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5201.

If1^(@)=alpha radians then the approximate value of cos(60^(@) 1') is

Answer»

`1/2+(alphasqrt3)/120`
`1/2-alpha/120`
`1/2-(alphasqrt3)/120`
`1/2+alpha/120`

ANSWER :C
5202.

Differentiate each of the following from the first principle . (i) sin sqrt(x) , (ii) cos sqrt(x) (iii) tan sqrt(x)

Answer»

Solution :(i) Let `y = sin sqrt(X)`
Let `deltay` be an increment in y, corresponding to an increment `deltax` in x.
Then, `y +deltay=sinsqrt(x+deltax)`
`rArr deltay=sinsqrt(x+deltax)-sinsqrt(x)`
`rArr (deltay)/(deltax)=(sinsqrt(x+deltax)-sinsqrt(x))/(deltax)`
`rArr(dy)/(dx) = underset(deltaxrarr0)("lim")(deltay)/(deltax)`
`= underset(deltaxrarr0)("lim")({sinsqrt(x+deltax) - sinsqrt(x)})/(deltax)`
`= underset(deltararr0)("lim") (2cos((sqrt(x+deltax) + sqrt(x))/(2)) sin ((sqrt(x+deltax) - sqrt(x))/(2)))/({(x+deltax) - x})`
`["USING" sinX - sinD = 2 cos' ((C+D))/(2) sin ((C-D)/(2))]`
`= underset(deltaxrarr0)("lim") (2cos((sqrt(x+deltax) + sqrt(x))/(2))sin ((sqrt(x+deltax) - sqrt(x))/(2)))/((sqrt(x+deltax) + sqrt(x)) (sqrt(x+deltax) - sqrt(x)))`
`[ :'(x+deltax) -x = (sqrt(x+deltax) + sqrt(x))(sqrt(x+deltax)-sqrt(x))]`
`= underset(deltaxrarr0)("lim")(sin((sqrt(x+deltax) - sqrt(x))/(2)))/(((sqrt(x+deltax)-sqrt(x))/(2))). underset(deltaxrarr0)("lim")cos((sqrt(x+deltax)+sqrt(x))/(2)). underset(deltaxrarr0)("lim") (1)/((sqrt(x+deltax) + sqrt(x)))`
`= underset(thetararr0)("lim")(sintheta)/(theta).underset(deltaxrarr0)("lim")(cos(sqrt(x+deltax) + sqrt(x))/(2)).underset(deltaxrarr0)("lim")(1)/((sqrt(x+deltax)+sqrt(x)))`,
where `((sqrt(x+deltax) - sqrt(x)))/(2) = theta` and `deltax rarr 0 rArr theta rarr 0`
`= 1 xx cos sqrt(x) xx 1/(2sqrt(x))= (cossqrt(x))/(2sqrt(x))`.
HENCE , `d/(dx) (sinsqrt(x)) = (cossqrt(x))/(2sqrt(x))`.
(i) Let `y = cos sqrt(x)`.
Let `deltay` be an increment in y, corresponding to an increment `deltax`in x.
Then, `y+deltay = cossqrt(x + deltax) - cos sqrt(x)`
`rArr(deltay)/(deltax) = ({cossqrt(x+deltax)-cossqrt(x)})/(deltax)`
`rArr (dy)/(dx) = underset(deltaxrarr0)("lim")(deltay)/(deltax)`
`= underset(deltax rarr 0)("lim") ({cossqrt(x+deltax)- cos(x)})/(deltax)`
` = underset(deltaxrarr0)("lim") (-2sin((sqrt(x+deltax) + sqrt(x))/(2)) sin ((sqrt(x+deltax) - sqrt(x))/(2)))/({(x+deltax) - x})`

` = underset(deltaxrarr0)("lim") (-2sin((sqrt(x+deltax) + sqrt(x))/(2)) sin ((sqrt(x+deltax) - sqrt(x))/(2)))/((sqrt(x+deltax)+sqrt(x))(sqrt(x+deltax)-sqrt(x)))`
`=-underset(deltaxrarr0)("lim")sin((sqrt(x+deltax)+sqrt(x))/(2)).underset(deltararr0)("lim") (sin '((sqrt(x+ deltax) - sqrt(x))/(2)))/(((sqrt(x+deltax)-sqrt(x))/(2))) . underset(deltararr0)("lim")(1)/((sqrt(x+deltax)+sqrt(x)))`
where `((sqrt(x+deltax)-sqrt(x)))/(2) = theta`
`= - sin((2sqrt(x))/(2)) xx 1 xx 1/(2sqrt(x) ) = (-sinsqrt(x))/(2sqrt(x))`
Hence , d/(dx) (cossqrt(x)) = (-sinsqrt(x))/(2sqrt(x))`.
(iii) Let `y = TAN sqrt(x)`.
Let `deltay` be an increment in y, corresponding to an increment `deltax` in x.
Then, `y + deltay = tan sqrt(x+deltax)`
`rArr deltay = tan sqrt(x+deltax) - tan sqrt(x)`
`rArr (deltay)/(deltax) = (tan(sqrt(x+deltax)- tan sqrt(x)))/(deltax)`
`rArr (dy)/(dx) = underset(deltax rarr 0)("lim") (deltay)/(deltax)`
`= underset(deltax rarr 0)("lim") ((tan sqrt(x+deltax) - tan sqrt(x)))/(deltax)`
`= underset(deltaxrarr0)("lim")({(sinsqrt(x+deltax))/(cos sqrt(x+deltax))-(sinsqrt(x))/(cossqrt(x))})/(deltax)`
`= underset(deltaxrarr0)("lim") ((sinsqrt(x+deltax) cos sqrt(x) - cos sqrt(x+deltax) sinsqrt(x)))/(deltax.cos sqrt(x+deltax).cossqrt(x))`
` = underset(deltaxrarr0)("lim") {(sin(sqrt(x+deltax) - sqrt(x)))/(|(x+deltax)-x|) . (1)/(cossqrt(x+deltax).cossqrt(x))}`
`[ :' deltax = (x+deltax )- x]`
`= underset(deltaxrarr0)("lim") {(sin(sqrt(x+deltax)-sqrt(x)))/((sqrt(x+deltax) - sqrt(x))(sqrt(x+deltax) + sqrt(x))).(1)/(cos sqrt(x+deltax).cossqrt(x))}`
`=underset(deltararr0)("lim")(sin(sqrt(x+deltax) - sqrt(x)))/((sqrt(x+deltax) - sqrt(x))).underset(deltaxrarr0)("lim") (1)/((sqrt(x+deltax)+sqrt(x))).underset(deltaxrarr0)("lim") (1)/(cossqrt(x+deltax).cossqrt(x))`
`= (underset(thetararr0)("lim") (sintheta)/(theta)).1/(2sqrt(x)).(1)/(cos^(2)sqrt(x))=1xx 1/(2sqrt(x))xx(1)/(cos^(2)sqrt(x))`
`= (sec^(2)sqrt(x))/(2sqrt(x))`.
Hence, `d/(dx) (tansqrt(x)) = (sec^(2)sqrt(x))/(2sqrt(x))`,
5203.

D(2,1,0) E(2,0,0) F(0,1,0) are mid points of the sides BC,CA,AB of Delta ABC respectively. The centroid of Delta ABC is ……..

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`(1/3,1/3,1/3)`
`(4/3,2/3,0)`
`((-1)/3,1/3,1/3)`
`(2/3,1/3,1/3)`

ANSWER :B
5204.

Middle term in the expansion (x+(1)/(x))^(2n) is ........... .

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ANSWER :(n+1)
5205.

For arithmetic sequence

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ANSWER :`2a_(m)`
5206.

A ladder of 20 mt long reaches a point 20 mt below the top of a flag staff. If the angle of elevation of the top of the flag staff at the foot of the ladder is60 ^(@)then the height of the flag staff is

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` 10 (sqrt3-1) `
` 300( sqrt3+1) MT`
` 120 ( sqrt3-1) `
` 30 mt `

Answer :D
5207.

Three cards are drawn at a time at random from a well shuffled [ack of 52 cards . Find the probability that all the three cardshave same number .

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ANSWER :`(1)/(425)`
5208.

If x = h + p Sec alpha, y = k + q cosec alpha then ((p)/(x-h))^(2) +((q)/(y-k))^(2)=

Answer»

1
-1
0
`1//2`

ANSWER :A
5209.

If for an ellipse length of minor axis and distance between two focii are equal then its eccentricity e = ……….

Answer»

`(1)/(SQRT(2))`
`(sqrt(2))/(3)`
`(sqrt(3))/(2)`
`(2)/(sqrt(3))`

ANSWER :A
5210.

The side of an equilateral triangle increases at a uniform rate of 0.05 cm/sec. Find the rate of increase in the area of the triangle when the side is 20cm.

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ANSWER :`sqrt3/2 CM^(2)`
5211.

The product of r consecutive positiveintegersis divisibleby

Answer»

R!
r! + 1
(r + 1)
NONE of these

ANSWER :D
5212.

If the orthocentre and the centorid of a triangle are (-3, 5, 1), (3, 3, -1) then find the Circumcentre

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ANSWER :(6, 2, -2)
5213.

Find the locus of point whose distance from the origin is 5.

Answer»


ANSWER : `X^(2)+y^(2)=25`
5214.

The voluime of the largest possible right circular cylinder that can be in inscribed in a sphereof radius =sqrt(3) is

Answer»

`(4)/(3) sqrt(3) pi`
`(8)/(3) sqrt(3) pi`
`4PI`
`2PI`

ANSWER :C
5215.

Find the value of p so that the three lines3x + y - 2 = 0, px + 2y - 3 = 0 " and " 2x - y - 3 = 0may intersect at one point.

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ANSWER :5
5216.

Find lim_(nrarrinfty) (1^3 + 2^# +…..n^3)/[(2n^2 +1)(2n^4 +4)].

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ANSWER :`1/16`
5217.

Let f(x) = x -1 andg(x) = (1)/(x). Then the set of point where going is continuous is

Answer»

R -{0}
R- {1}
`(-OO, oo)`
`(-oo, oo) - { 0,1}`

ANSWER :D
5218.

f(x) {{:((sqrt(1 + kx)-sqrt(1-kx))/(x)" for " - 1 le x lt 0),(2x^(2) + 3x - 2 " for " 0 le x lt 1 ):} is continuous at x = 0 k =

Answer»

-4
-3
-2
-1

Answer :C
5219.

Find the acute angle between pair of lines represented by following equations. x^(2)+2xy cot alpha-y^(2)=0

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ANSWER :`(PI)/(2)`
5220.

A and B are two events such that P(A) =0.54,P(B) =0.69 andP( A nnB) =0.35 . Find (a) P( A uuB) (ii) P( A 'nn B') (iii) P( A nn B') ( iv) P( B nnA ')

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ANSWER :(I) 0.88 (II) 0.12 (III) 0.19 (V) 0.34
5221.

If a and b are chosen randomly from the set {2,3,4,5} with replacement, then the probability of the real roots of the equation x^(2)+ax+b=0 is:

Answer»

`(11)/(26)`
`(1)/(8)`
`(1)/(2)`
`(1)/(4)`

ANSWER :C
5222.

A tower subtends an angle alpha at a point A on the same level as the foot of the tower B is a point vertically above A and AB = h metres. The angle of depression of the foot of the tower from B is beta. The height of the tower is

Answer»

`H tan ALPHA COT beta`
`h tan alpha tan beta`
`h cot alpha cot beta`
`h cot alpha tan beta`

ANSWER :A
5223.

The locus of the point P such that PA^(2)+PB^(2)=10 where A=(2,3,4), B=(2,3,4) is

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`X^(2)+y^(2)+Z^(2)-4x-6y-8z+24 =0`
`x^(2)+y^(2)+z^(2)-6y-8z+24 =0`
`2(x^(2)+y^(2)+z^(2))-6y-8z+12 =0`
`x^(2)+y^(2)+z^(2)-6y-8z+12 =0`

ANSWER :A
5224.

No of solutions of theta in [0,2pi] satisfying the equation log_(sqrt(3))tan theta)(sqrt((log_(tan theta)3+log_(sqrt(3))3sqrt(3)))=-1 is

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ANSWER :2
5225.

Let |{:(tan^(-1)x, tan^(-1)2x, tan^(-1)3x), (tan^(-1)3x, tan^(-1)x, tan^(-1)2x), (tan^(-1)2x, tan^(-1)3x, tan^(-1)x):}|=0, then the number of values of x satisfying the equation is

Answer»


ANSWER :1
5226.

Use the first derivative test to find local extrema of f(x) = x^(2) - 6x + 8 on R.

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Answer :Point of LOCAL MINIMUM X = 3, local minimum= -1
5227.

The points which divides the join of (3,-2,1 ) and (-2,3,11 ) in the ratio 2 : 3 is :

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`(33//5, 2//5, 9)`
`(4, 0, 7)`
`(32//5, -12//5, 17//5)`
`(20, 0, 35)`

ANSWER :b
5228.

Solve the following equations sin 3 theta = (sqrt(3))/(2), 3theta is an acute angle

Answer»


ANSWER :`(-5 PI)/(9), (-4PI)/(9), (pi)/(9), (2PI)/(9), (7pi)/(9), (8pi)/(9)`
5229.

If tan alpha = m//(m+1), tan beta =1//(2m + 1), then alpha + beta =

Answer»

`NPI+(PI)/(2)`
`npi+(pi)/(3)`
`npi-(pi)/(4)`
`npi+(pi)/(4)`

Answer :D
5230.

Solve the equations: 3^(2x+1)+3^(2)=3^(x+3)+3^(x).

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ANSWER :2 or -1.
5231.

Find n if : "" ^(2n)P_(n+1) : ""^(2n-2) P_n = 56 : 3

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ANSWER :` 4`
5232.

If (cosx)/(cosy)=2 and cos (x-y)=(sqrt(3))/(2) then tan y=

Answer»

`SQRT(3)+4`
`sqrt(3)+1`
`sqrt(3)-4`
`sqrt(3)-1`

Answer :C
5233.

If x = sin (alpha - beta) sin (gamma - delta) y = sin (beta - gamma) sin (alpha - delta ) and z = sin (gamma - alpha ) sin (beta - delta ) then

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`x + y + Z =0`
`x + y - z =0`
`y + z -x =0`
`x ^(2) + y ^(3) + z ^(3) = 3 XYZ`

Answer :A::D
5234.

Show that lim_(n to infty)1/1.2+1/2.3+1/3.4+...+1/(n(n+1))=1

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ANSWER :1
5235.

The period of sin x sin(120^(@) + x) sin(120^(@) -x) is

Answer»

`(PI)/(3)`
`(2PI)/(3)`
`(pi)`
`(pi)/(2)`

Answer :B
5236.

A pole of height h stands at one corner of a park in the shape of an equilateral triangle. If alpha is the angle which the pole subtends at the midpoint of the opposite side, the length of each side of the park is

Answer»

`((sqrt(3))/(2))h COT ALPHA`
`((2)/(sqrt(3)))h cot alpha`
`((sqrt(3))/(2)) h TAN alpha`
`((2)/(sqrt(3))) h tan alpha`

Answer :B
5237.

If 3bari+4barj+5bark is P.V. of one end of adiameter of the sphere abs(barr-(4bari+3barj+4bark))=sqrt3 then other end is

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1)`2bari+3barj+5bark`
2)`bari+2barj+3bark`
3)`5bari+2barj+3bark`
4)`5bari-4barj-3bark`

ANSWER :C
5238.

Using binomial theorem, expand [ ( x+y)^(5) + (x-y)^(5) ] and hence find the value of [ ( sqrt(3) + 1)^(5) - ( sqrt(3) - 1)^(5) ].

Answer»


ANSWER :152
5239.

Computethe derivative of sinx using first principal method ?

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ANSWER :`= 2COS((2x+0)/2)xx1/2 = COS X`
5240.

If A=[(0,2,1),(-2,0,-2),(-1,x,0)] is skew symmetric matrix, then find x.

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ANSWER :`x=2`
5241.

The value of (2 sinx)/(sin3x) +(tanx)/(tan3x) is ______ .

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ANSWER :1
5242.

If sec^(6) theta - tan^(6) theta = a sec^(4) theta + b sec^(2) theta + c then a+b+c=

Answer»

0
-1
3
1

Answer :D
5243.

lim_(x to 0) ( sqrt(1 - cos 2x))/ x

Answer»

0
1
`SQRT2`
does not exist

Answer :C
5244.

If f(x)=|x|-{x} where {x} denotes the fractional part of x , then f(x) is dreasing in

Answer»

`(-(1)/(2),0)`
`(-(1)/(2),2)`
`(-(1)/(2),2)`
`((1)/(2),OO)`

ANSWER :A
5245.

How many different words. Beinging and ending with aconsonant can be made out of the letters of the word ' EQUATIONS' ?

Answer»

<P>

ANSWER :` ""^(3)p_2 XX ""^(6)P_ = 4320 `
5246.

Consider a pendulum of length 50 cms If the tip of the pendulum describes an arc of length 10 cms find the anglethrough which the pendulum swings. Find this angle in degrees.

Answer»

SOLUTION :`11^@27.`
5247.

Vector equation of the plane to which the vector bari+barj " normal, and which contains the line " barr=bari+barj+bark+t(bari-barj-bark) is

Answer»

`BARR.(bari+barj)=2`
`barr.(bari+barj+bark)=2`
`barr.(bari+barj)=1`
`barr.(bari+barj+bark)=1`

ANSWER :A
5248.

If sin theta+ cos theta= a and than theta+cot theta=b then

Answer»

`a=B`
`(1)/(b)=(a^(2)-1)/(2)`
`a^(2)=2b+1`
`ab=1`

ANSWER :B
5249.

One cards is drawn from a pack of 52 cards , each of the 52 cards beingequally like to be drawn . Find the probability that the card drawn is red and a king.

Answer»


ANSWER :`(""^(2)C_(1))/(""^(52)C_(4))=(1)/(26)`
5250.

One cards is drawn from a pack of 52 cards , each of the 52 cards beingequally like to be drawn . Find the probability that the card drawn is either red or king

Answer»


Answer :`(""^(28)C_(1))/(""^(52)C_(1))=(7)/(13)`