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5451.

If 3 + 4i is a root of x^(2) + px + q = 0, where p, q in R , then (1)/(13) (2p + q)is equal to ________

Answer»


ANSWER :`1 . 00`
Discussion
5452.

Given P(x)=x^(4)+ax^(3)+ + bx^(2)+cx+d such that x =0 is the only real root of p'' (x)=0. If P(-1) lt P(1), then in the interval [-1, 1] :

Answer»

<P>P(-1) is the minimum and P(1) is the MAXIMUM of P
P(-1) is not the minimum and P(1) is the maximum of P
P(-1) is the minimum but P(1) is not the maximum of P
NEITHER `P(-1)` is the minimum nor P(1) is the maximum of P.

Answer :B
Discussion
5453.

I : If a, b are two points then r = (1 - s)a + sb represents a line. II : If a, b, c are three noncollinear points then r = (1 - s - t)a + sb + tc represents a plane

Answer»

only I is TRUE
only II is true
both I and II are true
NEITHER I nor II are true

Answer :C
Discussion
5454.

Find (dy)/(dx) ifx-y=pi

Answer»

Solution :`x-y=pi` Differentiating through outw.r.t.x we GET `1-(DY)/(dx)=0iff(dy)/(dx)=1` OR `x-y=piy=x-pi(dy)/(dx)=1`
Discussion
5455.

Evaluate the following integrals int_0^(pi/4)tan^2xdx

Answer»

SOLUTION :`int_0^(pi/4)TAN^2xdx=int_0^(pi/4)(sec^2 x-1)DX`
`[tanx-x]_o^(pi/4)`
`tan(pi/4)-(pi/4)=(4-pi)/4`
Discussion
5456.

If |vec(a)|=3,|vec(b)|=4and|vec(c)|=5, then

Answer»

RANGE of `|VEC(a)-vec(B)|" is "[1,7]`
range of `|vec(b)-vec(c)|" is "[1,9]`
range of `|vec(c)-vec(a)|" is "[2,8]`
NONE of these

Answer :A::B::C
Discussion
5457.

If all the letters of the word 'KRISHNA' are arranged in all possible ways and these words are written out as in a dictionary, then the rank of the word 'KRISHNA' is :

Answer»

324
341
1359
2710

Solution :
Discussion
5458.

If intx/((x-1)(2x-1))=int[A/(x-1)+B/(2x-1)]dx find A and B.

Answer»

SOLUTION :`INT(XDX)/((x-1)(2x-1))=int[1/(x-1)+1/(2x-1)]dxA=B=1`
Discussion
5459.

Transformations of inverse trigonometric functions need to be handled with care. Consider the identity sin2theta=(2tantheta)/(1+tan^2theta) in its domain of definition. Suppose we set tantheta=x, we have sin2theta=(2x)/(1+x^2) Taking sin^(-1)of both sides yields 2theta=sin^(-1).(2x)/1+x^2i.e.,2tan^(-1)x=sin^(-1).(2x)/(1+ x^2). But we will discover that the above identity is not valid for all x. Choose x=sqrt3, LHS =2tan^(-1)sqrt3=2xxpi/3=(2pi)/3, RHS=sin.(2sqrt3)/(1+3)=sin^(-1).(sqrt3)/2=pi/3. And so left hand and right hand side don't match. The reason is that we have disregarded the principal values of inverse functions. So it is well to remember that the iverse trigonometric formmulae have restrictions attached to the argument. When the values ofx lie outside the interval of validity then the formula needs to be corrected. tan^(-1). (pi/2)equals

Answer»

`pi+sin^(-1)((4PI)/(4+pi^2))`
`pi/2-1/2sin^(-1)((4pi)/(4+pi^2))`
`pi/2+1/2sin^(-1)((4pi)/(4+pi^2))`
`-pi+sin^(-1)((4pi)/(4+pi^2))`

ANSWER :B
Discussion
5460.

Transformations of inverse trigonometric functions need to be handled with care. Consider the identity sin2theta=(2tantheta)/(1+tan^2theta) in its domain of definition. Suppose we set tantheta=x, we have sin2theta=(2x)/(1+x^2) Taking sin^(-1)of both sides yields 2theta=sin^(-1).(2x)/1+x^2i.e.,2tan^(-1)x=sin^(-1).(2x)/(1+ x^2). But we will discover that the above identity is not valid for all x. Choose x=sqrt3, LHS =2tan^(-1)sqrt3=2xxpi/3=(2pi)/3, RHS=sin.(2sqrt3)/(1+3)=sin^(-1).(sqrt3)/2=pi/3. And so left hand and right hand side don't match. The reason is that we have disregarded the principal values of inverse functions. So it is well to remember that the iverse trigonometric formmulae have restrictions attached to the argument. When the values ofx lie outside the interval of validity then the formula needs to be corrected. Let f(x)=sin^(-1)/(2x)/(1+x^2),g(x)=2tan^(-1)x. Then the largest interval in R on which f and g both are agree

Answer»

`[-1,1]`
`(-OO,-1]`
`[1,oo)`
`(-oo,1)`

ANSWER :A
Discussion
5461.

Transformations of inverse trigonometric functions need to be handled with care. Consider the identity sin2theta=(2tantheta)/(1+tan^2theta) in its domain of definition. Suppose we set tantheta=x, we have sin2theta=(2x)/(1+x^2) Taking sin^(-1)of both sides yields 2theta=sin^(-1).(2x)/1+x^2i.e.,2tan^(-1)x=sin^(-1).(2x)/(1+ x^2). But we will discover that the above identity is not valid for all x. Choose x=sqrt3, LHS =2tan^(-1)sqrt3=2xxpi/3=(2pi)/3, RHS=sin.(2sqrt3)/(1+3)=sin^(-1).(sqrt3)/2=pi/3. And so left hand and right hand side don't match. The reason is that we have disregarded the principal values of inverse functions. So it is well to remember that the iverse trigonometric formmulae have restrictions attached to the argument. When the values ofx lie outside the interval of validity then the formula needs to be corrected. If x in (-oo,-1), then which of the relation holds?

Answer»

`2TAN^(-1)x=sin^(-1).(2X)/(1+x^2)-pi/2`
`2tan^(-1)x=-pi-sin^(-1).(2x)/(1+x^2)`
`2tan^(-1)x=pi-sin^(-1).(2x)/(1+x^2)`
`2tan^(-1)x=pi/2+sin^(-1).(2x)/(1+x^2)`

ANSWER :B
Discussion
5462.

Ifthe roots ofx^3 - 9x^2+kx +l=0arein A.Pwithcommondifference2 then(k,l) =

Answer»

`(15-15)`
`(23,-15)`
`(15,-23)`
`(-15 ,23)`

ANSWER :B
Discussion
5463.

The orthogonal trajectories of the family of semicubical parabola is given by

Answer»

`X^(2)+ 3Y^(2) =C^(2)`
`3x^(2) + y^(2) =c^(2)`
`x+3y^(2)=c^(2)`
`3y^(2) + 2x^(2)=c^(2)`

ANSWER :D
Discussion
5464.

Write minors and co-factors of the elements of following determinants:|[a, c],[b,d]|

Answer»

SOLUTION :`|[a, C],[B,d]|`
MINORS: `M_11=d, M_12=b`,
`M_21=c` and `M_22=a`
Co-factors: `A_11=d, A_12=-b, A_21=-c`,
`A_22=a`
Discussion
5465.

If int (x^(2010)+x^(804)+x^(402)) ...............

Answer»


Solution :`INT x(x^(2009)+x^(803)+x^(401))(2x^(1608)+5x^(402)+10)^(1//402) dx`
Put `2x^(2010)+5x^(804)+10x^(402)=t`,
`4020 (x^(2009)+x^(803) +x^(401))dx=DT`
`(x^(2009)+x^(803)+x^(401))dx=(dt)/4020`,
`=1/4020 int t^(1//402) dt =1/4020 (t^(1/402+1)/(1/402 +1))`
`=1/4020xx402/403 t^(403/402)`
`a=403`
Discussion
5466.

Food X contains 6 units of vitamin A per gm and 7 units of vitamin B per gm and costs ₹ 2.00 per gm.Food Y contains 8 units of Vitamin A per gm and 12 units of vitamin B per gm and costs ₹ 2.50 per gm.The daily minimum requirement of vitamin A and vitamin V are 100 units and 120 units respectively.Formulate the above as a linear programming problem to minimize the cost.

Answer»


ANSWER :Minimum COST = ₹ 31.25 when 12.5 GMS of FOOD Y is used.
Discussion
5467.

For any two real numbers, an operation * defined by a *b = 1 + ab is

Answer»

NEITHER COMMUTATIVE nor ASSOCIATIVE
Commutative but not associative
both commutative and associative
Associative but not commutative

ANSWER :B
Discussion
5468.

If int (dx)/((x^(2)+a^(2))^(3))=(x)/(4a^(2)(x^(2)+a^(2)))+(m)/(na^(2)){(x)/(2a^(2)(x^(2)+a^(2)))+(1)/(2a^(3))tan^(-1)(x/a)}+C. Then |m-n| is equal to

Answer»

4
3
2
1

Answer :D
Discussion
5469.

If a!=p,b!=q,c!=r and the system of equations px+by+cz=0 ax+qy+cz=0 ax+by+rz=0 has non zero solution, then value of (p+a)/(p-a)+(q+b)/(q-b)+(r+c)/(r-c) is

Answer»

2
`-3`
`1`
1

Answer :D
Discussion
5470.

Evaluate the following integrals int_0^(pi/2)cos2xdx

Answer»

SOLUTION :`int_0^(pi/4)COS2X DS =1/2[SIN2X]_0^(pi/4)=1/2`
Discussion
5471.

Determine order and Degree(if defined) of differential equations given (d^(2)y)/(dx^(2)) = cos3x + sin 3x

Answer»


ANSWER :ORDER 2; DEGREE 1
Discussion
5472.

Prove that |{:(a+bx,c+dx,p+qx),(ax+b,cx+d,px+q),(mu,vartheta,omega):}|=(1-x^2)|{:(a,c,p),(b,d,q),(mu,vartheta,omega):}|

Answer»


ANSWER :`(1-x^2)|{:(a,C,p),(B,d,Q),(U,v,w):}|`
Discussion
5473.

Which of the following curves is concave down ?

Answer»

`y=-X^(2)`
`y=x^(2)`
`y=e^(x)`
`y=x^(2)+2x-3`

ANSWER :B
Discussion
5474.

Translate" If the boy is poor, then he will be hungry and if he is hungry, then he cannot be honest" propositions into symbolic form, stating the prime components

Answer»

<P>

Solution :Let p: The BOY is poor.
q :He will be HUNGRY.
R: He can be honest.
`:.` Answer is (prereq) ^^(qaraar~~r)
Discussion
5475.

In the interval [-2, 4], the absolute maximum of f(x) = 2x^(3) - 3x^(2) - 12x + 5 occurs x =

Answer»

A4
B`-2`
C`-1`
D2

Answer :A
Discussion
5476.

PQ is the double ordinate of the parabola y^(2)=4x which passes through the focus S. DeltaPQA is an isosceles right angle triangle, where A is on the axis of the parabola to the right of focus. Line PA meets the parabola at C and QA meets the parabola at B. The circumradius of trapezium PBCQ is

Answer»

<P>`6sqrt(5)`
`3sqrt(6)`
`sqrt(10)`
`5sqrt(3)`

Solution :(3)
For `y^(2)=4x`, the coordinates of the ends of latus rectum are P(1,2) and Q(1,-2).

`DeltaPAQ` is isosceles right - angled. Therefore, the slope of PA is -1 and its equation is y-2=-(x-1), i.e., x+y-3=0.
Similarly, the equation of line QB is x-y-3=0.
Solving x+y-3=0 with the parabola `y^(2)=4x`, we have
`(3-x)^(2)=4x,i.e.,x^(2)-10x+9=0`
`:.x=1,9`
Therefore, the coordinates of B and C are (9,-6) and (9,6), respectively.
AREA of trapezium `PBCQ=(1)/(2)xx(12+4)xx8=64` SQ. units
Let the circumcenter of trapezium PBCQ be T(h,0). Then,
PT=BT
`orsqrt((h-1)^(2)+4)=sqrt((h-9)^(2)+36)`
`or-2h+=-18h+81+36`
`or16h=112`
`orh=7`
Hence, radius is `sqrt(40)=2sqrt(10`.
Let the inradius of `DeltaAPQ` be `r_(1)`. Then,
`r_(1)=(Delta_(1))/(s_(1))`
`((1)/(2)xx4xx2)/((1)/(2)(4+2sqrt(4+4)))`
`=(2)/(1+sqrt(2))=2(sqrt(2)-1)`
Let the inradius of `DELTAABC` be `r_(2)`. Then,
`r_(2)=(Delta_(2))/(s_(2))`
`=((1)/2xx12xx6)/((1)/(2)(12+2sqrt(36+36)))`
`=(6)/(1+sqrt(2))=6(sqrt(2)-1)`
`:." "(r_(2))/(r_(1))=3`
Discussion
5477.

PQ is the double ordinate of the parabola y^(2)=4x which passes through the focus S. DeltaPQA is an isosceles right angle triangle, where A is on the axis of the parabola to the right of focus. Line PA meets the parabola at C and QA meets the parabola at B. The of trapezium PBCQ is

Answer»

96 sq. UNITS
64sq. units
72sq. units
48sq. units

Solution :(2)
For `y^(2)=4X`, the coordinates of the ends of latus rectum are P(1,2) and Q(1,-2).

`DeltaPAQ` is isosceles RIGHT - angled. Therefore, the slope of PA is -1 and its equation is y-2=-(x-1), i.e., x+y-3=0.
Similarly, the equation of line QB is x-y-3=0.
Solving x+y-3=0 with the parabola `y^(2)=4x`, we have
`(3-x)^(2)=4x,i.e.,x^(2)-10x+9=0`
`:.x=1,9`
Therefore, the coordinates of B and C are (9,-6) and (9,6), respectively.
Area of trapezium `PBCQ=(1)/(2)xx(12+4)xx8=64` sq. units
Let the circumcenter of trapezium PBCQ be T(h,0). Then,
PT=BT
`orsqrt((h-1)^(2)+4)=sqrt((h-9)^(2)+36)`
`or-2h+=-18h+81+36`
`or16h=112`
`orh=7`
Hence, radius is `sqrt(40)=2sqrt(10`.
Let the INRADIUS of `DeltaAPQ` be `r_(1)`. Then,
`r_(1)=(Delta_(1))/(s_(1))`
`((1)/(2)xx4xx2)/((1)/(2)(4+2sqrt(4+4)))`
`=(2)/(1+sqrt(2))=2(sqrt(2)-1)`
Let the inradius of `DeltaABC` be `r_(2)`. Then,
`r_(2)=(Delta_(2))/(s_(2))`
`=((1)/2xx12xx6)/((1)/(2)(12+2sqrt(36+36)))`
`=(6)/(1+sqrt(2))=6(sqrt(2)-1)`
`:." "(r_(2))/(r_(1))=3`
Discussion
5478.

PQ is the double ordinate of the parabola y^(2)=4x which passes through the focus S. DeltaPQA is an isosceles right angle triangle, where A is on the axis of the parabola to the right of focus. Line PA meets the parabola at C and QA meets the parabola at B.The ratio of the inradius of DeltaABCand that of DeltaPAQ is

Answer»

`2:1`
`3:2`
`4:3`
`3:1`

Solution :(4)
For `y^(2)=4x`, the coordinates of the ends of LATUS rectum are P(1,2) and Q(1,-2).

`DeltaPAQ` is isosceles right - angled. Therefore, the slope of PA is -1 and its EQUATION is y-2=-(x-1), i.e., x+y-3=0.
SIMILARLY, the equation of line QB is x-y-3=0.
Solving x+y-3=0 with the parabola `y^(2)=4x`, we have
`(3-x)^(2)=4x,i.e.,x^(2)-10x+9=0`
`:.x=1,9`
Therefore, the coordinates of B and C are (9,-6) and (9,6), respectively.
Area of trapezium `PBCQ=(1)/(2)xx(12+4)xx8=64` sq. units
Let the circumcenter of trapezium PBCQ be T(h,0). Then,
PT=BT
`orsqrt((h-1)^(2)+4)=sqrt((h-9)^(2)+36)`
`or-2h+=-18h+81+36`
`or16h=112`
`orh=7`
Hence, radius is `sqrt(40)=2sqrt(10`.
Let the inradius of `DeltaAPQ` be `r_(1)`. Then,
`r_(1)=(Delta_(1))/(s_(1))`
`((1)/(2)xx4xx2)/((1)/(2)(4+2sqrt(4+4)))`
`=(2)/(1+sqrt(2))=2(sqrt(2)-1)`
Let the inradius of `DeltaABC` be `r_(2)`. Then,
`r_(2)=(Delta_(2))/(s_(2))`
`=((1)/2xx12xx6)/((1)/(2)(12+2sqrt(36+36)))`
`=(6)/(1+sqrt(2))=6(sqrt(2)-1)`
`:." "(r_(2))/(r_(1))=3`
Discussion
5479.

A fair die is rolled. Consider the events E={1,3,5}, F = {2,3} and G={2,3,4,5}.Find P(E/F) and P(F/E)

Answer»

<P>

Solution :Here, S = {1,2,3,4,5,6}
`EnnF`=`QUAD`3
`P(EnnF)`=`(N(EnnF))/(n(S))`=1/6
Also,P(F)=2/6
THEREFORE P(E/F)=`(P(EnnF))/(P(F))`=1/6/2/6=1/2.
We have P (E)=3/6,
P(F/E)=`(P(EnnF))/(P(E))`
1/6/3/6=1/3
Discussion
5480.

Find the angle between the following pair of lines :(i) (x-2)/(2)=(y-1)/(5)=(z+3)/(-3) and (x+2)/(-1)=(y-4)/(8)=(z-5)/(4)(ii) x/2=y/2=z/1 and (x-5)/(4)=(y-2)/(1)=(z-3)/(8)

Answer»


ANSWER :(i) `THEREFORE theta=cos^(-1)((26)/(9sqrt(38))`
(ii) `therefore theta=cos^(-1)(2/3)`
Discussion
5481.

The negation of the statement ' 72 is divisible by 2 and 3' is

Answer»

72 is not DIVISIBLE by 2 or 72 is not divisible by 3
72 is divisible by 2 or 72 ISDIVISIBLE by 3
72 is divisible by 2 and72 isdivisible by 3
72 is not divisible by 2 and 3

Answer :A
Discussion
5482.

If f(x)=x^(2)-2xthen find f(A) where A=[{:(0,1,2),(4,5,0),(0,2,3):}].

Answer»


ANSWER :`F(A)=[{:(4,7,2),(12,19,8),(8,12,3):}]`
Discussion
5483.

Let f(x) = (x)/(sqrt(a^(2)+x^(2))=((d-x))/(sqrt(b^(2)+(d-x)^(2)) x in R wherea band d are non-zero real constant. Then

Answer»


ANSWER :B
Discussion
5484.

Let ABC be a triangle with verticesat points A( 2,3,5 ), B ( -1,3,2)and ( lambda , 5, mu )in three dimensional space. If the median through A is equally inclined with the axes , then(lambda , mu )is equal to

Answer»

`(0,7) `
`(7, 5) `
` (7, 10 )`
` (5, 7) `

ANSWER :A
Discussion
5485.

Show that the following number are equidistant from the origin :sqrt2 +i ,1 +isqrt2, isqrt3

Answer»

Solution :`|SQRT2+i|=sqrt((sqrt2)^2+1^2)=SQRT3`
`|1+isqrt2|=sqrt(1^2+(sqrt2)^2)=sqrt3"and"|isqrt3|=sqrt3`
`:.` The points are equidistant from the origin.
Discussion
5486.

On analysis, a certain compound was found to contain 254 g of iodine (at. Mass 127) and 80 g oxygen (at. Mass 16). What is the formula of the compound ?

Answer»

IO
`I_(2)O`
`I_(5)O_(3)`
`I_(2)O_(5)`

ANSWER :A
Discussion
5487.

Incircle of DeltaABC touches the sides BC, AC and AB at D, E and F, respectively. Then answer the following question Area of DeltaDEF is

Answer»

`2r^(2) sin(2A) sin(2B) sin(2C)`
`2r^(2) cos.(A)/(2) cos.(B)/(2) cos.(C)/(2)`
`2r^(2) sin (A -B) sin (B -C) sin(C -A)`
none of these

Solution :
Points I, D, C, E are concylic. Therefore,
`angleEID = pi -C`
Also, I is CIRCUMCENTER of `DeltaDEF`. Hence,
`angleDFE = (1)/(2) angleDIE = (pi -C)/(2)`
Similarly, `angleFDE = (pi -A)/(2)`
`angleFED = (pi -B)/(2)`
`:.` Area of `DeltaDEF = 2r^(2) sin (ANGLED) sin (angleE) sin (angleF)`
`= 2r^(2) "cos"(A)/(2) "cos"(B)/(2) "cos"(C)/(2)`
Also by sine rule, in `DeltaDEF`,
`(EF)/(sin angleD) = 2r`
or `EF = 2r cos.(A)/(2)`
Discussion
5488.

Incircle of DeltaABC touches the sides BC, AC and AB at D, E and F, respectively. Then answer the following question angleDEF is equal to

Answer»

`(pi -B)/(2)`
`pi - 2B`
`A -C`
none of these

Solution :
Points I, D, C, E are concylic. THEREFORE,
`angleEID = pi -C`
Also, I is circumcenter of `DeltaDEF`. Hence,
`angleDFE = (1)/(2) angleDIE = (pi -C)/(2)`
Similarly, `angleFDE = (pi -A)/(2)`
`angleFED = (pi -B)/(2)`
`:.` Area of `DeltaDEF = 2r^(2) sin (angleD) sin (angleE) sin (angleF)`
`= 2r^(2) "cos"(A)/(2) "cos"(B)/(2) "cos"(C)/(2)`
Also by sine rule, in `DeltaDEF`,
`(EF)/(sin angleD) = 2r`
or `EF = 2r cos.(A)/(2)`
Discussion
5489.

Incircle of DeltaABC touches the sides BC, AC and AB at D, E and F, respectively. Then answer the following question The length of side EF is

Answer»

`r sin.(A)/(2)`
`2R sin.(A)/(2)`
`r cos.(A)/(2)`
`2r cos.(A)/(2)`

SOLUTION :
Points I, D, C, E are concylic. Therefore,
`angleEID = pi -C`
Also, I is CIRCUMCENTER of `DeltaDEF`. Hence,
`angleDFE = (1)/(2) angleDIE = (pi -C)/(2)`
SIMILARLY, `angleFDE = (pi -A)/(2)`
`angleFED = (pi -B)/(2)`
`:.` Area of `DeltaDEF = 2r^(2) sin (angleD) sin (angleE) sin (angleF)`
`= 2r^(2) "cos"(A)/(2) "cos"(B)/(2) "cos"(C)/(2)`
Also by sine RULE, in `DeltaDEF`,
`(EF)/(sin angleD) = 2r`
or `EF = 2r cos.(A)/(2)`
Discussion
5490.

Consider the family of planes x+y+z=c where c is a parameter intersecting the coordinate axes P, Q andR and alpha, beta and gamma are the angles made by each member of this family with positive x, y and z-axes. Which of the following interpretations hold good got this family?

Answer»

Each member of this family is EQUALLY inclined withcoordinate axes.
`sin^2(alpha)+sin^2(gamma)+sin^2(BETA)=1`
`cos^2(alpha)+cos^2(beta)+cos^2(gamma)=2`
For c=3 AREA of the `trianglePQR is 3sqrt(3)` sq. units.

Answer :(a, b, c)
Discussion
5491.

If the value of "^(n)C_(0)+2*^(n)C_(1)+3*^(n)C_(2)+...+(n+1)*^(n)C_(n)=576,then n is

Answer»

`7`
`5`
`6`
`9`

Solution :`(a)` `S=^(N)C_(0)+2.^(n)C_(1)+3.(n)C_(2)+...+(n+1).^(n)C_(n)`
Here `T_(r )=(r+1)^(n)C_(r )=n^(n-1)C_(r-1)+^(n)C_(r )`
`:.S=n^(2n-1)+2^(n)=576` (given)
`:.n=7`
Discussion
5492.

Evaluate the following integrals. int(1)/(2+3cosx)dx

Answer»


Answer :`(1)/(SQRT(5))log|(sqrt(5)+tan(x//2))/(sqrt(5)-tan(x//2))|+c`
Discussion
5493.

If ( x_1, y_1) , (x_2, y_2) ,(x_3y_3)are feet of the three normals drawn from a point to the parabola y^(2) =4ax "then"sum( x_1 -x_2) /( y_3)

Answer»

`4A`
` 2A`
` 1`
`0`

ANSWER :D
Discussion
5494.

Differentiate the functions given in Exercises 1 to 11 w.r.t. x. x^x(cos x)+(x^(2)+1)/(x^(2)-1).

Answer»


Answer :`X^(x COS x)[cos x. (1+ log x)-x SIN xlog x]-(4x)/((x^(2)-1)^2)`.
Discussion
5495.

If H_(1),H_(2),H_(3),…..H_(2n+1) are in H.P then sum_(i=1)^(2n)(-1)^(j)((H_(i)+H_(i+1))/(H_(i)-H_(i+1))) is equal to

Answer»


ANSWER :C
Discussion
5496.

The sumof the numbersformedby takingall thedigits{1,2,5,7,9} is

Answer»

56,67,850
63,65,458
66,66,6600
76,61,523

Answer :C
Discussion
5497.

Let z_(1) and z_(2) be complex numbers such that z_(1)nez_(2) and abs(z_(1)) and abs(z_(2)). If Re(z_(1))gt0 and Im(z_(2))lt0, then (z_(1)+z_(2))/(z_(1)-z_(2)) is

Answer»

one
real and positive
real and negative
purely imaginary

Answer :D
Discussion
5498.

Find the inverse of each of the following matrices[{:(-1,2),(-3,5):}]

Answer»


ANSWER :`[{:((1)/(2),(1)/(4)),((1)/(2),(3)/(4)):}]`
Discussion
5499.

Statement I : If theopen half plane representedby ax +by gtM has no pointcommon with the unbounded feasible regions then M is the maximum value of z otherwisez has no maximum value Statement II : If the half plane ax+byltm has no pointcommonwith the unboundedfeasible region then m is the minimumvalue of z otherwise z hasno minimum value

Answer»

STATEMENTI is trueStatement II is TRUE , StatementII isa correctexplanationfor Statement I
Statement I is ture STATEMNET II is true, StatementII ISNOT a correct explanationfor Statement I
Statement I is trueStatementII is false
Statement I is false , StatementII is true

ANSWER :B
Discussion
5500.

Find the minor of elements 6 in the determinant Delta=|{:(1,2,3),(4,5,6),(7,8,9):}|

Answer»


ANSWER :-6
Discussion