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201.

Evaluate:`|(a+i b, c+i d),(-c+i d, a-i b)|`A. `(a^(2) +b^(2)-c^(2)-d^(2))`B. `(a^(2)-b^(2)+c^(2)-d^(2))`C. `(a^(2)+b^(2) +c^(2)+d^(2))`D. none of these

Answer» Correct Answer - C
`Delta = (a+ib)(a-ib) + (c-"id")(c+"id") = (a^(2) + b^(2) +c^(2) +d^(2))`
Discussion
202.

`|["cos"15^(@), "sin"15^(@)], ["sin"15^(@), "cos"15^(@)]|=?`A. 1B. `(1)/(2)`C. `(sqrt(3))/(2)`D. none of these

Answer» Correct Answer - C
`Delta = ("cos"^(2)15^(@) - "sin"^(2)15^(@)) = "cos"( 2 xx 15^(@)) = "cos"30^(@) = (sqrt(3))/(2)`
Discussion
203.

`|[x+1, 3, 5], [2, x+2, 5], [2, 3, x+4]|=0`

Answer» Apply `C_(1) to C_(1) + C_(2) + C_(3)` and take (x+9) common from `C_(1).`
Discussion
204.

Prove that`|[b+c,a,b],[c+a,c,a],[a+b,b,c]|=(a+b+c)(a-c)^2`

Answer» Let the given determinant be `Delta`. Then,
`Delta = |{:(b+c, a, b), (c+a, c, a), (a+b, b, c):}|`
`= |{:(b, a, b), (c, c, a), (a, b, c):}| + |{:(c, a, b), (a, c, a), (b, b, c):}|`
`= |{:(a+b+c, a+b+c, a+b+c), (" "c, " "c, " "a), (" "a, " "b, " "c):}| + |{:(a+b+c, a+b+c, a+b+c), (" "a, " "c, " "a), (" "b, " "b, " "c):}| [R_(1) to (R_(1) + R_(2) + R_(3))" in each determinant"]`
`=(a+b+c) *|{:(1, 1, 1), (c, c, a), (a, b, c):}| + (a+b+c) * |{:(1, 1, 1), (a, c, a),(b, b,c):}| ["taking out(a+b+c) common from"R_(1) "in each determinant"]`
`=(a+b+c) * |{:(1, " "0, " "0), (c, " "0, a-c), (a, b-a, c-a):}| +(a+b+c)* |{:(1, " "0, 0), (a, c-a, 0), (b, " "0, c-b):}| [C_(1) to (C_(2) - C_(1)) "and "C_(3) to (C_(3)-C_(1))"in each"]`
`=(a+b+c) *1* |{:(0, a-c), (b-a, c-a):}| +(a+b+c)*1* |{:(c-a, " "0), (" "0, c-b):}| ["each det. expanded by"R_(1)]`
`=(a+b+c)*[0-(b-a)(a-c)] + (a+b+c)(c-a)(c-b)`
`=(a+b+c)(a-b)(a-c) - (a+b+c) (a-c)(c-b)`
`=(a+b+c)(a-c){(a-b)-(c-b)}`
`=(a+b+c)(a-c)^(2)`,
Hence, `Delta = (a+b+c)(a-c)^(2)`
Discussion
205.

The solution set of the equation `|[x, 3, 7], [2, x, 2], [7, 6, x]|=0` is

Answer» Apply `R_(1) to R_(1) + R_(2) + R_(3)` and take (x+9) common from `R_(1).`
Discussion
206.

Prove that ` |[a,b-c,c+b],[a+c,b,c-a],[a-b,a+b,c]|=(a+b+c)(a^2+b^2+c^2) `

Answer» Apply `C_(1) to aC_(1) "and divide"Delta "by a`
`therefore Delta = (1)/(a) * |[a^(2), b-c, c+b], [a^(2)+ac, b, c-a], [a^(2)-ab, b+a, c]|`
` = (1)/(a) |[a^(2)+b^(2)+c^(2), b-c, c+b], [a^(2) +b^(2)+c^(2), b, c-a], [a^(2) +b^(2)+c^(2), b+a, c]| ["applying "C_(1) to C_(1) + bC_(2) +cC_(3)]`
Discussion
207.

The number of real roots of the equation ` | (x,-6,-1), (2,-3x,x-3), (-3, 2x, x-2) | =0 ` is(i) `0`(ii) `1`(iii) `2`(iv) `3`

Answer» Apply `R_(1) to R_(1) -R_(2)` and take (x-2) common from `R_(1).`
Discussion
208.

`=|a a^2 0 1 2a+b(a+b)0 1 2a+3b|`is divisible by`a+b`b. `a+2b`c. `2a+3b`d. `a^2`A. `a+b `B. `a+2b`C. `2a+3b`D. `a^(2)`

Answer» Correct Answer - A::B
`Delta = |{:(a,,a^(2),,0),(1,,2a+b,,(a+b)),(0,,1,,2a+2b):}|`
`=a |{:(1,,a,,0),(1,,(2a+b),,(a+b)),(0,,1,,2a+3b):}|`
`=a(a+b) |{:(1,,a,,0),(0,,1,,(a+b)),(0,,1,,2a+3b):}|" "(R_(2) to R_(2) -R_(1))`
`=a(a+b) (2a+3b-a-b)`
`=a(a+b) (a+2b)`
Discussion
209.

the roots of the equations `|{:(.^(x)C_(r),,.^(n-1)C_(r),,.^(n-1)C_(r-1)),(.^(x+1)C_(r),,.^(n)C_(r),,.^(n)C_(r-1)),(.^(x+2)C_(r),,.^(n+1)C_(r),,.^(n+1)C_(r-1)):}|=0`A. `x=n`B. `x=n+1`C. `x=n-1`D. `x=n-2`

Answer» Correct Answer - A::C
`|{:(.^(x)C_(r),,.^(n-1)C_(r),,.^(n)C_(r)),(.^(x+1)C_(r),,.^(n)C_(r),,.^(n+1)C_(r)),(.^(x+2)C_(r),,.^(n+1)C_(r),,.^(n+2)C_(r)):}|=0`
`|{:((x!)/(r!(x-r!)),,((n-1)!)/(r!(n-r-1)!),,(n!)/(r!(n-r)!)),(((x+1)!)/(r!(x+1-r)!),,(n!)/(r!(n-r)!),,((n+1)!)/(r!(n-r+1)!)),(((x+2)!)/(r!(x+2-r)!),,((n+1)!)/(r!(n+1-r)!),,((n+2)!)/(r!(n-r+2)!)):}|=0`
Taking `(x!)/(r!(x-r)!)` common from `C_(1)` we have quadratic equations in x.
Now in (1) if we put `x=n-1,C_(1)` and `C_(2)` are the same hence x=n-1 is one root of the equation.
If we put x=n then `C_(1) `and `C_(3)` are same .hence x=n is the other root.
Discussion
210.

If `|(a^2,b^2,c^2),((a+b)^2 ,(b+1)^2,(c+1)^2),((a-1)^2 ,(b-1)^2,(c-1)^2)| =k(a-b)(b-c)(c-a)` then the value of k is a. 4 b. -2 c.-4 d. 2

Answer» Correct Answer - `k=-4`
`" Let " |{:(a^(2),,b^(2),,c^(2)),((a+1)^(2),,(b+1)^(2),,(c+1)^(2)),((a-1)^(2),,(b-a)^(2),,(c-1)^(2)):}|`
`= 4 |{:(a^(2),,b^(2),,c^(2)),(a,,b,,c),((a-1)^(2),,(b-1)^(2),,(c-1)^(2)):}|`
[Applying `R_(2) to R_(2)-R_(3),` then taking 4 common from `R_(2)]`
`=4 |{:(a^(2),,b^(2),,c^(2)),(a,,b,,c),(1,,1,,1):}| [R_(3) to R_(3)-R_(1)+2R_(2)]`
`=-4 |{:(1,,1,,1),(a,,b,,c),(a^(2),,b^(2),,c^(2)):}|`
`=- 4(a-b)(b-c)(c-a)`
Hence `k=-4.`
Discussion
211.

If `a ,b ,c in R ,`then findthe number of real roots of the equation `=|x c-b-c x a b-a x|=0`

Answer» from the symmetry of the determinant it is easier to expand by sarrus rule.
`Delta =x^(3) +abc -abc + (b^(2)x +a^(2) + a^(2)x+ c^(2)x )=0`
` " or " x^(3) + x( a^(2) + b^(2) + c^(2))=0`
`rArr x=0 " or " x^(2) = -(a^(2) +b^(2) + c^(2))`
` rArr x=0 " or " x= +- i sqrt( a^(2) + b^(2) +c^(2))`
Discussion
212.

Suppose f(x) is a function satisfying the following conditions : (i) f(0)=2,f(1)=1 (ii) f has a minimum value at `x=5//2` (iii) for all `x,f (x) = |{:(2ax,,2ax-1,,2ax+b+1),(b,,b+1,,-1),(2(ax+b),,2ax+2b+1,,2ax+b):}|` The value of f(2) isA. `1//4`B. `1//2`C. `-1`D. `3`

Answer» Correct Answer - B
`f(x) = |{:(2ax,,2ax-1,,2ax+b+1),(b,,b+1,,-1),(2(ax+b),,2ax+2b+1,,2ax+b):}|`
Applying `(C_(1) to C_(1)-C_(3) ,C_(2) to C_(2) -C_(3)`
`f(x) =|{:(-(b+1),,-(b+2),,2ax+b+1),((b+1),,(b+2),,-1),(b,,b+1,,2ax+b):}|`
Applying `R_(1) to R_(1)+R_(2)" and " R_(3) to R_(3)-R_(2)` we get
`f(x) = |{:(0,,0,,2ax+b),(b+1,,b+2,,-1),(-1,,-1,,2ax+b+1):}|`
`=(2ax +b) [-b-1+b+2]`
`:. f(x) =2ax+b`
`:. f(x) = ax^(2) +bx +c`
`f(0) =2 rArr c=2`
`f(1) =1rArr a+b+2=1rArr a+b =1`
`f(5//2) =0 rArr 5a+ b=0`
`rArr a=1//4 ,b=-5//4`
hence `f(x) =(1)/(4) x^(2) -(5)/(4)x+2`
Clearly . discriminant (D) of the equation f(x) =0 is less than 0.
hence f(x)=0 has imaginary roots .Also f(2) `=1//2` .and minimum value of f(x) is
`-((25)/(16)-4.(1)/(4) (2))/(4.(1)/(4)) =(7)/(16)`
Hence range of the f(x) is `[(7)/(16),oo)`
Discussion
213.

Suppose f(x) is a function satisfying the following conditions : (i) f(0)=2,f(1)=1 (ii) f has a minimum value at `x=5//2` (iii) for all `x,f (x) = |{:(2ax,,2ax-1,,2ax+b+1),(b,,b+1,,-1),(2(ax+b),,2ax+2b+1,,2ax+b):}|` f(x)=0 hasA. both roots positiveB. both roots negativeC. roots of opposite signD. imaginary roots

Answer» Correct Answer - D
`f(x) = |{:(2ax,,2ax-1,,2ax+b+1),(b,,b+1,,-1),(2(ax+b),,2ax+2b+1,,2ax+b):}|`
Applying `(C_(1) to C_(1)-C_(3) ,C_(2) to C_(2) -C_(3)`
`f(x) =|{:(-(b+1),,-(b+2),,2ax+b+1),((b+1),,(b+2),,-1),(b,,b+1,,2ax+b):}|`
Applying `R_(1) to R_(1)+R_(2)" and " R_(3) to R_(3)-R_(2)` we get
`f(x) = |{:(0,,0,,2ax+b),(b+1,,b+2,,-1),(-1,,-1,,2ax+b+1):}|`
`=(2ax +b) [-b-1+b+2]`
`:. f(x) =2ax+b`
`:. f(x) = ax^(2) +bx +c`
`f(0) =2 rArr c=2`
`f(1) =1rArr a+b+2=1rArr a+b =1`
`f(5//2) =0 rArr 5a+ b=0`
`rArr a=1//4 ,b=-5//4`
hence `f(x) =(1)/(4) x^(2) -(5)/(4)x+2`
Clearly . discriminant (D) of the equation f(x) =0 is less than 0.
hence f(x)=0 has imaginary roots .Also f(2) `=1//2` .and minimum value of f(x) is
`-((25)/(16)-4.(1)/(4) (2))/(4.(1)/(4)) =(7)/(16)`
Hence range of the f(x) is `[(7)/(16),oo)`
Discussion
214.

`|[a-b, b-c, c-a], [b-c, c-a, a-b], [c-a, a-b, b-c]|=?`A. (a+b+c)B. 3(a+b+c)C. 3abcD. 0

Answer» Correct Answer - D
`R_(1) to (R_(1) + R_(2) + R_(3))` gives all zeros in `R_(1) "and hence"Delta = 0`
Discussion
215.

Evaluate `|(1,omega,omega^2),(omega,omega^2,1),(omega^2,omega,omega)|` where `omega` is cube root of unity.A. 1B. `-1`C. 0D. none of these

Answer» Correct Answer - C
Apply `R_(1) to (R_(1) + R_(2) +R_(3)) " and use the result"(1+omega +omega^(2)) = 0`
Discussion
216.

Which of the following values of `alpha`satisfying the equation `|(1+alpha)^2(1+2alpha)^2(1+3alpha)^2(2+alpha)^2(2+2alpha)^2(2+3alpha)^2(3+alpha)^2(3+2alpha)^2(3+3alpha)^2|=-648alpha?``-4`b. `9`c. `-9`d. `4`A. `-4`B. `9`C. `-9`D. 4

Answer» Correct Answer - 2,3
Wen have `|{:((1+alpha^(2)),,(1+2alpha)^(2),,(1+3alpha)^(2)),((2+alpha)^(2),,(2+2alpha)^(2),,(2+3alpha)^(2)),((3+alpha)^(2),,(3+2alpha)^(2),,(3+3alpha)^(2)):}|=-648alpha`
Applying `R_(3)to R_(3) -R_(2),R_(2) to R_(2)-R_(1)`
`rArr |{:((1+alpha)^(2),,(1+2alpha)^(2),,(1+3alpha)^(2)),(3+2alpha,,3+4alpha,,3+6alpha),(5+2alpha,,5+4alpha,,5+6alpha):}|=-648alpha`
Applying `R_(3) to R_(3)-R_(2)`
`rArr |{:((1+alpha)^(2),,(1+2alpha)^(2),,(1+3alpha)^(2)),(3+2alpha,,3+4alpha,,3+6alpha),(2,,2,,2):}|=-648alpha`
Applying `C_(3)to C_(3)-C_(2),C_(2)to C_(2)-C_(1)`
`|{:((1+alpha)^(2),,alpha(2+3alpha),,alpha(2+5alpha)),(3+2alpha,,2alpha,,2alpha),(2,,0,,0):}|=-648alpha`
`rArr 2alpha^(2)(2+3alpha)-2alpha^(2)(2+5alpha)=-324alpha`
`rArr -4alpha^(3)=-324alpha`
`rArr alpha=0,+-9`
Discussion
217.

If `omega` is a complex cube root of unity then the value of the determinant `|[1,omega,omega+1] , [omega+1,1,omega] , [omega, omega+1, 1]|` isA. 2B. 4C. 0D. `-3`

Answer» Correct Answer - B
`1+omega +omega^(2) =0 rArr (1+omega) = -omega^(2) "Put"(1+omega) = -omega^(2)` and expand.
Discussion
218.

If a, b, c are non zero complex numbers satisfying `a^(2) + b^(2) + c^(2) = 0 and |(b^(2) + c^(2),ab,ac),(ab,c^(2) + a^(2),bc),(ac,bc,a^(2) + b^(2))| = k a^(2) b^(2) c^(2)`, then k is equal toA. 3B. 2C. 4D. 1

Answer» Correct Answer - C
Let `Delta = |(b^(2) + c^(2),ab,ac),(ab,c^(2) + a^(2),bc),(ac,bc,a^(2) b^(2))|`
Applying `R_(1) rarr R_(1) (a), R_(1) rarr R_(2) (b) and R_(3) (c)`, we get
`Delta = (1)/(abc) |(a(b^(2) + c^(2)),a^(2)b,a^(2)c),(ab^(2),b(c^(2) + a^(2)),b^(2)c),(ac^(2),bc^(2),c(a^(2) + b^(2)))|`
Taking a, b and c common from `C_(1), C_(2) and C_(3)` respectively, we get
`Delta = (abc)/(abc) |(b^(2) + c^(2),a^(2),a^(2)),(b^(2),c^(2) + a^(2),b^(2)),(c^(2),c^(2),a^(2) + b^(2))|`
`rArr Delta = |(2(b^(2) + c^(2)),2(c^(2) + a^(2)),2(a^(2) + b^(2))),(b^(2),c^(2) + a^(2),b^(2)),(c^(2),c^(2),a^(2) + b^(2))|`
`rArr Delta = 2 |(b^(2) + c^(2),c^(2) + a^(2),a^(2) + b^(2)),(b^(2),c^(2) + a^(2),b^(2)),(c^(2),c^(2),a^(2) + b^(2))|` [Taking 2 common from `R_(1)`]
Applying `R_(2) rarr R_(2) - R_(1) and R_(3) rarr R_(3) - R_(1)`, we get
`Delta = 2 |(b^(2) + c^(2),c^(2) + a^(2),a^(2) + b^(2)),(-c^(2),0,-a^(2)),(-b^(2),-a^(2),0)|`
`rArr Delta = 2 |(0,c^(2),b^(2)),(-c^(2),0,-a^(2)),(-b^(2),-a^(2),0)| " Applying "R_(1) rarr R_(1) + R_(2) + R_(3)`
`rArr Delta = 2 {0 |(0,-a^(2)),(-a^(2),0)| -c^(2) |(-c^(2),-a^(2)),(-b^(2),0)|+ b^(2) |(-c^(2),0),(-b^(2),-a^(2))|}`
`rArr Delta = 2 (a^(2) b^(2) c^(2) + a^(2) b^(2) c^(2)) = 4a^(2) b^(2) c^(2)`
But, it is given that `Delta = ka^(2) b^(2) c^(2)`
`:. k = 4`
Discussion
219.

`|((1+alpha)^2,(1+ 2alpha)^2, (1+3alpha)^2),((2+alpha)^2,(2+ 2alpha)^2, (2+3alpha)^2),((3+alpha)^2,(3+ 2alpha)^2, (3+3alpha)^2)| =-648alpha` FInd the value of `alpha`A. `-4`B. 9C. `-9`D. 4

Answer» Correct Answer - B::C
We have,
`|((1 + alpha)^(2),(1 + 2 alpha^(2)),(1 + 3 alpha)^(2)),((2 + alpha)^(2),(2 + 2 alpha)^(2),(2 + 3 alpha)^(2)),((3 + alpha)^(2),(3 + 2 alpha)^(2),(3 + 3 alpha)^(2))| = -648 alpha`
`rArr|((1 + alpha)^(2),(1 + 2 alpha)^(2),(1 + 3 alpha)^(2)),(3 + 2 alpha,3 + 4 alpha,3 + 6 alpha),(2(4 + 2 alpha),2(4 + 4 alpha),2(4 + 6 alpha))| = -648 alpha " " ["Applying" R_(2) - R_(2) - R_(1), R_(3) rarr R_(3) - R_(1)]`
`rArr 2 |((1 + alpha)^(2),(1 + 2 alpha)^(2),(1 + 3 alpha)^(2)),(3 + 2 alpha,3 + 4 alpha,3 + 6 alpha),(4 + 2 alpha,4 + 4 alpha,4 + 6 alpha)| = -648 alpha` [Taking 2 common from `R_(3)`]
`rArr 2|((1 + alpha)^(2),(1 + 2 alpha)^(2),(1 + 3 alpha)^(2)),(3 + 2 alpha,3 + 4 alpha,3 + 6 alpha),(1,1,1)| = -648 alpha " " [ "Applying " R_(3) rarr R_(3) - R_(2)]`
`rArr 2 |((1 + alpha)^(2),alpha (2 + 3 alpha),2alpha (2 + 4alpha)),(3 + 2 alpha,2 alpha,4 alpha),(1,0,0)| = -648 alpha " " ["Applying " C_(2) rarr C_(2) - C_(1), C_(3) rarr C_(3) - C_(1)]`
`rArr 2|(alpha (2 + 3 alpha),2 alpha (2 + 4 alpha)),(2 alpha,4 alpha)| = -648 alpha` [Expanding along `R_(3)`]
`rArr 4 alpha^(2) |(2 + 3 alpha,2 + 4 alpha),(2,2)| = -648 alpha`
`rArr 4 alpha^(2) (4 + 6 alpha - 4 - 8 alpha) = -648 alpha`
`rArr - 8 alpha^(3) = -648 alpha`
`rArr - 8 alpha (al[ha^(2) - 81) = 0`
`rArr alpha = 0, alpha +- 9`
Discussion
220.

`a!=p , b!=q,c!=r` and `|(p,b,c),(a,q,c),(a,b,r)|=0` the value of `p/(p-a)+q/(q-b)+r/(r-c)=`

Answer» Correct Answer - D
Discussion
221.

If `Delta = |(cos alpha,- sin alpha,1),(sin alpha,cos alpha,1),(cos (alpha + beta),- sin (alpha + beta),1)|`, thenA. `Delta in [1 -sqrt2, 1 + sqrt2]`B. `Delta in [-1, 1]`C. `Delta in [-sqrt2 sqrt2]`D. none of these

Answer» Correct Answer - A
From example 15, we have
`Delta = 1 + sin beta - cos beta`.
Now,
`- sqrt2 le sin beta - cos beta le sqrt2`
`rArr 1 - sqrt2 le 1 + sin beta - cos beta le 1 + sqrt2`
`rArr Delta in [1 -sqrt2, 1 + sqrt2]`
Discussion
222.

If the system of equations `x+a y=0,a z+y=0`and `a x+z=0`has infinite solutions, then the value of `a`is`-1`(b) 1(c) 0 (d)no real valuesA. `-1`B. 1C. 0D. no real values

Answer» Correct Answer - A
The given system of equations will have infinitely many solutions, if
`|(1,a,0),(0,1,a),(a,0,1)| = 0 rArr 1 + a^(3) = 0 rArr a = -1`
Discussion
223.

In a `Delta ABC " if " |(1,a,b),(1,c,a),(1,b,c)| =0`, then `sin^(2) A + sin^(2) B + sin^(2) C` isA. `(3 sqrt3)/(2)`B. `(9)/(4)`C. `(5)/(4)`D. 2

Answer» Correct Answer - B
We have,
`|(1,a,b),(1,c,a),(1,b,c)| =0`
`rArr |(1,a,b),(0,c -a,a -b),(0,b -a,c -a)| = 0 " " [("Applying" R_(2) rarr R_(2) - R_(1)),(R_(3) rarr R_(3) - R_(1))]`
`rArr (c -a) (c -b) + (a-b)^(2) = 0`
`rArr a^(2) + b^(2) + c^(2) - ab - bc - ca = 0`
`rArr 2a^(2) + 2b^(2) + 2c^(2) - 2ab - 2bc - 2ca = 0`
`rArr 2a^(2) + 2b^(2) + 2c^(2) - 2ab - 2bc - 2ca = 0`
`rArr (a -b)^(2) + (b -c)^(2) + (c -a)^(2) = 0`
`rArr a = b = c`
`rArr Delta ABC` is equilateral
`rArr A = B = C = (pi)/(3)`
`:. sin^(2) A + sin^(2) B + sin^(2) C = 3 sin^(2) (pi)/(3) = (9)/(4)`
Discussion
224.

`|(x,p,q),(p,x,q),(p,q,x)| =`A. `(x + p) (x +q) (x -p -q)`B. `(x -p) (x -q) (x + p +q)`C. `(x -p) (x -q) (x - p -q)`D. `(x + p) (x + q) (x + p + q)`

Answer» Correct Answer - B
Discussion
225.

If `p+q+r=0=a+b+c ,`then the value of the determnalnt `|p a q b r c q c r a p b r b p c q a|i s``0`b. `p a+q b+r c`c. `1`d. none of these

Answer» Correct Answer - A
Discussion
226.

The value of the determinant `|(cos alpha, -sin alpha,1),(sin alpha,cos alpha,1),(cos(alpha+beta),-sin(alpha+beta),1)|` is equalA. independent of `alpha`B. independent of `beta`C. independent of `alpha and beta`D. none of these

Answer» Correct Answer - A
We have, `|(cos alpha,- sin alpha,1),(sin alpha,cos alpha,1),(cos (alpha + beta),- sin (alpha + beta),1)|`
`= |(cos alpha,- sin alpha,1),(sin alpha,cos alpha,1),(0,0,1 + sin beta - cos beta)| " " ["Applying "R_(3) rarr R_(3) - R_(1) (cos beta) + R_(2) (sin beta)]`
`= (1 + sin beta - cos beta) (cos^(2) alpha + sin^(2) alpha)`
`= 1 + sin beta - cos beta`, which is independent of `alpha`
Discussion
227.

The value of `Delta = |(1,1 + ac,1 +bc),(1,1 + ad,1 + bd),(1 ,1 + ae,1 + be)|`, isA. 1B. 0C. 3D. `a + b + c`

Answer» Correct Answer - B
We have,
`Delta =|(1,1 + ac,1 +bc),(1,1 + ad,1 + bd),(1 ,1 + ac,1 + be)|`
`rArr Delta = |(1,ac,bc),(1,ad,bd),(1,ae,be)|["Applying " C_(2) rarr C_(2) - C_(1),C_(3) rarr C_(3) - C_(1)]`
`rArr Delta = ab |(1,c,c),(1,d,d),(1,e,e)| " " [("Taking a and b common from"),(C_(2) and C_(3) " respectively")]`
`rArr Delta = 0`
Discussion
228.

In a `Delta ABC " if " |(1,a,b),(1,c,a),(1,b,c)| =0`, then `sin^(2) A + sin^(2) B + sin^(2) C` isA. `(9)/(4)`B. `(4)/(9)`C. `(3 sqrt3)/(2)`D. 1

Answer» Correct Answer - A
We have,
`|(1,a,b),(1,c,a),(1,b,c)| =0`
`rArr a^(2) + b^(2) + c^(2) - ab - bc - ca = 0`
`rArr (1)/(2) {(a -b)^(2) + (b-c)^(2) + (c -a)^(2)} = 0`
`rArr a - b = 0, b -c =0, c -a = 0 rArr a = b = c`
`rArr Delta ABC` is equilateral
`rArr A = B = C = pi//3`
`:. sin^(2) A + sin^(2) B + sin^(2) C = (3)/(4) + (3)/(4) + (3)/(4) = (9)/(4)`
Discussion
229.

The factors of `|(x,a,b),(a,x,b),(a,b,x)|`, areA. `x - a, x -b, and x + a + b`B. `x + a, x + b and x + a + b`C. `x + a, x + b and x - a -b`D. `x - a, x - b and x - a -b`

Answer» Correct Answer - A
Discussion
230.

If `a + b + c = 0` one root of `|(a -x,c,b),(b,b -x,a),(b,a,c -x)|= 0`, isA. `x = 1`B. `x = 2`C. `x = a^(2) + b^(2) + c^(2)`D. `x = 0`

Answer» Correct Answer - D
Discussion
231.

If `omega` is a complex cube root of unity, then a root of the equation `|(x +1,omega,omega^(2)),(omega,x + omega^(2),1),(omega^(2),1,x + omega)| = 0`, isA. x = 1B. `x = omega`C. `x = omega^(2)`D. `x = 0`

Answer» Correct Answer - D
We have,
`|(x +1,omega,omega^(2)),(omega,x + omega^(2),1),(omega^(2),1,x + omega)| = 0`
`rArr |(x + 1 + omega + omega^(2),omega,omega^(2)),(x + 1 + omega + omega^(2),x + omega^(2),1),(x + 1 + omega + omega^(2),1,x + omega)| = 0 [("Applying"),(C_(1) rarr C_(1) + C_(2) + C_(3))]`
`rArr (x + 1 + omega + omega^(2)) |(1,omega,omega^(2)),(1,x + omega^(2),1),(1,1,x + omega)| =0`
`rArr x |(1,omega,omega^(2)),(0,x + omega^(2) - omega,1 - omega^(2)),(0,1 - omega,x + omega - omega^(2))| = 0 [("Using " R_(2) rarr R_(2) - R_(1)),(" "R_(3) rarr R_(3) - R_(1))]`
`rArr x |(x + omega - omega^(2)) (x + omega^(2) - omega) - (1 - omega) (1 - omega^(2))| = 0`
`rArr x = 0` is a root of the given equation
Discussion
232.

Prove that `|{:(a,,a^(2),,bc),(b ,,b^(2),,ac),( c,,c^(2),,ab):}|``|{:(1,,1,,1),(a^(2) ,,b^(2),,c^(2)),( a^(3),, b^(3),,c^(3)):}|`

Answer» we have `Delta =|{:(a,,a^(2),,bc),(b,,b^(2),,ac),(c^(2),,c^(3),,ab):}|`
Multiplying `R_(1),R_(2) " and " R_(3) by a, b," and " c ` respectively we get
`Delta =(1)/(abc) |{:(a^(2),,a^(3),,abc),(b^(2),,b^(3),,abc),(c^(2),,c^(3),,abc):}|`
`=|{:(a^(2),,a^(3),,1),(b^(2),,b^(3),,1),(c^(2),,c^(3),,1):}|` (Taking abc common from `C_(3))`
`=- |{:(1,,a^(3),,a^(2)),(1,,b^(3),,b^(2)),(1,,c^(3),,c^(2)):}|" "(C_(1) hArr C_(3))`
`=|{:(1,,a^(2),,a^(3)),(1,,b^(2),,b^(3)),(1,,c^(2),,c^(3)):}|" "(C_(2) hArr C_(2))`
`=|{:(1,,1,,1),(a^(2),,b^(2),,c^(2)),(a^(3),,b^(3),,c^(3)):}| " " "(Taking transpose)"`
Discussion
233.

If `alpha,beta and gamma` are such that `alpha+beta+gamma=0`, then `|(1,cos gamma,cosbeta),(cosgamma,1,cos alpha),(cosbeta,cos alpha,1)|`A. `cos alpha cos beta cos gamma`B. `cos alpha + cos beta + cos gamma`C. 1D. none of these

Answer» Correct Answer - D
Let A, B and C be three number such that `alpha = B -C, beta = C-A and gamma = A-B`. Clearly, `alpha + beta + gamma = 0`
`:. |(1,cos (A-B),cos (C-A)),(cos (A-B),1,cos (B-C)),(cos (C -A) ,cos (B -C),1)|`
`= |(cos^(2) A + sin^(2) A,cos A cos B + sin A sin B,cos A cos C + sin A sin C),(cos A cos B + sin A sin B,cos^(2) B + sin^(2) B,cos B cos C + sin B sin C),(cos C cos A + sin C sin A,cos B cos C + sin B sin C,cos^(2) C + sin^(2) C)|`
`= |(cos A ,sin A,0),(cos B,sin B,0),(cos C,sin C,0)||(cos A ,sin A,0),(cos B ,sin B,0),(cos C,sin C,0)| = 0 xx 0 = 0`
Discussion
234.

If `1 + (1)/(a) + (1)/(b) + (1)/(c) = 0`, then `Delta = |(1 +a,1,1),(1,1 +b,1),(1,1,1 +c)|` is equal to

Answer» Correct Answer - A
Discussion
235.

`|[x+a, b, c], [a, x+b, c], [b, b, x+c]|=0`

Answer» Apply `C_(1) to C_(1) + C_(2) + C_(3)` and take (x +a+b+c) common from `C_(1).`
Discussion
236.

Prove that the value of determinant `|{:(1,,omega,,omega^(2)),(omega ,,omega^(2),,1),( omega^(2),, 1,,omega):}|=0` where `omega` is complex cube root of unity .

Answer» `Delta =|{:(1,,omega,,omega^(2)),(omega,,omega^(2),,1),(omega^(2),,1,,omega):}|`
Applying `R_(1) to omega R_(1)` we get
`Delta =(1)/(omega) |{:(omega,,omega^(2),,omega^(3)),(omega,,omega^(2),,1),(omega^(2),,1,,omega):}|`
`=(1)/(omega)|{:(omega,,omega^(2),,omega^(3)),(omega,,omega^(2),,1),(omega^(2),,1,,omega):}|`
`(As R_(1) " and " R_(2) "are identical )"`
`=0`
Discussion
237.

Let `omega = - (1)/(2) + i (sqrt3)/(2)`, then the value of the determinant `|(1,1,1),(1,-1- omega^(2),omega^(2)),(1,omega^(2),omega^(4))|`, isA. `3 omega`B. `3 omega (omega -1)`C. `3 omega^(2)`D. `3 omega (1 - omega)`

Answer» Correct Answer - D
Discussion
238.

If `|(a +x,a -x,a -x),(a -x,a +x,a -x),(a -x,a -x,a +x)| = 0`, then x is equal toA. `0, 2a`B. `a, 2a`C. `0, 3a`D. none of these

Answer» Correct Answer - C
Discussion
239.

`|[1,x,x^3],[1,b,b^3],[1,c,c^3]|`=0; `b!=c`

Answer» Apply `R_(1) to R_(1)-R_(2) "and "R_(3) to R_(3) -R_(2).`
Take (x-b) common from`R_(1).`
Discussion
240.

Given `a_(i)^(2) + b_(i)^(2) + c_(i)^(2) = 1, i = 1, 2, 3 and a_(i) a_(j) + b_(i) b_(j) + c_(i) c_(j) = 0 (i !=j, i, j =1, 2, 3)`, then the value of the determinant `|(a_(1),a_(2),a_(3)),(b_(1),b_(2),b_(3)),(c_(1),c_(2),c_(3))|`, isA. `(1)/(2)`B. 0C. 2D. 1

Answer» Correct Answer - D
We have, `|(a_(1),a_(2),a_(3)),(b_(1),b_(2),b_(3)),(c_(1),c_(2),c_(3))|^(2)`
`= |(a_(1),a_(2),a_(3)),(b_(1),b_(2),b_(3)),(c_(1),c_(2),c_(3))||(a_(1),a_(2),a_(3)),(b_(1),b_(2),b_(3)),(c_(1),c_(2),c_(3))| = |(Sigmaa_(i)^(2),Sigmaa_(i)b_(j),Sigma a_(i) c_(j)),(Sigma a_(i) b_(j),Sigma b_(i)^(2),Sigma b_(i) c_(j)),(Sigma a_(i) c_(j),Sigma b_(i) c_(j),Sigma c_(i)^(2))|`
`= |(1,0,0),(0,1,0),(0,0,1)| =1`
`:. |(a_(1),a_(2),a_(3)),(b_(1),b_(2),b_(3)),(c_(1),c_(2),c_(3))| = +- 1`
Discussion
241.

If `a != 0, b!= 0, c!= 0`, then `|(1 +a,1,1),(1,1 +b,1),(1,1,1 +c)|` is equal toA. abcB. `abc (1 + (1)/(a) + (1)/(b) + (1)/(c))`C. 0D. `1 + (1)/(a) + (1)/(b) + (1)/(c)`

Answer» Correct Answer - B
Discussion
242.

The value of the determinant `|(x+2,x+3,x+5),(x+4,x+6,x+9),(x+8,x+11,x+15)|` isA. 2B. `-2`C. 3D. `x -1`

Answer» Correct Answer - B
Applying `R_(2) rarr R_(2) - R_(1) and R_(3) rarr R_(3) - R_(2)`, we get
`Delta = |(x +2,x +3,x +5),(2,3,4),(4,5,6)|`
`rArr Delta = 2|(x,x,x +1),(2,3,4),(1,1,1)| " " [("Applying" R_(1) rarr R_(1) - R_(2)),(and R_(3) rarr R_(3) - R_(2))]`
`rArr Delta = 2 |(x,0,1),(2,1,1),(1,0,0)| " " [("Applying " C_(2) rarr C_(2) - C_(1)),(and C_(3) rarr C_(3) - C_(2))]`
`rArr Delta = -2` [Expanding along `R_(3)`]
Discussion
243.

If `f(x)=|(a,-1,0),(ax,a,-1),(ax^2,ax,a)|`, then `f(2x)-f(x)` equalsA. `a(2a + 3x)`B. `ax(2x + 3a)`C. `ax(2a + 3x)`D. `x(2a + 3x)`

Answer» Correct Answer - C
Applying `R_(2) rarr R_(2) - xR_(1) and R_(3) rar R_(3) - xR_(3) - xR_(2)`, we get
`f(x) = |(a,-1,0),(0,a +x,-1),(0,0,a +x)| = a (a + x)^(2)`
`:. f(2x) - f(x) = a (a +2x)^(2) -a(a +x)^(2) = ax (2a + 3x)`
Discussion
244.

Coefficient of x in `f(x)=|(x,(1+sinx)^3,cosx),(1,log(1+x),2),(x^2,(1+x)^2,0)|` isA. `Delta_(2)^(3)`B. `Delta_(2)^(2)`C. `Dleta_(2)^(4)`D. none of these

Answer» Correct Answer - A
Discussion
245.

If `|(x,2,3),(2,3,x),(3,x,2)|=|(1,x,4),(x,4,1),(4,1,x)|=|(0,5,x),(5,x,0),(x,0,5)|=0,` then the value of x equals `(x in R):`

Answer» Correct Answer - C
Discussion
246.

If `a^(2) + b^(2) + c^(2) = -2 and f(x) = |(1 + a^(2)x,(1 + b^(2))x,(1 + c^(2)) x),((1 + a^(2))x,1 + b^(2)x,(1 + c^(2))x),((1 + a^(2)) x,(1 + b^(2))x,1 + c^(2)x)|`, then f(x) is a polynomial of degree

Answer» Correct Answer - C
Discussion
247.

If `a!=6,b,c` satisfy`|[a,2b,2c],[3,b,c],[4,a,b]|=0` ,then abc =A. `a +b +c`B. 0C. `b^(3)`D. `ab + bc`

Answer» Correct Answer - C
Discussion
248.

Show that `x = 2` is a root of the equation `|[x, -6, -1], [2, -3x, x-3], [-3, 2x, 2+x]|=0`

Answer» Apply `R_(1) to (R_(1) - R_(2)) " and take (x-2) common from"R_(1)`
Discussion
249.

If `f(x) =|(1,x,(x+1)),(2x,x(x-1),(x+1)x),(3x(x-1), x(x-1)(x-2),x(x-1)(x+1))|` then f(50)+f(51)..f(99) is equal to

Answer» Correct Answer - A
We have, `f(x) |(1,x,x +1),(2x,x (x-1),(x+1) x),(3x(x -1),x (x -1) (x -2),(x +1) x (x -1))|` Applying `C_(3) rarr C_(3) - C_(2)`, we get
`f(x) |(1,x,1),(2x,x(x -1),2x),(3x(x -1),x(x -1) (x -2),3x (x -1))| =0`
`:. f(100) = 0
Discussion
250.

If a,b,c be respectively the `p^(th),q^(th)andr^(th)` terms of a H.P., then `Delta=|{:(bc,ca,ab),(p,q,r),(1,1,1):}|` equalsA. `p + q + r`B. `(a +b +c)`C. 1D. none of these

Answer» Correct Answer - D
Discussion