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.
| 26501. |
Which of the following propositions is a contradiction ? |
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Answer» <P>`(~p VV ~Q) vv (p vv ~q)` |
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| 26502. |
Which of the following is not a statement in logic? |
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Answer» X+0=x, x`in` R |
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| 26503. |
The number of ways can 5 red and 4 white balls be drawn from a bag containing 10 red and 8 white balls is |
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Answer» `""^(8)C_(5)XX""^(10)C_(4)` |
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| 26504. |
Let AP and BQ be two vertical poles at points A and B respectively. If AP = 16 m, BQ = 22m and AB = 20 m, then find the distance of a point R on AB from the point A such that RP^(2)+RQ^(2) is minimum. |
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| 26505. |
What is common amylase, rennin and trypsin ? |
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Answer» They are produce in stomach |
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| 26506. |
Show that the circles x^(2) + y^(2)-6x -2y + 1=0 , x ^(2) + y^(2) + 2x -8y + 13 =0touch each other find the point of contact and the equation of the common tangent at their point of contact. |
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| 26507. |
If alpha, beta are roots of equation x^(2)-(a-1)x+(a^(2))/(4)=0 and delta, gamma are roots of equation x^(2)-(a+1)x+1=0, A quadrilateral is constructed in real plane by taking vertices as (alpha, beta), (beta, alpha), (delta, gamma), (gamma, delta). Find the value of 'a' such that quadrilateral PQRS never cuts the circle (x-1)^(2)+(y-1)^(2)=a^(2). |
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| 26508. |
Let Delta = |{:(-bc,,b^(2)+bc,,c^(2)+bc),(a^(2)+ac,,-ac,,c^(2)+ac),(a^(2)+ab,,b^(2)+ab,,-ab):}| and the equation px^(3) +qx^(2) +rx+s=0 has roots a,b,c where a,b,c in R^(+) The value of Delta is |
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Answer» `le 9r^(2)//p^(2)` `Delta = |{:(-bc,,ab+AC,,ac+ab),(ab+bc,,-ac,,bc+ab),(ac+bc,,bc+ac,,-ab):}|` Nowusing `C_(2) to C_(2)-C_(1)" and " C_(3) to C_(3) -C_(1) ` and thentaking(ab+bc+ca) common from `C_(2) " and" C_(3)` we get `Delta =|{:(-bc,,1,,1),(ab+bc,,-1,,0),(ac+bc,,0,,-1):}|xx (ab +bc +ca)^(2)` Nowapplying `R_(2) to R_(2) +R_(1)` we get `Delta = |{:(-bc,,1,,1),(ab,,0,,1),(ac+bc,,0,,-1):}| (ab+bc+ca)^(2)` Expandingalong `c_(2)` we get `Delta =(ab+bc+ca)^(2)[ac+bc+ca)^(2)` `=(ab+bc+ca)^(2)` `=(r//p)^(3) =r^(3)//p^(3)` Now givena,b,care allpositivethen `A.M ge G.M.` `rArr (ab+bc+ac)/(3) ge (abxx bcxx ac)^(1//3)` `" or" (ab+bc+ac)^(3) ge 27A^(2)b^(2)c^(2)` `"or" (ab+bc+ca)^(3) ge 27(s^(2)//p^(2))` if `Delta =27` thenab+bc+ca =3 and GIVENTHAT `a^(2) +b^(2)+c^(2)=3` From `(a+b+c)^(2) =a^(2) +b^(2)+c^(2) +2 (ab+bc+ca)` we have `a+b+c = ne 3` `rArr a+b+c =3` `rArr 3p+q=0` |
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| 26509. |
A(3,2,-1),B(4,1,1,)C(6,2,5) and D(3,3,3) are four points G_(1)G_(2)G_(3)" and "G_(4)respectively are the centroids of the triangle DeltaBCD, DeltaCDA,DeltaDAB,DeltaABC. The point of concurrence of the lines AG_(1),BG_(2),CG_(3)," and "DG_(4) is |
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Answer» (4,2,2) |
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| 26510. |
Which of the following is not a statementin logic? |
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Answer» The SUM of ANGLES of a quadrilateral is `180^@`. |
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| 26511. |
Match the following |
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Answer» c,d,a,b |
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| 26512. |
Let Delta = |{:(-bc,,b^(2)+bc,,c^(2)+bc),(a^(2)+ac,,-ac,,c^(2)+ac),(a^(2)+ab,,b^(2)+ab,,-ab):}| and the equation px^(3) +qx^(2) +rx+s=0 has roots a,b,c where a,b,c in R^(+) the valueof Deltais |
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Answer» <P>`r^(2)//p^(2)` `Delta = |{:(-bc,,ab+AC,,ac+ab),(ab+bc,,-ac,,bc+ab),(ac+bc,,bc+ac,,-ab):}|` Nowusing `C_(2) to C_(2)-C_(1)" and " C_(3) to C_(3) -C_(1) ` and thentaking(ab+bc+ca) common from `C_(2) " and" C_(3)` we get `Delta =|{:(-bc,,1,,1),(ab+bc,,-1,,0),(ac+bc,,0,,-1):}|xx (ab +bc +ca)^(2)` Nowapplying `R_(2) to R_(2) +R_(1)` we get `Delta = |{:(-bc,,1,,1),(ab,,0,,1),(ac+bc,,0,,-1):}| (ab+bc+ca)^(2)` Expandingalong `c_(2)` we get `Delta =(ab+bc+ca)^(2)[ac+bc+ca)^(2)` `=(ab+bc+ca)^(2)` `=(r//p)^(3) =r^(3)//p^(3)` Now GIVENA,b,CARE allpositivethen `A.M ge G.M.` `rArr (ab+bc+ac)/(3) ge (abxx bcxx ac)^(1//3)` `" or" (ab+bc+ac)^(3) ge 27a^(2)b^(2)c^(2)` `"or" (ab+bc+ca)^(3) ge 27(s^(2)//p^(2))` if `Delta =27` thenab+bc+ca =3 and giventhat `a^(2) +b^(2)+c^(2)=3` From `(a+b+c)^(2) =a^(2) +b^(2)+c^(2) +2 (ab+bc+ca)` we have `a+b+c = ne 3` `rArr a+b+c =3` `rArr 3p+q=0` |
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| 26514. |
int "log " sqrt(x + 1 ) dx = |
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Answer» `(1)/(2)` [ (X - 1) LOG (x + 1) - x ] + C |
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| 26515. |
Integrate the following intsin3xdx |
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Answer» SOLUTION :`INTSIN3XDX` [PUT 3x=t then 3dx=dt or dx=1/3dt `intsintcdot(1/3)dt` 1/3(-cost+C =1/3(COS3X)+C |
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| 26516. |
Construct the switching circuit for the following statement : [p vv (~p^^q)]vv[(~q^^r) vv ~ p]. |
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Answer» `(##NVT_21_MAT_XII_C01_E07_001_A02##)` |
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| 26517. |
Find a vector in the direction of the vector (3, -2, 2) which has magnitude 2sqrt(17) units. |
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| 26518. |
A ladder rests against the side of a wall and reaches a point that is h meters above the ground. The angle formed by the ladder and the ground is theta^(@). A point on the ladder is k meters from the wall.What is the vertical distance, in meters, from this point on the ladder to the ground? |
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Answer» `(h-k)TAN THETA^(@)` |
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| 26519. |
LetA and B be obtuse angles such thatsinA=(4)/(5)andcosB=-(12)/(13). |
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Answer» `-(63)/(65)` It is given that A and B are obtuse angle `rArrcosA=+-sqrt(1-sin^(2))A=+-sqrt(1-(16)/(25))=-(3)/(5)` Negative sign is taken for cos A SINCE A being obtuse lies in second quadrant. `sinB=+-sqrt(1-cos^(2)B)=+-sqrt(1-((-12)/(13))^(2))` `=sqrt((169-144)/(169))=(5)/(13)` Positive sign is taken since , sin B is positive in second quadrant . `rArrcosA=(-3)/(5)andsinB=(3)/(13)` `thereforesin(A+B)=sinAcosB+cosAsinB` `=(4)/(5)XX((-12)/(13))+(-3)/(5)xx((5)/(13))=-(48)/(65)-(15)/(65)` `=(-48-15)/(65)=(-63)/(65)` |
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| 26520. |
S is sample space if P(A)= (1)/(3)P(B) and S= A cup B Then P(A) = ....... Where A and B are mutually exclusive events. |
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Answer» `(1)/(4)` |
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| 26521. |
The solution of (dy)/(dx) + y = e^(x) is |
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Answer» `E^(y//x) = CX` |
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| 26522. |
Differentiate.5^(ln sinx) |
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Answer» SOLUTION :`y=5^(In SIN X)` `dy/dx=5^(In sin xcdot)In5d/dx(In sin x)` `5^(In sin x).cdot In 5cdot x` |
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| 26523. |
Find the equation of the circle for which the point given below are the end points of a diameter. (1,1) (2,-1) |
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| 26524. |
Find the number of seven digited palindromes that can be formed using 0,1,2,3,4. |
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| 26525. |
C_(1) is a circle with centre at the origin and radius equal to r and C_(2) is a circle with centre at (3r, 0)and radius t 2r. The number of common tangents that can be drawn to the two circles are |
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Answer» 1 |
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| 26526. |
If sin alpha/2 + cos alpha/2=1.4, then: sin alpha cos alpha= |
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| 26527. |
Find the order and degree of the following D.E's (i) (d^(2)y)/(dx^(2)) + 2((dy)/(dx))^(2) + 5y = 0 (ii) 2(d^(2)y)/(dx^(2)) = (5+(dy)/(dx))^((5)/(3)) (iii) 1+((d^(2)y)/(dx^(2)))^(2) = [2+((dy)/(dx))^(2)]^((3//2)) (iv) [(d^(2)y)/(dx^(2))+((dy)/(dx))^(3)]^((6/(5)) = 6y (v) [((dy)/(dx))^(2) + (d^(2)y)/(dx^(2))]^((7)/(3)) = (d^(3y))/(dx^(3)) (vi) [((dy)/(dx))^((1)/(2)) + ((d^(2)y)/(dx^(2)))^((1)/(2))]^((1)/(4)) = 0 (vii) (d^(2)y)/(dx^(2)) + p^(2)y = 0 (viii) ((d^(3)y)/(dx^(3)))^(2) -3((dy)/(dx))^(2) - e^(x) = 4 |
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Answer» (iii) order 2, degree 4 (iv) order 2, degree 6 (v) order 3, degree 3 (VI) order 2, degree 1/2 (vii) order 2, degree 1 (viii) order 3, degree 2 |
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| 26528. |
Using differentials, find the approximate value of each of the up to 3 places of decimal. (0.009)^((1)/(3)) |
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| 26529. |
The number of values of x int [-2pi, 2pi] satisfying tanx + cotx2" cosec x" is |
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Answer» 2 |
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| 26530. |
If bar(a),bar(b) and bar( c ) are unit vectors such that bar(a)+bar(b)+bar( c )=bar(0), then the value of bar(a).bar(b)+bar(b).bar( c )+bar( c ).bar(a)=………… |
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Answer» 1 |
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| 26531. |
The right hand derivative of f(x)= [x] tan (pi x) at a point x=7 is k pi then find the value of k. where [.] is the greatest integer function. |
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| 26532. |
Find the area of the triangle with vertices A(3,-1,2),B(1,-1,-3) and C(4,-3,1). |
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| 26534. |
Show that the opeation * on Q -{1} defined gby a*b=a+b-ab for all a, b in Q- {1} Satifies (i) the closure property (ii) the associative law (iii) the commutative alw (iv) what is theidentity element ? (v)for each a in Q-{1} find the inverse of a |
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Answer» Solution :(i) closure porperty Let a `in` Q-{1} and b in Q-{1} We know that Q is closed for addition substraction and MULTIPLICATION `therefore`a+b-ab in Q but a*b =1 `rarr` a+b-ab=1 `rarr` a(1-b)=(1-b)`rarra`=1 which is a contradication since 1 `in` Q-{1} therefore a*b `ne` 1 Thus a `in` Q-{1} ,b `in` Q-{1} `rarr` a*b in Q-{1} `therefore` * is a binary operation on Q-{1} (II) Associative law Let a,b,c =(a+b-ab)*c =(a+b-ab)+c-(a+b-ab)c =(a+b+c)-(ab+BC+ac)+bc And a*(b*c)=a*(b+c-bc) =a(b+c-bc)-a(b+c-bc) =(a+b+c)-(ab+bc+ac)+abc `therefore` (a*b)*c=a*(b*c) Hence * is associative (iii) commutative Law Let a,b `in`Q-{1} Then a*b=a+b-ab ltrbgt =b+a-ab =b*a (iv)Existence of identity element Let e be the identity element Then for all a in Q-{1} we have a* e =a `rarr` a + e - ae =a `rarr` e(1-a)=x `rarre=0 in` Q-{1} Nowa*0=a+0-ax0=a And 0*a=0+a-0xa=a ltrbgt Thus o is the identity element in Q-{1} Existence of inverse Let a in Q-{1} and `le t a^(-1)=b`then a*b=0 `rarra`+b-ab=0 `rarr` a=ab-b=(a-1)b `RARRB=(a)/(a-1)in Q-{1}` `therefore a^(-1)=(a)/(a-1)in Q-{1}` Thus each a in Q-{1} has its iverse in Q-{1} |
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| 26535. |
If A= {1,2,3...,n}and B ={ a,b}Then , the number of subjection from A into B is ......... |
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Answer» `""^(N)P_2` |
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| 26536. |
If x = (2)/(5) + (1.3)/(2!)((2)/(5))^(2) + (1.3.5)/(3!)((2)/(5))^(3) +…, then x + (1)/(x) = |
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Answer» `(1+sqrt(5))/(4)` |
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| 26538. |
If f(x) and g(x) are of degrees 7 and 4 respectively such that f(x)= g(x) q(x)+ r(x) then find possible degrees of q(x) and r(x). |
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| 26539. |
Let ** be the binary opertion on N defined by a **b=H.C.F. of a and b. Is ** commutative ? Is ** associative ? Does there exist identity for this binary opertion on N ? |
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| 26540. |
(i) Find the equation of the circle passing through (0,0) and making intercepts 4,3 on X axis and Y-axis respectively. (ii) Find the equation of the circle passing through (0,0) and making intercepts 6 units on X-axis and intercepts 4 units on Y-axis. |
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| 26541. |
Evaluate the following integrals: int_0^(pi/4) sin2xdx |
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Answer» Solution :`int_0^(pi/4) SIN 2x DX = ((-COS 2x)/2)_0^(pi/4)` = `-1/2(cos2x)_0^(pi/4)` =`-1/2 (cos pi/2 - cos 0)` =-1/2(0-1) = 1/2 |
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| 26542. |
On a stormy night ten guests came to dinner party and left their shoes outside the room in order to keep the carpet clean. After the dinner there was a blackout, and the guests leaving one by one, put on at random, any pair of shoes big enough for their feet, (Each pair of shoes stays together). Any guest who could not find a pair big enough spent the night there. What is the largest number of guests who might have had to spend the night there ? |
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| 26543. |
If a, b. c in R and 3b^(2)-8ac lt 0, then the equation ax^(4)+bx^(3)+cx^(2)+5x-7=0 has |
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Answer» all real roots `f'(x)=4ax^(3)+3BX^(2)+2cx+5` `f''(x)=12ax^(2)+6bx+2c=0` `implies "disc"=36b^(2)-4(12a)(2c)` `=12(3b^(2)-8ac) lt 0` `(:' 3b^(2) lt 8ac)` `f'(x)=0` has only one real root `implies f(x)=0` has exactly two real roots. |
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| 26544. |
If y = tan^(-1) [(5 cos x - 12 sin x)/(12 cos x + 5 sin x)] , then (dy)/(dx) is equal to |
| Answer» ANSWER :B | |
| 26545. |
Prove the following: int_0^1 x e^x dx = 1 |
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Answer» SOLUTION :`int_0^1 X E^x DX = [x e^x]_0^1 - int_0^1 1xxe^x dx` =`(1e^1-0)-(e^x)_0^1` = `e-(e^1-e^0) = e^0 = 1` |
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| 26546. |
If a gt 0" then " underset(x to oo)"Lt" ([ax+b])/(x)= |
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Answer» 0 |
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| 26547. |
The set of all real values of lambdafor which exactly two common tangents can be drawn to the circlex^(2) +y^(2) -4x -4y +6=0and x^(2) +y^(2) -10 x-10 y +lambda =0 is the interval |
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Answer» `( 12,32) ` |
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| 26548. |
If f(x) = (4x+3)/(6x-4) , x ne 2/(3) , show that fof (x) = x, for all x ne 2/(3) . What is the inverse of f ? |
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| 26549. |
Write the value of inte^x(In sin x+cotx)dx. |
| Answer» SOLUTION :`inte^x(In SIN x+COTX)dx=e^xIn(SINX)+C(1d/dx(Insinx)=cotx)` | |
| 26550. |
IFalpha, betaand gammaareanglessuchthattan alpha+ tanbeta+ tangamma=tanalphatan betatangammaandx=cosalpha+ i sinalpha, y = cosbeta +i sinbetaandz= cosgamma+ i sin gamma, thenxyz = |
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Answer» 1, butnot -1 |
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