JEE Advance - Mathematics (1983)
- 2A vector $$\overrightarrow A $$ has components $${A_1},{A_2},{A_3}$$ in a right -handed rectangular Cartesian coordinate system $$oxyz.$$ The coordinate system is rotated about the $$x$$-axis through an angle $${\pi \over 2}.$$ Find the components of $$A$$ in the new coordinate system in terms of $${A_1},{A_2},{A_3}.$$Svare(C)A₁' = A₁, A₂' = A₃, A₃' = -A₂
- 7$$A, B, C$$ are events such that
$$P\left( A \right) = 0.3,P\left( B \right) = 0.4,P\left( C \right) = 0.8$$
$$P\left( {AB} \right) = 0.08,P\left( {AC} \right) = 0.28;\,\,P\left( {ABC} \right) = 0.09$$If $$P\left( {A \cup B \cup C} \right) \ge 0.75,$$ then show that $$P$$ $$(BC)$$ lies in the interval $$0.23 \le x \le 0.48$$
Svare(B)0.23 ≤ P(BC) ≤ 0.48 - 8Cards are drawn one by one at random from a well - shuffled full pack of $$52$$ playing cards until $$2$$ aces are obtained for the first time. If $$N$$ is the number of cards required to be drawn, then show that $${P_r}\left\{ {N = n} \right\} = {{\left( {n - 1} \right)\left( {52 - n} \right)\left( {51 - n} \right)} \over {50 \times 49 \times 17 \times 13}}$$ where $$2 \le n \le 50$$Svare(C)The probability that the nth card is an ace and exactly one of the first n-1 cards is an ace.
- 23m men and n women are to be seated in a row so that no two women sit together. If $$m > n$$, then show that the number of ways in which they can be seated is $$\,{{m!(m + 1)!} \over {(m - n + 1)!}}$$Svare(A)Arrange the men in m! ways, then choose n spaces from m+1 spaces and arrange women in n! ways, giving m! * (m+1)Cn * n! = m! * (m+1)! / (m-n+1)!
- 24Use mathematical Induction to prove : If $$n$$ is any odd positive integer, then $$n\left( {{n^2} - 1} \right)$$ is divisible by 24.Svare(A)Base Case: n=1, 1(1^2-1) = 0, divisible by 24. Inductive Step: Assume k(k^2-1) is divisible by 24 for odd k. Then (k+2)((k+2)^2-1) = (k+2)(k^2+4k+3) = k^3 + 6k^2 + 11k + 6 = (k^3-k) + 6k^2 + 12k + 6 = k(k^2-1) + 6(k^2 + 2k + 1) = k(k^2-1) + 6(k+1)^2. Since k is odd, k+1 is even, so (k+1)^2 is divisible by 4. Thus 6(k+1)^2 is divisible by 24. Therefore, (k+2)((k+2)^2-1) is divisible by 24.
- 25If $${\left( {1 + x} \right)^n} = {C_0} + {C_1}x + {C_2}{x^2} + ..... + {C_n}{x^n}$$ then show that the sum of the products of the $${C_i}s$$ taken two at a time, represented $$\sum\limits_{0 \le i < j \le n} {\sum {{C_i}{C_j}} } $$ is equal to $${2^{2n - 1}} - {{\left( {2n} \right)!} \over {2{{\left( {n!} \right)}^2}}}$$Svare(A)2^(2n-1) - (2n)! / (2 * (n!)^2)
- 27The vertices of a triangle are
$$\left[ {a{t_1}{t_2},\,\,a\left( {{t_1} + {t_2}} \right)} \right],\,\,\left[ {a{t_2}{t_3},a\left( {{t_2} + {t_3}} \right)} \right],\,\,\left[ {a{t_3}{t_1},\,a\left( {{t_3} + {t_1}} \right)} \right]$$. Find the orthocentre of the triangle.Svare(D)$$\left( { - a,a\left( {{t_1} + {t_2} + {t_3}} \right) + a{t_1}{t_2}{t_3}} \right)$$ - 44The function $$y = 2{x^2} - In\,\left| x \right|$$ is monotonically increasing for values of $$x\left( {x \ne 0} \right)$$ satisfying the inequalities ......... and monotonically decreasing for values of $$x$$ satisfying the inequalities ............Svare(C)$$x \in ( -\frac{1}{2}, 0 ) \cup ( \frac{1}{2}, \infty )$$; $$x \in ( -\infty, -\frac{1}{2} ) \cup ( 0, \frac{1}{2} )$$
- 55The end $$A, B$$ of a straight line segment of constant length $$c$$ slide upon the fixed rectangular axes $$OX, OY$$ respectively. If the rectangle $$OAPB$$ be completed, then show that the locus of the foot of the perpendicular drawn from $$P$$ to $$AB$$ is $${x^{{2 \over 3}}} + {y^{{2 \over 3}}} = {c^{{2 \over 3}}}$$Svare(A)The locus of the foot of the perpendicular drawn from P to AB is x^(2/3) + y^(2/3) = c^(2/3)
- 56The coordinates of $$A, B, C$$ are $$(6, 3), (-3, 5), (4, -2)$$ respectively, and $$P$$ is any point $$(x, y)$$. Show that the ratio of the area of the triangles $$\Delta $$ $$PBC$$ and $$\Delta $$$$ABC$$ is $$\left| {{{x + y - 2} \over 7}} \right|$$Svare(A)The problem asks to prove a relationship between the areas of two triangles, PBC and ABC, given the coordinates of their vertices.
