)$$ and the slope at $$(x,y|)$$ as $$\,{{{{\left( {x + 1} \right)}^2} + \left( {y - 3} \right)} \over {x + 3}}$$. Find the equation of the curve. Find the area bounded by curve and $$x$$-axis in fourth quadrant.回答(B)Area = 4/3 sq. units, Equation: y = (x+1)^2 + 3ln|x+3| - 1 - 3ln5
- 12Using Rolle's theorem, prove that there is at least one root
in $$\left( {{{45}^{1/100}},46} \right)$$ of the polynomial
$$P\left( x \right) = 51{x^{101}} - 2323{\left( x \right)^{100}} - 45x + 1035$$.回答(C)P(45^(1/100)) = P(46) = 0, hence by Rolle's theorem, there exists at least one root in (45^(1/100), 46).
