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JEE Advanced

The hyperconjugative stabilities of tert-butyl cation and 2-butene, respectively, are due to

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Let $O$be the origin and let PQR be an arbitrary triangle. The point S is such that$OPO˙Q+ORO˙S=ORO˙P+OQO˙S=OQ$.$OR+OPO˙S$Then the triangle PQ has S as its:circumcentre (b) orthocentre (c) incentre (d) centroid

Let $f:[−21 ,2]→R$ and $g:[−21 ,2]→R$ be functions defined by $f(x)=[x_{2}−3]$ and $g(x)=∣x∣f(x)+∣4x−7∣f(x)$, where [y] denotes the greatest integer less than or equal to y for $y∈R$. Then,

Six cards and six envelopes are numbered 1, 2, 3, 4, 5, 6 and cards are to be placed in envelopes so that each envelope contains exactly one card and no card is placed in the envelope bearing the same number and moreover cards numbered 1 is always placed in envelope numbered 2. Then the number of ways it can be done isa.$264$ b. $265$ c. $53$ d. $67$

Let $a$ and $b$ be two unit vectors such that $a.b=0$ For some $x,y∈R$, let $c=xa+yb+(a×b$ If $(∣c∣=2$ and the vector $c$ is inclined at same angle $α$ to both $a$ and $b$ then the value of $8cos_{2}α$ is

Consider the cube in the first octant with sides OP,OQ and OR of length 1, along the x-axis, y-axis and z-axis, respectively, where $O(0,0,0)$ is the origin. Let $S(21 ,21 ,21 )$ be the centre of the cube and T be the vertex of the cube opposite to the origin O such that S lies on the diagonal OT. If $p =SP,q =SQ ,r=SR$ and $t=ST$ then the value of $∣(p ×q )×(r×(t)∣is$

The following integral $∫_{4π}(2cosecx)_{17}dx$is equal to$(a)∫_{0}2(e_{u}+e_{−u})_{16}du$$(b)∫_{0}2(e_{u}+e_{−u})_{17}du$$(c)∫_{0}2(e_{u}−e_{−u})_{17}du$$(d)∫_{0}2(e_{u}−e_{−u})_{16}du$

Let $X$be the set consisting of the first 2018 terms of the arithmetic progression $1,6,11,,¨ $and $Y$be the set consisting of the first 2018 terms of the arithmetic progression $9,16,23,¨$. Then, the number of elements in the set $X∪Y$is _____.

Let w = ($3 +2ι )$ and $P={w_{n}:n=1,2,3,…..},$ Further $H_{1}={z∈C:Re(z)>21 }andH_{2}={z∈c:Re(z)<−21 }$ Where C is set of all complex numbers. If $z_{1}∈P∩H_{1},z_{2}∈P∩H_{2}$ and O represent the origin, then $∠Z_{1}OZ_{2}$ =