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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.
| 99201. |
The point on the line `(x-2)/1=(y+3)/(-2)=(z+5)/(-2)`at a distance of 6 from the point `(2,-3,-5)`isa. `(3,-5,-3)`b. `(4,-7,-9)`c. `0,2,-1`d. none of theseA. `(3,-5,-3)`B. `(4,-7,-9)`C. `(0,1,-1)`D. `(-3,5,3)` |
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Answer» Correct Answer - B::C Direction cosines of the given line are `(1)/(3),-(2)/(3)-(2)/(3)` Hence, the equation of line can be a point in the form `(x-2)/((1)/(3))=(y+3)/((-2)/(3))=(z+5)/((-2)/(3))=r` Therefore, any point of the line is `(2+(r)/(3),-3-(2r)/(3),-5-(2r)/(3))` where `r=+-6` Points aer `(4,j-7,-9)` and `(0,1,-1)` |
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| 99202. |
If the lines `(x-2)/(1)=(y-3)/(1)=(z-4)/(lamda)` and `(x-1)/(lamda)=(y-4)/(2)=(z-5)/(1)` intersect thenA. `lamda=-1`B. `lamda=2`C. `lamda=-3`D. `lamda=0` |
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Answer» Correct Answer - A::D The two lines will intersect only if the shortest distance between them is zero. `implies(veca_(2)-veca_(1)).(vecb_(1)xxvecb_(2))=0` `|{:(1,-1,-1),(1,2,lamda),(lamda,2,1):}|=0` `implies(1-2lamda)+(1-lamda^(2))-1(2-lamda)=0` `implies-lamda^(2)-lamda=0` `implieslamda=0` or `lamda=-1` |
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| 99203. |
in the given figure `y=x^(2)+bx+c` is a quadratic polynomial which meets x-axis at A and B and y-axis at C Q is vertex of quadratic polynomial and foot of perpendicular from Q to x-axis and y-axis are P and R respectively. Then answer the following questions. Q. if two circles passing through A and B touches y-axis at C and R respectively thenA. `b^(2)=2c`B. `b^(2)=3c`C. `b^(2)=4c`D. `b^(2)=8c` |
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Answer» Correct Answer - C `OA.OB=OC^(2)` `alpha.beta=c^(2)` & `((b^(2)-4c)/(4))^(2)=alphabeta` `(b^(2)-4)^(2)=16` `c=c^(2)` `b^(2)-4=4` `c=1` `b^(2)=8` |
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| 99204. |
Cosider `f(x)=1-e^((1)/(x)-1)` Q. If `D` is the set of all real `x` such that `f(x)ge0` then `D` is equal toA. `(-infty,infty)`B. `(2,infty)`C. `(-infty,0)cup(1,infty)`D. `(-1,infty)` |
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Answer» Correct Answer - B `1-e^((1)/(x)-1)gt0` i.e., `e^((1)/(x)-1)lt1` i.e., `(1)/(x)-1lt0` i.e., `(x-1)/(x)gt0` i.e., `x in (-infty,0)cup(1,infty)` |
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| 99205. |
The quadratic equation `ax^(2)+bx+c=0` has real roots if:A. `alt-1,0ltclt1,bgt0`B. `alt-1,-1ltclt0,0ltblt1`C. `alt-1,clt0,bgt1`D. `agt0,bgt0,cgt0` |
| Answer» Correct Answer - A | |
| 99206. |
For a real number x let `[x]` denote the largest integer less than or equal to x and `{x}=x-[x]`. The possible integer value of n for which `int_(1)^(n)[x]{x}dx` exceeds 2013 isA. 63B. 64C. 90D. 91 |
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Answer» Correct Answer - D `int_(1)^(n)[x]{x}dx=int_(1)^(n)[x](x-[x])dx` `int_(1)^(2)(x-1)dx+int_(2)^(3)2(x-2)dx+int_(3)^(4)3(x-3)dx+int_(4)^(3)4(x-4)dx+…+int_(n-1)^(n)(n-1)(x-n+1)dx` `=(1)/(2)+2(1)/(2)+3((1)/(2))+4((1)/(2))+…..+((n-1))/(2)` `=(n(n-1))/(4)` Check by options as to which will make it exceed 2013. |
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| 99207. |
A homogeneous polynomial of the second degree in `n` variables i.e., the expression `phi=sum_(i=1)^(n)sum_(j=1)^(n)a_(ij)x_(i)x_(j)` where `a_(ij)=a_(ji)` is called a quadratic form in `n` variables `x_(1),x_(2)`….`x_(n)` if `A=[a_(ij)]_(nxn)` is a symmetric matrix and `x=[{:(x_(1)),(x_(2)),(x_(n)):}]` then `X^(T)AX=[X_(1)X_(2)X_(3) . . . .X_(n)][{:(a_(11),a_(12) ....a_(1n)),(a_(21),a_(22)....a_(2n)),(a_(n1),a_(n2)....a_(n n)):}][{:(x_(1)),(x_(2)),(x_(n)):}]` `=sum_(i=1)^(n)sum_(j=1)^(n)a_(ij)x_(i)x_(j)=phi` Matrix A is called matrix of quadratic form `phi`. Q. If number of distinct terms in a quadratic form is 10 then number of variables in quadratic form isA. 4B. 3C. 5D. can not found uniquely |
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Answer» Correct Answer - B If number of variable is `n` then there are `n` terms of the type `x_(i)^(2)` and `.^(n)C_(2)` terms of the type `x_(ij)`. Hence total number of distinct terms `=n+.^(n)C_(2)=10` `impliesn+(n(n-1))/(2)=10impliesn(n+1)=20impliesn=4` |
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| 99208. |
For the A.P given by `a_(1),a_(2). . .a_(n)` . . . ., the equation satisfied is/areA. `a-(1)+2a_(2)+a_(3)=0`B. `a_(1)+2a_(2)+a_(3)=0`C. `a_(1)+3a_(2)-3a_(3)-a_(4)=0`D. `a_(1)-4a_(2)+6a_(3)-4a_(4)+a_(5)=0` |
| Answer» Correct Answer - B::D | |
| 99209. |
All possible roots of polynomial with integral coefficients can be identified by "RATIONAL ROOT TEST" according to rational root test if a plynomial ltbr. `a_(n)p^(n)+a_(n-1)p^(n-1)q+a_(n-2)p^(n-2)q^(2)+`…..`+a_(2)p^(2)q^(n-2)+a_(1)pq^(n-1)+a_(0)q^(n)=0` Every term in above equation except possible the last one is divisible by `p` hence `p` should also divide `a_(0)q^(n)`, since `q` and `p` are relatively prime `p` must divide `a_(0)`. Similarly `q` also divides `a_(n)`. Now consider an equation `6x^(5)-19x^(4)-9x^(3)-16x^(2)+9x-1=0` and answer the following questions Q. if given equation and equation `x^(3)+ax^(2)+ax+1=0` have two roots common then possible value(s) of is/are?A. `0`B. `-2`C. `-3`D. `3` |
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Answer» Correct Answer - D (13 to 14) `6x^(5)-19x^(4)-9x^(3)-16x^(2)+9x-1=0` rational roots of equation `x=+-1,+-(1)/(2),+-(1)/(2),+-(1)/(6)` are possible `x=(1)/(6)` satisfies the equation `6x^(5)-x^(4)-18x^(4)+3x^(3)-12x^(3)+2x^(2)-18x^(2)+3x+6x-1=0` `(6x-1)(x^(4)-3x^(3)-2x^(2)-3x+1)=0` `x^(4)-3x^(3)-2x^(2)-3x+1)=0` `x^(4)-3x^(3)-2x^(2)-3x+1=0` `x^(2)+(1)/(x^(2))-3(x+(1)/(x))-2=0` `(x+(1)/(x))^(2)-3(x+(1)/(x))-4=0` `t^(2)-3t-4=0` `(t-4)(t+1)=0` `x+(1)/(x)=4` or `x+(1)/(x)+1=0` `x^(2)-4x+1=0` `x^(2)+x+1=0` sum of real roots will be `4+(1)/(6)=(25)/(6)` for rational values of a roots of cubic `x^(2)+ax^(2)+ax+1=0` will be in conjugate pair so either `x^(2)-4x+1=0` or `x^(2)+x+1=0` is a factor of `x^(3)+ax^(2)+1=0` if `x^(2)-4x+1=0` is a factor of `x^(3)+ax^(2)+ax+1=0` then `a=-3` and `x^(2)+x+1=0` is a factor of `x^(3)+ax^(2)+ax+1=0` then `a=2` |
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| 99210. |
A homogeneous polynomial of the second degree in `n` variables i.e., the expression `phi=sum_(i=1)^(n)sum_(j=1)^(n)a_(ij)x_(i)x_(j)` where `a_(ij)=a_(ji)` is called a quadratic form in `n` variables `x_(1),x_(2)`….`x_(n)` if `A=[a_(ij)]_(nxn)` is a symmetric matrix and `x=[{:(x_(1)),(x_(2)),(x_(n)):}]` then `X^(T)AX=[X_(1)X_(2)X_(3) . . . . .X_(n)][{:(a_(11),a_(12) ....a_(1n)),(a_(21),a_(22)....a_(2n)),(a_(n1),a_(n2)....a_(n n)):}][{:(x_(1)),(x_(2)),(x_(n)):}]` `=sum_(i=1)^(n)sum_(j=1)^(n)a_(ij)x_(i)x_(j)=phi` Matrix A is called matrix of quadratic form `phi`. Q. The quadratic form of matrix `A[{:(0,2,1),(2,3,-5),(1,-5,8):}]` isA. `3x_(2)^(2)+8x_(3)^(2)+2x_(1)x_(2)+x_(1)x_(3)-5x_(2)x_(3)`B. `3x_(2)^(2)+8x_(3)^(2)+4x_(1)x_(2)+2x_(1)x_(3)-10x_(3)x_(2)`C. `x_(1)^(2)+2x_(2)^(2)+x_(3)^(2)+3x_(1)x_(2)-5x_(2)x_(3)+8x_(1)x_(2)`D. `3x_(1)^(2)+8x_(2)^(2)+4x_(1)x_(2)+2x_(1)x_(3)+10x_(3)x_(2)` |
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Answer» Correct Answer - A `=[x_(1)x_(2)x_(3)][{:(2x_(2)+x_(3)),(2x_(1)+3x_(2)-5x_(3)),(x_(1)-5x_(2)+8x_(3)):}]` `=2x_(1)x_(2)+x_(1)x_(3)+2x_(1)x_(2)+3x_(2)^(2)-5x_(3)x_(2)+x_(3)x_(1)+5x_(3)x_(2)+8x_(3)^(2)` `=3x_(2)^(3)+8x_(3)^(2)+4x_(1)x_(2)+2x_(1)x_(3)-10x_(2)x_(3)` |
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| 99211. |
Suppose a series of `n` terms given by `S_(n)=t_(1)+t_(2)+t_(3)+`. . ..`+t_(n)` then `S_(n-1)=t_(1)+t_(2)+t_(3)+` . . . `+t_(n-1),nge1` subtracting we get `S_(n)-S_(n-1)=t_(n),nge2` surther if we put `n=1` is the first sum then `S_(1)=t_(1)` thus w can write `t_(n)=S_(n)-S_(n-1),nge2` and `t_(1)=S_(1)` Q. if the sum of `n` terms of a series `a.2^(n)-b` then the sum `sum_(r=1)^(infty)(1)/(t_(r))` isA. `a`B. `(a)/(2)`C. `(2)/(a)`D. `(1)/(a)` |
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Answer» Correct Answer - C `t_(1)=S_(1)=2a-b` `t_(n)=S_(n)-S_(n-1)=a.2^(n)-a.2^(n-1)=a.2^(n-1),ngt1` Now `(t_(n+1))/(t_(n))=2,nge2` `sum_(r=2)^(infty)(1)/(t_(r))=(1)/(a)sum_(r=2)^(infty)(1)/(2^(r-1))=(1)/(a)(1//2)/(1-1//2)=(1)/(a)` |
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| 99212. |
Which of the following figure represents are correct solution of the equation `sin^(-1)xcos^(-1)y=0`?A. B. C. D. |
| Answer» Correct Answer - A | |
| 99213. |
All possible roots of polynomial with integral coefficients can be identified by "RATIONAL ROOT TEST" according to rational root test if a plynomial ltbr. `a_(n)p^(n)+a_(n-1)p^(n-1)q+a_(n-2)p^(n-2)q^(2)+`…..`+a_(2)p^(2)q^(n-2)+a_(1)pq^(n-1)+a_(0)q^(n)=0` Every term in above equation except possible the last one is divisible by `p` hence `p` should also divide `a_(0)q^(n)`, since `q` and `p` are relatively prime `p` must divide `a_(0)`. Similarly `q` also divides `a_(n)`. Now consider an equation `6x^(5)-19x^(4)-9x^(3)-16x^(2)+9x-1=0` and answer the following questions Q. Sum of real roots of equation will beA. 4B. `(25)/(6)`C. `(1)/(6)`D. can not determine |
| Answer» Correct Answer - C | |
| 99214. |
Suppose a series of `n` terms given by `S_(n)=t_(1)+t_(2)+t_(3)+`. . . .`+t_(n)` then `S_(n-1)=t_(1)+t_(2)+t_(3)+`. . . .`+t_(n-1),nge1` subtracting we get `S_(n)-S_(n-1)=t_(n),nge2` surther if we put `n=1` is the first sum then `S_(1)=t_(1)` thus w can write `t_(n)=S_(n)-S_(n-1),nge2` and `t_(1)=S_(1)` Q. The sum of `n` terms of a series is `a.2^(n)-b`. where a and b are constant then the series isA. A.PB. G.PC. A.G.PD. G.P from second term onwards |
| Answer» Correct Answer - D | |
| 99215. |
Give an example of co-prime numbers. |
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Answer» We know that two number a and b are called co-prime numbers if HCF (a,b) = 1. Since, 9 and 16 have HCF (9,16) = 1, therefore 9 and 16 are co-prime numbers. |
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| 99216. |
The probability that the graph of `y=16x^2 +8(a +5)x-7a-5=0,` is strictly above the x-axis, If `a in [-20,0]`A. `14/20`B. `14/21`C. `13/21`D. `13/20` |
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Answer» Correct Answer - D `16x^(2)+8(a=5)x-7a-5gt0AAxepsilonR` `:.Dlt0` `implies64(a+5)^(2)-4.16(-7a-5)lt0` `implies(a+15)(a+2)lt0` `implies-15ltalt-2` `implies` Required probability `=13/20` |
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| 99217. |
The number of values of `k` for which the system of equations: `kx+(3k+2)y=4k` `(3k-1)x+(9k+1)y=4(k+1)` has no solution, areA. `0`B. `1`C. `2`D. `3` |
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Answer» Correct Answer - A `cancel (9k^(2))+k=cancel(9k^(2))+3k-2` `k=1` but it is not possible because at `k=1` there will be infinite solution. |
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| 99218. |
Equation of a plane parallel to the plane z = 3 is :(A) x = 3 (B) y = 3 (C) z = 0 (D) y = -3 |
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Answer» Option : (C) z = 0 |
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| 99219. |
A function `f` from integers to integers is defined as `f(x)={(n+3, n "is odd"),(n/2 , n "is even"):}`. If `k` is an odd integer and `f(f(f(k)))=27` then the sum of digits of `k` isA. `3`B. `6`C. `9`D. `12` |
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Answer» Correct Answer - B `kepsilon` odd `f(k)=k+3` even `f(f(k))=(k+3)/2` Case 1 if `(k+3)/2epsilon` odd `27=(k+3)/2+3impliesk=45` (not possible) Case 2 if `(k+3)/2 epsilon` even `27=f(f(f(k)))=f((k+3)/2)=(k+3)/4` `implies k=105` sum digits `=6` |
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| 99220. |
In a city no two persons have identical set ofteeth and there is no person without a tooth. Also no person has more than 32teeth. If we disregard the shape and size of tooth and consider only thepositioning of the teeth, the maximum population of the city isa. `2^(32)`b. `(32)^2-1`c. `2^(32)-1`d. `2^(32-1)`A. `2^(32)`B. `2^(32)+1`C. `2^(32)-1`D. `2^(32)-2` |
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Answer» Correct Answer - C There are `32` places for the teeth in the mouth for each place we have two choices either there is a tooth or there is on tooth. `implies` choices of `32` teeth `e=2^(32)` which include without teeth (each choice has no tooth) but given no person without a tooth `implies` maximum population of the city `=2^(32)-1` |
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| 99221. |
Equation of a plane parallel to the plane 2x + y – z = 11 is :(A) x + y – z = 11 (B) 2x + y –z = 7 (C) 2x – y + z = 1 (D) None of these |
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Answer» Option : (B) 2x + y –z = 7 |
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| 99222. |
Number of normal(s) that can be drwan through the point `(sqrt(2,2))` to the ellipse `(x^(2))/(2)+(y^(2))/(1)=1` is / are |
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Answer» Correct Answer - A `T=S_(1)rarr3rarr3xh-2yk+2(x+h)-3(y+k)=3h^(2)-2k^(2)+4h-6k` Slope `=2implies(-3h-2)/(-2k-3)=2implies-3n-2=-4y-6implies3n-4y=14` |
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| 99223. |
Line `x-y=1` intersect the parabola `y^(2)=4x` at A and B. Normals at A and B intersect at C. If D is the point at which line CD is normal to the parabola, then coordinate of point D isA. `(4,-4)`B. (4,4)C. `(-4,-4)`D. None of these |
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Answer» Correct Answer - D `(Sigmax_(i))/(10)=2" and "sqrt((Sigma(x_(i)-2)^(2))/(10))=3` `(Sigma(x_(i)+1)^(2))/(10)=(1)/(10).Sigma(x_(1)-2+3)^(2)=(1)/(10).[Sigma(x_(i)-2)^(2)+6.Sigma(x_(i)-2)+3^(2).(10)]` `=(1)/(10).[90+6.(20-20)+90]=18` |
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| 99224. |
The extremities of the diagonal of a rectangle are `(-4, 4)` and `(6,-1)`. A circle circumscribes the rectangle and cuts inercept of length `AB` on the `y`-axis. The length of `AB` isA. `11`B. `12`C. `13`D. `14` |
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Answer» Correct Answer - A Since circle circumscribes the rectangle and so the its diagonal with vertices `(-4,4)` and `(6,-1)` will act as diameter of the circle. Hence its equation is `implies(x+4)(x-6)+(y-4)(y+1)=0` `impliesx^(2)+y^(2)-2x-3y-2=0` length of `AB=2sqrt((3/2)^(2)+28)=11` |
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| 99225. |
Equation of a plane perpendicular to the plane x + 5y + 11z = 7 is :(A) x + y + z = 3 (B) x + 5y + 11z = 1 (C) x + 2y – z = 3 (D) None of these |
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Answer» Option : (C) x + 2y – z = 3 |
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| 99226. |
Length of common chord of the curve `y^(2)-4x-4=0 " and "4x^(2)+9y^(2)=36` isA. 1B. 2C. 4D. 8 |
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Answer» Correct Answer - B Intersection point of `x+y+4=0 & 2x-y+7=0`. |
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| 99227. |
\(\begin{vmatrix}2& 5 & 8 \\[0.3em]-7& 13 & 19 \\[0.3em]14 &-26 & -38\end{vmatrix}\) =(A) 143 (B) -298 (C) 0 (D) 1|(2,5,8),(-7,13,19),(14,-26,-38)| = |
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Answer» Option : (C) 0 |
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| 99228. |
\(\begin{vmatrix} 3& 5 & 17 \\[0.3em] 6& 7 & 31 \\[0.3em] 2&3 & 11 \end{vmatrix}\) =(A) 1025 (B) -1940 (C) 0 (D) 2160|(3,5,7),(6,7,31),(2,3,11)| = |
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Answer» Option : (C) 0 |
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| 99229. |
\(-2\begin{bmatrix}1&-1 \\[0.3em]3 & 5\end{bmatrix}=\)(A) \(\begin{bmatrix}-2&-1 \\[0.3em]3 & 5\end{bmatrix}\) (B) \(\begin{bmatrix}-2&-2 \\[0.3em]3 & 5\end{bmatrix}\) (C) \(\begin{bmatrix}-2&2 \\[0.3em]3 & 5\end{bmatrix}\)(D) \(\begin{bmatrix}-2&2 \\[0.3em]-6 & -10\end{bmatrix}\) -2[(1,-1),(3,5)| = |
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Answer» Option : (D) \(\begin{bmatrix}-2&2 \\[0.3em]-6 & -10\end{bmatrix}\) |
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| 99230. |
\(4\begin{vmatrix} 2&-2 \\[0.3em] 1 & 0 \end{vmatrix}=\) (A) \(\begin{vmatrix} 8&-2 \\[0.3em] 1 & 0 \end{vmatrix}\) (B) \(\begin{vmatrix} 8&-8 \\[0.3em] 4 & 0 \end{vmatrix}\) (C) \(\begin{vmatrix} 8&-2 \\[0.3em] 4 &1 0 \end{vmatrix}\) (D) \(\begin{vmatrix} 8&-8 \\[0.3em] 1 & 0 \end{vmatrix}\) 4|(2,-2),(1,0)| = |
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Answer» Option : (D) \(\begin{vmatrix} 8&-8 \\[0.3em] 1 & 0 \end{vmatrix}\) |
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| 99231. |
The equation of common tangent to the parabola `y^2 =8x` and hyperbola `3x^2 -y^2=3` is |
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Answer» `a^2=1` so, `a=+-1` `b^2=3` so,`b=+-sqrt3` equation of tangent to hyperbola `y=mx+-sqrt(a^2m^2-b^2)` `y=mx+-sqrt(m^2-3)` -(1) equation of parabola`y^2=8x` so, 4a=8 a=2 equation of tangent to parabola `=y=mx+a/m` `=y=mx+2/m` -(2) tangent is common so equation(1) and (2) will be equal `mx+-sqrt(m^2-3)=mx+2/3` `+-2/m=sqrtm^2-3` squaring both side `4/m^2=m^2-3` `4=m^4-3m2` (t-4)(t+1) t=4,-1 but -1 is not possible `m^2=4` `m=+-2` equation of tangent `y=mx+2/m` when m=2 `y=2x+1` whrn m=2 y+-2x-1 |
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| 99232. |
State the two de'morgans laws for all sets A and B and proof them |
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Answer» The complement of the union of two sets is equal to the intersection of their complements and the complement of the intersection of two sets is equal to the union of their complements. These are called De Morgan's laws. For any two finite sets A and B; (i) (A U B)' = A' ∩ B' (which is a De Morgan's law of union). |
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| 99233. |
\(\begin{bmatrix}1 & 0 \\[0.3em]0 &1 \\[0.3em]\end{bmatrix}\)\(\begin{bmatrix}11 \\[0.3em]17\\[0.3em]\end{bmatrix}\)=(A) \(\begin{bmatrix}11 & 0 \\[0.3em]17 &1 \\[0.3em]\end{bmatrix}\) (B) [11 17](C) \(\begin{bmatrix}11 \\[0.3em]17\\[0.3em]\end{bmatrix}\)(D) \(\begin{bmatrix}19 \\[0.3em]25\\[0.3em]\end{bmatrix}\)[(1,0),(0,1)] [11 17] = |
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Answer» Option : (C) \(\begin{bmatrix}11 \\[0.3em]17\\[0.3em]\end{bmatrix}\) |
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| 99234. |
\(\begin{vmatrix}2&-3 & 5 \\[0.3em]6 & 0 &4 \\[0.3em]3 & 15&-21\end{vmatrix}\) =(A) 12 (B) 84 (C) -12 (D) -84|(2,-3,5),(6,0,4),(3,15,-21) = |
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Answer» Option : (D) -84 |
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| 99235. |
d/dx[cot-1(tanx)] =(A) 1 (B) -1(C) \(\frac{\pi}{2}-x\) (D) \(-\frac{tanx}{1+x^2}\) |
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Answer» Option : (B) -1 |
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| 99236. |
If \(\begin{bmatrix}3 & 5 \\[0.3em]7 & 9\end{bmatrix}\) then adjoint of X = (A) \(\begin{bmatrix}-3 & 5 \\[0.3em]-7 & 9\end{bmatrix}\)(B) \(\begin{bmatrix}9 & -5 \\[0.3em]-7 & 3\end{bmatrix}\) (C) \(\begin{bmatrix}-9 & 5 \\[0.3em]7 & -3\end{bmatrix}\)(D) None of theseX = [(3,5),(7,9)] |
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Answer» Option : (B) \(\begin{bmatrix}9 & -5 \\[0.3em]-7 & 3\end{bmatrix}\) |
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| 99237. |
If x = acos2θ, y = asin2θ then dy/dx is equal to :(A) 1 (B) -1(C) tan2θ(D) - tan2θ |
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Answer» Option : (B) -1 |
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| 99238. |
d2/dx2(20x3 + 7x) =(A) 60x2 (B) 60x2 + 7 (C) 120 (D) 120x |
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Answer» Option : (D) 120x |
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| 99239. |
If `(1!)^(2) + (2!)^(2) + (3!)^(2) + "…….." + (99!)^(2)` is divided by `100`, the remainder isA. 27B. 28C. 17D. 14 |
| Answer» Correct Answer - A | |
| 99240. |
`cot16^0cot44^0+cot44^0cot76^0-cot76^0cot16^0=`1 (b) 2(c) 3 (d)4 |
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Answer» Correct Answer - 3 Value of cot ………… `cot 16^(@)cot44^(@)`………. `(tan76^(@)+tan16^(@)-tan44^(@))/(tan16^(@)tan44^(@)tan76^(@))` `= (3tan48^(@))/(tan48^(@)) = 3` `(cos16cos44)/(sin16.sin44)-1+(cos44.cos76)/(sin44sin76)` Aliter `-1-(cos76cos16)/(sin76.sin16)-1+3` `= (cos60)/(sin16.sin44)+(cos120)/(sin44sin76) - (cos60)/(sin76.sin16)+3` `=(1)/(2)((sin76-sin16)/(sin16.sin44.sin76))-(1)/(2sin76.sin16)+3=3`. |
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| 99241. |
If x = at4, y = 2at2 then dy/dx is equal to :(A) t2(B) \(\frac{1}{t^2}\) (C) t (D) None of these |
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Answer» Option : (B) \(\frac{1}{t^2}\) |
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| 99242. |
What is Disporportionation reaction |
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Answer» A reaction in which both oxidation and reduction occur simultaneously is called a disproportionation rxn
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| 99243. |
The output current of a 60 percent modulated AM generator is 1.5 A. To what value will this current rise if the generator is modulated additionally by another audio wave whose modulation index is 0.7? What will be the percentage power saving if the carrier and one of the sidebands are now suppressed? |
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Answer» We know that Ic = \(\cfrac{I}{\sqrt{1+\frac{m_1^2}2}}\) Ic = \(\cfrac{1.5}{\sqrt{1+\frac{(0.6)^2}2}}\) Ic = \(\frac{1.5}{\sqrt{1+0.18}}\) Ic = \(\frac{1.5}{\sqrt{1.18}}\) Inet = Ic\(\sqrt{1+\frac{m_1^2}2+\frac{m_2^2}2}\) Inet = \(\frac{1.5}{\sqrt{1.18}}\)\(\sqrt{1+0.18+0.245}\) Inet = \(\frac{1.5}{1.086}\)\(\sqrt{1.425}\) Inet = 1.38 x 1.19 Inet = 1.64 A |
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| 99244. |
An oscillation magnetometer is used to measure the magnetic moment of a bar magnet. Assume that the horizontal component of the earth’s magnetic field and the time period of oscillation are known. In addition, which of the following quantities need to be known: m = mass of magnet, l = length, b = width, h = height. (a) m only (b) m and l (c) m, l and b (d) m, l, b and h |
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Answer» Correct Answer is: (c) m, l and b The moment of inertia of the magnet has to be found. |
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| 99245. |
During classroom discussion a student said that sponges are more complex than cnidarians. Do you agree with him. Justify. |
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Answer» NO. Sponges are asymmetrical and body is formed of loose aggregate of cells. Cells are not organised to from tissues and organs. Cnidarians are radially symmetrical and tissue grade of organisation. So cnidarians are more complex than sponges. |
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| 99246. |
Suppose during your field visit for collection from a rocky seashore you have got some live specimens such as sea urchin, sea cucumber, sea anemone. Is it possible to keep them on an aquarium in your school. Give reasons for your answer. |
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Answer» No. It is not possible. Marine animals cannot live on freshwater because it leads to endosmosis and death occurs. |
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| 99247. |
In the circuit shown above, when the switch S is closed, (a) no charge flows through S (b) charge flows from A to B (c) charge flows from B to A (d) charge flows initially from A to B and later from B to A |
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Answer» Correct Answer is: (a) no charge flows through S For two conductors in series, their potential differences are proportional to their resistances. For two capacitors in series, their potential differences are inversely proportional to their capacitances. Hence, A and B are at the same potential and no charge will flow between them. |
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| 99248. |
1. Why nematocysts are more concentrated on the oral end and tentacles of cnidarians? 2. What are the difficulties that coelenterate have to face if nematocysts were absent in body. |
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Answer» 1. Nematocysts are concerned with defence and offence. Tentacles are usually used for defence, offence and food collection. Hence nematocysts are more concentrated in the oral end. 2. The major difficulties cnidarians has to face in the absence of nematocysts are for food collection and escaping from enemies. |
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| 99249. |
During a classroom discussion a student said that sponges are more complex than cnidarians. Do you agree with him justify. |
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Answer» 1. Sponges: Cellular grade of organisation and cell aggregate body plan. 2. Cnidarians: Tissue level of organisation and blind sac body plan. So Cnidarians are more complex than sponges. |
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| 99250. |
The horizontal range of a projectile is R and the maximum height attained by it is H. A strong wind now begins to blow in the direction of the motion of the projectile, giving it a constant horizontal acceleration = g/2. Under the same conditions of projection, the horizontal range of the projectile will now be(a) R + H/2.(b) R + H (c) R + 3H/2 (d) R + 2H |
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Answer» Correct Answer is: (d) R + 2H For the projectile, R = u2 sin 2θ/g, H = u2 sin2 θ/ 2g Time of flight = t = 2u sin θ/g. Initial horizontal velocity = u cos θ. When horizontal acceleration g/2 is present, the horizontal range = (u cos θ)t + 1/2 (g/2) t = (u cos θ) 2u sin θ/g + g/4 (4u2 sin2 θ/g2) = R + 2H. |
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