JEE Advance - Mathematics (1985)
- 2In a multiple-choice question there are four alternative answers, of which one or more are correct. A candidate will get marks in the question only if he ticks the correct answers. The candidate decides to tick the answers at random, If he is allowed upto three chances to answer the questions, find the probability that he will get marks in the questions.回答(A)1/5
- 3If $$\left| {\matrix{ a & {{a^2}} & {1 + {a^3}} \cr b & {{b^2}} & {1 + {b^3}} \cr c & {{c^2}} & {1 + {c^3}} \cr } } \right| = 0$$ and the vectors
$$\overrightarrow A = \left( {1,a,{a^2}} \right),\,\,\overrightarrow B = \left( {1,b,{b^2}} \right),\,\,\overrightarrow C = \left( {1,c,{c^2}} \right),$$ are non-coplannar, then the product $$abc=$$ .......回答(C)-1 - 4If $$\overrightarrow A \overrightarrow {\,B} \overrightarrow {\,C} $$ are three non-coplannar vectors, then -
$${{\overrightarrow A .\overrightarrow B \times \overrightarrow C } \over {\overrightarrow C \times \overrightarrow A .\overrightarrow B }} + {{\overrightarrow B .\overrightarrow A \times \overrightarrow C } \over {\overrightarrow C .\overrightarrow A \times \overrightarrow B }} = $$ ................回答(A)0 - 5If $$\overrightarrow A = \left( {1,1,1} \right),\,\,\overrightarrow C = \left( {0,1, - 1} \right)$$ are given vectors, then a vector $$B$$ satifying the equations $$\overrightarrow A \times \overrightarrow B = \overrightarrow {\,C} $$ and $$\overrightarrow A .\overrightarrow B = \overrightarrow {3\,} $$ ..........回答(A)$$\frac{5}{3}\widehat i + \frac{2}{3}\widehat j + \frac{2}{3}\widehat k$$
- 18Find the sum of the series : $$$\sum\limits_{r = 0}^n {{{\left( { - 1} \right)}^r}\,{}^n{C_r}\left[ {{1 \over {{2^r}}} + {{{3^r}} \over {{2^{2r}}}} + {{{7^r}} \over {{2^{3r}}}} + {{{{15}^r}} \over {{2^{4r}}}}..........up\,\,to\,\,m\,\,terms} \right]} $$$回答(B){{{2^{mn}} - 1} \over {{2^{mn}}left( {{2^n} - 1} ight)}}
- 26If $${f_r}\left( x \right),{g_r}\left( x \right),{h_r}\left( x \right),r = 1,2,3$$ are polynomials in $$x$$ such that $${f_r}\left( a \right) = {g_r}\left( a \right) = {h_r}\left( a \right),r = 1,2,3$$
and $$F\left( x \right) = \left| {\matrix{ {{f_1}\left( x \right)} & {{f_2}\left( x \right)} & {{f_3}\left( x \right)} \cr {{g_1}\left( x \right)} & {{g_2}\left( x \right)} & {{g_3}\left( x \right)} \cr {{h_1}\left( x \right)} & {{h_2}\left( x \right)} & {{h_3}\left( x \right)} \cr } } \right|$$ then $$F'\left( x \right)$$ at $$x = a$$ is ...........回答(B)$$0$$ - 30A ladder rests against a wall at an angle $$\alpha $$ to the horizintal. Its foot is pulled away from the wall through a distance $$a$$, so that it slides $$a$$ distance $$b$$ down the wall making an angle $$\beta $$ with the horizontal. Show that $$a = b\tan {1 \over 2}\left( {\alpha + \beta } \right)$$回答(E)There is no explanation provided.
- 33Let $$f\left( x \right) = {\sin ^3}x + \lambda {\sin ^2}x, - {\pi \over 2} < x < {\pi \over 2}.$$ Find the intervals in which $$\lambda $$ should lie in order that $$f(x)$$ has exactly one minimum and exactly one maximum.回答(B)\(\lambda \in \left( { - {3 \over 2},0} \right) \cup \left( {0,{3 \over 2}} \right)\)
