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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 12551. |
Find the velocity and acceleration at the end of 2 seconds of the particle moving according to the following rules. s = t^3 - 6t^2 + 15t + 12 |
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Answer» Solution :`s = t^3 - 6T^2 + 15t + 12` `ds/dt = 3t^2 - 12t + 15` `(d^2s)/dt^2 = 6t - 12, ds/dt](t=2) = 3` |
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| 12552. |
Find the velocity and acceleration at the end of 2 seconds of the particle moving according to the following rules. s = 3/2t + 1 |
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Answer» Solution :`s = 3/2t + 1 ` ds/dt = 6/(2t + 1)^2 (d^2s)/dt^2 = 24/ (2t + 1)^3` Now `ds/dt](t = 2) = 6/25` (d^2s)/dt^2](t = 2) = 24/125` `therefore` VELOCITY is - 6/25 units/sec and ACCELERATION is 24/125 `units/sec^2` . |
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| 12553. |
Find the velocity and acceleration at the end of 2 seconds of the particle moving according to the following rules.s = sqrt t + 1 |
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Answer» SOLUTION :`s = sqrt t + 1` ds/DT = 1/2 sqrt t` `(d^2s)/(d^2s)/dt^2= 1/4t^(3/2)` `(ds)/dt]_(t=2) =1/2 sqrt2` `(d^2s)/dt^2]_(t=2)= - 1/4 xx 2 sqrt 2=-1/8 sqrt 2` `THEREFORE` velocity is `1/2 sqrt 2` units/sec and ACCELERATION is `-1/8 sqrt 2 units /sec^2` |
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| 12554. |
Show that of all therectangles in a given fixed circle, the square has the maximum area. (##NTN_MATH_XII_C06_E04_252_Q01.png" width="80%"> |
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Answer» SOLUTION :Let r be the radius of givencircle. A rectangle ABCD is inscribed in the circle whose diagonal is AC= 2r. Let ` AB = X and BC = y` `:." In "Delta ABC`, `AB^(2)+BC^(2)=AC^(2)` ` rArrx^(2) + y^(2) = (2r)^(2)` ` rArry^(2) = 4r^(2) - x^(2)` ...(1) Now, area of rectangle `A = x * y` ` rArr A = x sqrt(4r^(2)-x^(2))` ` rArr A^(2) = x^(2) (4r^(2)-x^(2))` Let` Z=A^(2)` ` rArrZ = 4r^(2)x^(2) - x^(4)` ` rArr(dz)/(DX) = 8r^(2)x-4x^(3)` For maxima/minima ` (dZ)/(dx) = 0` ` rArr8r^(2)x-4x^(3) = 0rArrx^(2)=2r^(2)` and ` ((d^(2)Z)/(dr^(2))) = 8r^(2) - 12x^(2)` at ` x^(2) =2r^(2)` ` (d^(2)Z)/(dr^(2)) = 8r^(2) - 24r^(2) =- 16r^(2) lt 0 ` ` rArrat x^(2) = 2r^(2) , Z` is maximum. ` rArratx^(2) = 2r^(2), A^(2)` is maximum. ` rArratx^(2) = 2r^(2), A` is maximum. Now` x^(2) = 2r^(2)` ` rArr 2x^(2) = 4r^(2) = x^(2)+y^(2)` [From equation (1)] ` rArrx^(2) = y^(2)` ` rArr x = y ` Therefore, the rectangle with maximum area, is a square. |
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| 12555. |
Find the velocity and acceleration at the end of 2 seconds of the particle moving according to the following rules. s = 2t^2 + 3t +1 |
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Answer» Solution :`s = 2t^2 + 3T + 1` `DS/dt = 4t + 3, d^2s/dt^2 = 4` `(ds)/dt]_(t=2) = 4 XX 2 +3 = 11`, `(d^2s)/dt^2] (t = 2) = 4` `therefore` Velocity is 11 units/sec and acceleration is 4 `units/sec^2`. |
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| 12556. |
Line x + 2y - 8 = 0 is the perpendicular bisector of AB. If B = (3, 5) the A is |
| Answer» ANSWER :D | |
| 12557. |
John Croxley, the mayor of Black Rock. NY, is counting the number of restaurants that have opened is his town per month for the last seven monts. He compiles the seven numbers into Set F, which contains the elements 4,5,11,13,16,18, and x. If both the median and average (arithmetic mean) of Set F equall 11, what must be the value of x, the unknown number of restaurants that opened in Mayor Croxley's town last month? |
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Answer» 9 |
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| 12558. |
If the equation of one asymptote of the hyperbola 14x^(2)+38xy+20y^(2)+x-7y-91=0is 7x+5y-3=0 then the other asymptote is |
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Answer» 2x-4y+1=0 |
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| 12559. |
Let A = [(5.1,-3.1,0),(-3.1,5.1,0),(0,0,2.2)] X be a non zero 3xx1 matrix and lambda is a real number . If A^(2)X = lambdaAX then sum of possible values of lambda is _________ |
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| 12560. |
Find the values of the following tan frac(1)(2) [sin^(-1)((2x)/(1+x^2))+ cos^(-1) ((1-y^2)/(1+y^2))],|x| lt 1, y gt 0 and x y lt 1 |
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Answer» Solution :GIVEN EXPRESSION `TAN frac(1)(2)[2 tan^(-1) x + 2 tan^(-1) y]` `tan(tan^(-1) x + tan^(-1) y)` `tan[tan^(-1) ((x+y)/(1-xy))] = (x+y)/(1_xy)` 
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| 12561. |
The maximum number of lines formed by the radical centres of 6 given circles is |
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Answer» 42 |
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| 12562. |
A coin tossed twice. Find the probability of getting exactly one head |
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Answer» `therefore S={HH,HT,TH,T T}, |S|=4` Let A be the event of getting EXACTLY one head. `thereforeA={HT,TH}implies|A|=2` `P(A)=|A|/|S|=2/4=1/2` |
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| 12563. |
Consider the sequence ('^(n)C_(0))/(1.2.3),("^(n)C_(1))/(2.3.4),('^(n)C_(2))/(3.4.5),...., if n=50 then greatest term is |
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Answer» `30^(th)` `implies("^(n)C_(r ))/((r+1)(r+2)(r+3)) ge ('^(n)C_(r-1))/((r )(r+1)(r+2))` `implies("^(n)C_(r ))/('^(n)C_(r-1)) ge (r+3)/(r )` `implies(n-r+1)/(r ) ge (r+3)/(r )` `impliesr le ((n-2)/(2))` For `n=50impliesr ge 24` So `24^(th)` and `25^(th)` TERMS of sequence are greatest. |
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| 12564. |
If P(A)= (4)/(5) and P(A cap B)= (7)/(10) then P(B//A) is equal to = …….. |
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Answer» `(1)/(10)` |
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| 12565. |
The ratio of 4^(th) term and 5^(th) term in the expansion of (x+(sinx)/(x))^(6) is (16)/(3pi^(2)), then x is equal to (1) (pi)/(2) (2) -(pi)/(2) (3) (pi)/(3) (4) Both (1) & (2) |
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Answer» Solution :Answer (1) `T_(4)=.^(6)C_(3)*x^(3)*((sinx)/(x))^(3)` `T_(5)=.^(6)C_(4)*x^(4)*((sinx)/(x))^(2)` Now, `(T_(4))/(T_(5))=(16)/(3PI^(2))` `implies(.^(6)C_(3)SIN^(3)x)/(.^(6)C_(4)x^(2)*sin^(2)x)=(16)/(3pi^(2))` `implies(20sinx)/(15x^(2))=(16)/(3pi^(2))` `implies(sinx)/(x^(2))=(16xx15)/(20xx3pi^(2))=(4)/(pi^(2))` `impliesx=(pi)/(2)` |
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| 12566. |
Find the number of ways in which 8 men and 4 ladies can sit around a round table so that all the ladies come together |
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| 12567. |
If the line x-6y-12=0 meets the circle S equiv x^(2)+y^(2)-4x+8y+6=0 at A and B, then the point of intersection of the tangents at A and B to S = 0 is |
| Answer» ANSWER :A | |
| 12568. |
Write down negations of Pen is mightier than sword. |
| Answer» SOLUTION :PEN is not MIGHTIER than SWORD. | |
| 12569. |
The vector equation of the plane 2x-z+1 = 0 is................ |
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Answer» `bar R.`(2,-1,0) + 1 = 0 |
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| 12570. |
Differentiate the following w.r.t. x : (e^x)/(sin x) |
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| 12572. |
If the foot of perpendicular from origin to the plane is (2,1,0) then the equation of the plane is .......... |
| Answer» Answer :B | |
| 12573. |
Integrationof trigonometricandhyperbolicfunctions I=int(sin^(4)x)/(cosx)dx. |
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| 12574. |
The longest smooth muscles are seen in :- |
| Answer» Answer :A | |
| 12575. |
If |(2a,x_(1),y_(1)),(2b,x_(2),y_(2)),(2c,x_(3),y_(3))| = (abc)/2 != 0, thentheareaof the trianglewhosevertices are ((x_(1))/a,(y_(1))/a),((x_(2))/b,(y_(2))/b)and ((x_(3))/c,(y_(3))/c) |
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Answer» `(1)/(8)` |
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| 12576. |
Find the distance of a point (2, 5, -3) from the plane vecr.(6hati-3hatj+2hatk)=4 |
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| 12577. |
Using integration, find the area of the region bounded by the line 2y=5x+7, x-axis and the lines x = 2 and x = 8. |
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| 12578. |
If f(x) = {((x^(2)-4x+3)/(x^(2)-1), ",","for"x != 1),(2, ",", "for"x = 1 ):} , then |
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Answer» f(x) is continuous at x = 0 |
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| 12579. |
If f (x) satisfies 97 f (x) + mf ((1)/(x)) =0, where f (x) = lim _( n to oo) n(x ^(1//n)-1). X gt 0, then the value of m is |
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Answer» `1/97` |
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| 12580. |
Find the are of the region bounded by y= log_(e )x and y= sin^(4) (pi x) |
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| 12581. |
Integrate the following functions : int(3x^(2))/(1+x^(6))dx |
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| 12582. |
Find the equation of line passing through the given points using determinants (3,-2),(-1,4) |
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| 12583. |
If overset(to)(a),overset(to)(b),overset(to)(c ) are threenon-zero , non-coplanar vectors and overset(to)(b)_(1)=overset(to)(b) -.(overset(to)b.overset(to)(a))/(|overset(to)(a)|) overset(to)(a) , overset(to)(b)_(2) +.(overset(to)(b).overset(to)(a))/(|overset(to)(a)|^(2))overset(to)(a) overset(to)(c)_(1) =overset(to)(c)-.(overset(to)(c).overset(to)a)/(|overset(to)(a)|^(2))overset(to)(a)-.(overset(to)(c).overset(to)(b))/(|overset(to)(b)|^(2))overset(to)(b),overset(to)c_(2)=overset(to)(c) -.(overset(to)(c).overset(to)(a))/(|overset(to)(a)|^(2))overset(to)(a)-.(overset(to)(c).overset(to)(b)_(1))/(|overset(to)(b)|^(2))overset(to)(b)_(1) overset(to)(c)_(3) =overset(to)(c) -.(overset(to)(c).overset(to)(a))/(|overset(to)(a)|^(2))overset(to)(a)-.(overset(to)(c).overset(to)(b)_(2))/(|overset(to)(b)_(2)|^(2))overset(to)(b)_(2).overset(to)(c)_(4)=overset(to)(a) -.(overset(to)(c).overset(to)(a))/(|overset(to)(a)|^(2))overset(to)(a) Thenwhichof the followingis aset ofmutually orthogonalvectors ? |
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Answer» `{overset(to)(a),overset(to)(b)_(1),overset(to)(c)_(1)}` `" and"vec( c)_(1) =vec(c ) - (vec( c) "."vec( a))/(|vec(a)|^(2)) vec(a) - (vec(c )"."vec(b))/(|vec(b)|^(2)) vec(b) vec( c)_(2) = vec( c) - (vec( c). vec(a))/(|vec(a)|^(2)) vec(a)- (vec(c ) "."vec(b)_(1))/(|vec(b)|^(2)) vec(b)_(1)` `vec(c)_(3) =vec(c) -(vec(c ).vec(a))/(|vec(a)|^(2))vec(a) -(vec(c ).vec(b)_(2))/(|vec(b)_(2)|^(2)) vec(b)_(2), vec(c )_(4) =vec(a) -(vec(c ). vec(a))/(|vec(a)|^(2))vec(a)` Whichshows `vec(a). vec(b)_(1) =0=vec(a) . vec(c )_(2)=vec(b)_(1).vec(c )_(2)` So `{vec(a),vec(b)_(1),vec(c )_(2)}` aremutually orthogonalvectors. |
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| 12584. |
Two cards are selected at random from 52 playing cards then the probability of selecting one king and one Queen is |
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Answer» `(4^(2))/(.^(52)C_(2))` |
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| 12585. |
Form the differential equation of the family of curve y^(2)=a(b^(2)-x^(2)). |
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| 12586. |
If n is a positive integer, then value of (3n+2)^n C_0+(3n -1)^n C_1 + (3n-4)^n C_2 + ….+2(""^n C_n) is |
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Answer» `(3N +4)^2(n-1)` |
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| 12587. |
Write the following functions in the simplest form : tan^(-1) (x/sqrt(a^2-x^2)), |x| lt a |
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Answer» Solution :PUT x = a `SIN theta, theta = sin^(-1)(x/a)` Then `TAN^(-1)(x/sqrt(a^2-x^2)) = tan^(-1)((a sin theta)/sqrt(a^2-a^2 sin^2 theta ))` `= tan^(-1)((a sin theta)/sqrt(a^2 cos^2 theta)) = tan^(-1) (tan theta) = theta = sin^(-1)(x/a)` |
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| 12588. |
The polar of a given point which respect to any one of the circles x^(2)+y^(2)-2kx+c^(2)=0, (k is a varaible) always passes through a fixed point whatever to be the value of k is |
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Answer» `(x_(1),(X^(2)-C^(2))/(y_(1)))` |
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| 12589. |
Find the area of the region bounded by x^(2)= 4y, y = 2, y = 4 and the y-axis in the first quadrant. |
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| 12590. |
Each of the statement ptoq,qtor and ~r is true. Then |
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Answer» <P>p is TRUE |
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| 12591. |
If omega is a complex cube root of unity , then (x + 1) ( x + omega) ( x- omega - 1) = |
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Answer» `X^(3) - 1` |
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| 12592. |
If int e^(2x) f^(1) (x) dx = g (x), then int (e)^(2x) f^(1) (x) + e^(2x) f(x) |
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Answer» `(1)/(2) [e^(2x) F(x) - G(x)] + C` |
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| 12593. |
Major product ............. |
Answer» 
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| 12594. |
If the point (2,k )lies outside the circles x^(2) +y^(2) =13and x^(2) +y^(2) +x-2y -14=0then |
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Answer» ` K in ( -infty ,-2)UU[3 ,infty ) ` |
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| 12595. |
Solve the following linear programming problems graphically : Maximize : Z = 100x + 170y subject to the constraints 3x+2y le 3600, x+4y le 1800, x, y ge 0. |
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| 12596. |
Find an angle theta, where 0 lt theta lt (pi)/(2) which increases twice as fast as its sine. |
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| 12598. |
int (1 + tan^(2) x)/(1 -tan^(2) x)dx = |
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Answer» `(sin 2X)/(2) + ` C |
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| 12599. |
int(x^(2)-1)/(xsqrt((x^(2)+alphax+1))(x^(2)+betax+1))dx is equal to |
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Answer» `LOG{(sqrt(X^(2)+alphax+1)+sqrt(x^(2)+betax+1))/(sqrt(x))}+C` |
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| 12600. |
Find (dy)/(dx) of the functions given in Exercises 12 to 15. x^(y)+y^(x)=1. |
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