Maths-
General
Easy

Question

L t subscript left parenthesis n rightwards arrow infinity right parenthesis invisible function application sum subscript left parenthesis n equals 1 right parenthesis to the power of n   left square bracket 1 divided by left parenthesis left parenthesis 2 n plus 1 right parenthesis left parenthesis 2 n plus 3 right parenthesis right parenthesis right square bracket

  1. 6
  2. 1
  3. 1 divided by 6
  4. 0

The correct answer is: 1 divided by 6

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L t subscript left parenthesis n rightwards arrow infinity right parenthesis invisible function application left parenthesis 1 cubed plus 2 cubed plus 3 cubed plus midline horizontal ellipsis. plus n cubed right parenthesis divided by left parenthesis n squared left parenthesis n squared plus 1 right parenthesis right parenthesis

L t subscript left parenthesis n rightwards arrow infinity right parenthesis invisible function application left parenthesis 1 cubed plus 2 cubed plus 3 cubed plus midline horizontal ellipsis. plus n cubed right parenthesis divided by left parenthesis n squared left parenthesis n squared plus 1 right parenthesis right parenthesis

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Let PQ and RS be tangents at the extremities of the diameter ‘PR’ of a circle of radius ‘r’. If PS and RQ intersect at a point ‘X’ on the circumference of the circle, then 2r equals :

Let PQ and RS be tangents at the extremities of the diameter ‘PR’ of a circle of radius ‘r’. If PS and RQ intersect at a point ‘X’ on the circumference of the circle, then 2r equals :

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L t subscript left parenthesis x rightwards arrow 0 right parenthesis invisible function application left parenthesis 6 to the power of x minus 3 to the power of x minus 2 to the power of x plus 1 right parenthesis divided by x squared

For such questions, we should be know different formulas of limit.

L t subscript left parenthesis x rightwards arrow 0 right parenthesis invisible function application left parenthesis 6 to the power of x minus 3 to the power of x minus 2 to the power of x plus 1 right parenthesis divided by x squared

Maths-General

For such questions, we should be know different formulas of limit.

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If 5 x minus 12 y plus 10 equals 0 blankand 12 y minus 5 x plus 16 equals 0 blankare two tangents to a circles then radius of the circle is

If 5 x minus 12 y plus 10 equals 0 blankand 12 y minus 5 x plus 16 equals 0 blankare two tangents to a circles then radius of the circle is

Maths-General
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The locus of center of a circle which passes through the origin and cuts off a length of 4 units from the lineblank x equals 3 blankis:

The locus of center of a circle which passes through the origin and cuts off a length of 4 units from the lineblank x equals 3 blankis:

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AB is a chord of the circle x to the power of 2 end exponent plus y to the power of 2 end exponent minus 7 x minus 4 equals 0. If (1, -1) is the mid point of the chord AB then the area of the triangle formed by AB and the coordinate axes is

AB is a chord of the circle x to the power of 2 end exponent plus y to the power of 2 end exponent minus 7 x minus 4 equals 0. If (1, -1) is the mid point of the chord AB then the area of the triangle formed by AB and the coordinate axes is

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The equation of a plane that passes through (1,2,3) and is at maximum distance from (-1,1,1) is

The equation of a plane that passes through (1,2,3) and is at maximum distance from (-1,1,1) is

Maths-General
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If the four faces of a tetrahedron are represented by the equations r with minus on top left parenthesis alpha i with minus on top plus beta j with minus on top right parenthesis equals 0 comma r with minus on top left parenthesis beta j with rightwards arrow on top plus gamma k with minus on top right parenthesis equals 0 comma r with minus on top left parenthesis gamma k with minus on top plus alpha i with rightwards arrow on top right parenthesis equals 0 and r with minus on top times left parenthesis alpha i with rightwards arrow on top plus beta j with minus on top plus gamma k with minus on top right parenthesis equals P then volume of the tetrahedron (in cubic units) is

If the four faces of a tetrahedron are represented by the equations r with minus on top left parenthesis alpha i with minus on top plus beta j with minus on top right parenthesis equals 0 comma r with minus on top left parenthesis beta j with rightwards arrow on top plus gamma k with minus on top right parenthesis equals 0 comma r with minus on top left parenthesis gamma k with minus on top plus alpha i with rightwards arrow on top right parenthesis equals 0 and r with minus on top times left parenthesis alpha i with rightwards arrow on top plus beta j with minus on top plus gamma k with minus on top right parenthesis equals P then volume of the tetrahedron (in cubic units) is

Maths-General
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If z subscript 1 comma z subscript 2 comma z subscript 3 are distinct non-zero complex numbers and a comma b comma c element of R to the power of plus end exponentsuch that fraction numerator a over denominator open vertical bar z subscript 1 end subscript minus z subscript 2 end subscript close vertical bar end fraction equals fraction numerator b over denominator open vertical bar z subscript 2 end subscript minus z subscript 3 end subscript close vertical bar end fraction equals fraction numerator c over denominator open vertical bar z subscript 3 end subscript minus z subscript 1 end subscript close vertical bar end fraction then fraction numerator a squared over denominator z subscript 1 minus z subscript 2 end fraction plus fraction numerator b squared over denominator z subscript 2 minus z subscript 3 end fraction plus fraction numerator c squared over denominator z subscript 3 minus z subscript 1 end fraction is always equal to

If z subscript 1 comma z subscript 2 comma z subscript 3 are distinct non-zero complex numbers and a comma b comma c element of R to the power of plus end exponentsuch that fraction numerator a over denominator open vertical bar z subscript 1 end subscript minus z subscript 2 end subscript close vertical bar end fraction equals fraction numerator b over denominator open vertical bar z subscript 2 end subscript minus z subscript 3 end subscript close vertical bar end fraction equals fraction numerator c over denominator open vertical bar z subscript 3 end subscript minus z subscript 1 end subscript close vertical bar end fraction then fraction numerator a squared over denominator z subscript 1 minus z subscript 2 end fraction plus fraction numerator b squared over denominator z subscript 2 minus z subscript 3 end fraction plus fraction numerator c squared over denominator z subscript 3 minus z subscript 1 end fraction is always equal to

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General
Chemistry-

6 CO subscript 2 plus 12 straight H subscript 2 straight O not stretchy ⟶ with text  Sun light  end text on top straight C subscript 6 straight H subscript 12 straight O subscript 6 plus 6 straight O subscript 2 plus 6 straight H subscript 2 straight O Equivalent weights of  and straight C subscript 6 straight H subscript 12 straight O subscript 6
respectively are

6 CO subscript 2 plus 12 straight H subscript 2 straight O not stretchy ⟶ with text  Sun light  end text on top straight C subscript 6 straight H subscript 12 straight O subscript 6 plus 6 straight O subscript 2 plus 6 straight H subscript 2 straight O Equivalent weights of  and straight C subscript 6 straight H subscript 12 straight O subscript 6
respectively are

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Let a with minus on top equals i with minus on top plus j with minus on top plus k with minus on top comma b with minus on top equals negative i with minus on top plus j with minus on top plus k with minus on top comma c with minus on top equals i with minus on top minus j with minus on top plus k with minus on top and d with minus on top equals i with minus on top plus j with minus on top minus k with minus on top. Then, the line of intersection of planes one determined by stack a with ‾ on top comma stack b with ‾ on top and other determined by stack c with ‾ on top comma stack d with ‾ on top is perpendicular to

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A non - zero vector stack a with rightwards arrow on top is parallel to the line of intersection of the plane P subscript 1 end subscript determined by i with ˆ on top plus j with ˆ on top and i plus 2 j with ˆ on top and plane P subscript 2 end subscript determined by vector 2 i with ˆ on top minus j with ˆ on top and 3 i with ˆ on top plus 2 k with ˆ on top, then angle between stack a with rightwards arrow on top and vector i with ˆ on top minus 2 j with ˆ on top plus 2 k with ˆ on top is

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Let a,b,c be distinct non - negative numbers and the vectors a i with ˆ on top blank plus a j with ˆ on top plus c k with ˆ on top comma i with ˆ on top plus k with ˆ on top comma i with ˆ on top plus j with ˆ on top plus b k with ˆ on top comma lie in a plane, then the quadratic equation a x to the power of 2 end exponent plus 2 c x plus b equals 0 has

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If stack a with rightwards arrow on top comma stack b with rightwards arrow on top and stack c with rightwards arrow on top are anythree vectors forming a linearly independent system then for all theta element of R open square brackets stack a with minus on top c o s invisible function application theta plus stack b with rightwards arrow on top s i n invisible function application theta plus stack c blank with rightwards arrow on top c o s invisible function application 2 theta comma stack a with minus on top c o s invisible function application open parentheses fraction numerator 2 pi over denominator 3 end fraction plus theta close parentheses plus stack b with minus on top s i n invisible function application open parentheses fraction numerator 2 pi over denominator 3 end fraction plus theta close parentheses plus stack c with rightwards arrow on top c o s invisible function application 2 open parentheses fraction numerator 2 pi over denominator 3 end fraction plus theta close parentheses close open stack a with ‾ on top c o s invisible function application open parentheses theta minus fraction numerator 2 pi over denominator 3 end fraction close parentheses plus stack b with ‾ on top s i n invisible function application open parentheses theta minus fraction numerator 2 pi over denominator 3 end fraction close parentheses plus stack c with rightwards arrow on top c o s invisible function application 2 open parentheses theta minus fraction numerator 2 pi over denominator 3 end fraction close parentheses close square brackets equals

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Maths-General
General
Chemistry-

Select the correct statement

Select the correct statement

Chemistry-General
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