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451.

Choose the correct answer from the following : The value of `|{:(1,a,a^(2)),(1,b,b^(2)),(1,c,c^(2)):}|is:`A. `(a -b) (b -c) (c -a)`B. `(a^(2) -b^(2)) (b^(2) -c^(2)) (c^(2) -a^(2))`C. `(a -b+ c) (b -c +a) (c -a +b)`D. none of these

Answer» Correct Answer - A
We have,
`Delta = |(1,a,a^(2)),(1,b,b^(2)),(1,c,c^(2))|`
`rArr Delta = |(1,a,a^(2)),(0,b -a,b^(2) -a^(2)),(0,c -a,c^(2) -a^(2))|` Applying `R_(2) rarr R_(2) - R_(1) and R_(3) rarr R_(3) - R_(1)`
`rArr Delta = (b -a) (c -a) |(1,a,a^(2)),(0,1,b +a),(0,1,c +a)|` Taking `(b -c) and (c -a)` common from `R_(2) and R_(3)` respectively
`rArr Delta = -(b -a) (c -a) (c-b) = (a -b) (b -c) (c -a)`
Discussion
452.

If `a^(-1)+b^(-1)+c^(-1)=0` such that `|{:(1+a," "1," "1),(" "1,1+b," "1),(" "1," "1,1+c):}|=lambda`, then what is `lambda` equal to ?

Answer» Correct Answer - B
Discussion
453.

The value of `|(a_(1) x_(1) + b_(1) y_(1),a_(1) x_(2) + b_(1) y_(2),a_(1) x_(3) + b_(1) y_(3)),(a_(2) x_(1) +b_(2) y_(1),a_(2) x_(2) + b_(2) y_(2),a_(2) x_(3) + b_(2) y_(3)),(a_(3) x_(1) + b_(3) y_(1),a_(3) x_(2) + b_(3) y_(2),a_(3) x_(3) + b_(3) y_(3))|`, isA. `a_(1) a_(2) a_(3) b_(1) b_(2) b_(3)`B. `x_(1) x_(2) x_(3) y_(1) y_(2) y_(3)`C. 0D. none of these

Answer» Correct Answer - C
We have,
`|(a_(1),b_(1),0),(a_(2),b_(2),0),(a_(3),b_(3),0)||(x_(1),y_(1),0),(x_(2),y_(2),0),(x_(3),y^(3),0)| = 0 xx 0 = 0`
`rArr |(a_(1) x_(1) + b_(1) y_(1) + 0,a_(1) x_(2) + b_(1) y_(2) + 0,a_(1) x_(3) + b_(1) y_(3) + 0),(a_(2) x_(1) + b_(2) y_(1) + 0,a_(2) x_(2) + b_(2) y_(2) + 0,a_(2) x_(3) + b_(2) y_(3) + 0),(a_(3) x_(1) + b_(3) y_(1) + 0,a_(3) x_(2) + b_(3) y_(2) + 0,a_(3) x_(3) + b_(3) y_(3) + 0)| = 0` [Using row by-row multiplication]
`rArr |(a_(1) x_(1) + b_(1) y_(1),a_(1) x_(2) + b_(1) y_(2),a_(1) x_(3) + b_(1) y_(3)),(a_(2) x_(1) + b_(2) y_(1),a_(2) x_(2) + b_(2) y_(2),a_(2) x_(3) + b_(2) y_(3)),(a_(3) x_(1) + b_(3) y_(1),a_(3) x_(2) + b_(3) y_(2),a_(3) x_(3) + b_(3) y_(3))| = 0`
Discussion
454.

`Delta = |(a =x,b,c),(b,x +c,a),(c,a,x +b)|`. Which of the following is a factor for the above determinant ?A. `x -(a + b +c)`B. `x + (a +b +c)`C. `a + b + c`D. `-(a +b +c)`

Answer» Correct Answer - B
We have,
`Delta = |(a +x,b,c),(b,x +c,a),(c,a,x + b)|`
`rArr Delta = |(x + a + b + c,b,c),(x +a + b + c,x +c,a),(x + a + b + c,a,x +b)|` Applying `C_(1) rarr C_(1) + C_(2) + C_(3)`
`rArr Delta = (x +a +b +c) |(1,b,c),(1,x +c,a),(1,a,x +b)|`
`rArr (x +a +b +c)` is a factor of `Delta`
Discussion
455.

If `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 the value of k isA. 2B. 1C. 4D. 3

Answer» Correct Answer - C
Discussion
456.

What is `|{:(-a^(2),ab,ac),(ab,-b^(2),bc),(ac,bc,-c^(2)):}|` equal to ?A. `4a^(2) b^(2)`B. `4b^(2) c^(2)`C. `4c^(2) a^(2)`D. `4a^(2) b^(2) c^(2)`

Answer» Correct Answer - D
We have,
`Delta = |(a^(2),-ab,-ac),(-ab,b^(2),-bc),(ca,bc,-c^(2))|`
`rArr Delta = abc |(a,-b,-c),(-a,b,-c),(a,b,-c)|` Taking a, b, c common from `T_(1), R_(2) and R_(3)` respectively
`rArr Delta = a^(2) b^(2) c^(2) |(1,-1,-1),(-1,1,-1),(1,1,-1)|` Taking a,b, c common from `C_(1), C_(2) and C_(3)` respectively
`rArr Delta = a^(2) b^(2) c^(2) |(1,0,0),(-1,0,-2),(1,2,0)|` Applying `C_(2) rarr C_(2) + C_(1), C_(3) rarr C_(3) + C_(1)
`rArr Delta = 4a^(2) b^(2) c^(2)`
Discussion
457.

The value of the determinant `|(b +c,a -b,a),(c +a,b -c,b),(a +b,c -a,c)|`, isA. `a^(3) + b^(3) + c^(3) - 3abc`B. `3abc - a^(3) -b^(3) - c^(3)`C. `3 abc + a^(3) + b^(3) + c^(3)`D. none of these

Answer» Correct Answer - B
We have,
`|(b +c,a -b,a),(c +a,b -c,b),(a +b,c -a,c)|`
`= |(a +b +c,-b,a),(b +c +a,-c,b),(c +a +b,-a,c)| " " ["Applying " C_(1) rarr C_(1) + C_(3), C_(2) rarr -C_(2) - C_(3)]`
`= - (a +b +c) |(1,b,a),(1,c,a),(1,a,c)|`
`= -(a +b +c) |(1,b,a),(0,c -b,b -a),(0,a -b,c -a)| [(" Applying"),(R_(2) rarr R_(2) - R_(1) ", "R_(3) rarr R_(3) - R_(1))]`
`= - (a + b +c) (a^(2) + b^(2) + c^(2)- ab - bc -ca)`
`= - (a^(3) + b^(3) + c^(3) - 3abc)`
Discussion
458.

`[[a,b,ax+by],[b,c,bx+cy],[ax+by,bx+cy,0]]=(b^2-ac)(ax^2+2bxy+cy^2)`A. zeroB. positiveC. negativeD. `b^(2) + ac`

Answer» Correct Answer - C
We have,
`Delta = |(a,b,ax + by),(b,c,bx + cy),(ax + by,bx + cy,0)|`
`rArr Delta = |(a,b,ax + by),(b,c,bx + cy),(0,0,-(ax^(2) + 2bxy + cy^(2)))| " " [("Applying " R_(3) rarr),(R_(3) - xR_(1) - yR_(2))]`
`rArr Delta = (b^(2) - ac) (ax^(2) + 2bxy + cy^(2))`
Now, `b^(2) - ac lt 0 and a gt 0`
`rArr` Discriminant of `ax^(2) + 2bxy + cy^(2)` is negative and `a gt 0`
`rArr ax^(2) + 2bxy+ cy^(2) gt 0` for all x,`y in R`
`rArr Delta = (b^(2) -ac) (ax^(2) + 2bxy + cy^(2)) lt 0`
Discussion
459.

If `x , y , z`are not all zero such that`a x+y+z=0`,`x+b y+z=0`,`x+y+c z=0`then prove that`1/(1-a)+1/(1-b)+1/(1-c)=1`

Answer» `D= | (a,1,1),(1,b,1),(1,1,c)| = 0`
`R_2 -> R_2 - R_1`
`R_3 -> R_3 - cR_1`
`D= |(a,1,1),((1-a),b-1,0,),(1-ac,1-c,0)|`
`=> (1-a)(1-c)- (b-1)(1-ac) = 0`
`= 1-a-c+ac- (b-1-abc+ac) = 0`
`= 1 - a-c+ac-b+1+abc-ac= 0`
`= 2-a-b-c+abc= 0 `
`abc = a+b+c - 2`
to prove :`1/(1-a) + 1/(1-b) + 1/(1-c) = 1`
LHS :`1/(1-a) + 1/(1-b) + 1/(1-c) = ((1-b)(1-c) + (1-a)(1-c) + (1-a)(1-b))/((1-a)(1-b)(1-c))`
`= (3-2(a+b+c) + (ab+bc+ac))/(1- (a+b+c) + (ab + bc + ac) - ab)`
`= 1`
Discussion
460.

If `f(x)=|[a,-1 ,0],[ax,a,-1],[a x^2,a x, a]|,`using properties of determinants, find the value of `f(2x)-f(x)dot`

Answer» Using the property`R_2rarrR_2-xR_1` and `R_3rarrR_3-xR_2`
we get=>`f(x)=|(a,-1,0),(0,a+x,-1),(0,0,a+x)|`
evaluating the determinant we get=>`f(x)=a(a+x)^2`
`f(2x)=a(a+2x)^2`
`f(2x)-f(x)=a((a+2x)^2-(a+x)^2)`
`f(2x)-f(x)=3ax^2+2a^2x`
Discussion
461.

Using properties of determinant show that: `|[1 , a , -bc] , [1 , b , -ca] , [1 , c , -a b]|=(a-b)(b-c)(c-a)`

Answer» `|(1,a,-bc),(1,b,-ca),(1,c,-ab)| `
`R_1 -> R_1 - R_2`
`|(0,a-b,c(a-b)), (1,b,-ca),(1,c,-ab)|`
`R_2 -> R_2 - R_3`
`|(0,a-b, c(a-b)),(0,b-c, a(b-c)),(1,c,-ab)|`
`(a-b)(b-c)|(0,1,c),(0,1,a),(1,c,-ab)|`
`(a-b)(b-c)(a-c)`
Answer
Discussion
462.

Show that :`|x y z x^2y^2z^2x^3y^3z^3|=x y z(x-y)(y-z)(z-x)dot`

Answer» `|(x,y,z),(x^2,y^2,z^2),(x^3 , y^3,z^3)|`
`= xyz|(1,1,1),(x,y,z),(x^2,y^2,z^2)|`
`c_2 -> c_2 - c_1`
`c_3 -> c_3 - c_1`
`=> |(1,0,0),(x, y-x, z-x),(x^2, y^2-x^2, z^2-x^2)|`
`= 1((y-x)(z^2-x^2) - (z-x)(y^2 - x^2))`
`= xyz(y-x)(z-x)(z+x-y-x)`
`= xyz(y-x)(z-x)(z-y)`= RHS
hence proved
Discussion
463.

If `f(x)=|a-1 0a x a-1a x^2a x a|,t h e nf(2x)-f(x)`is divisible by`a`b. `b`c.`c ,d ,e`d. none of theseA. xB. aC. `2a+3x`D. `x^(2)`

Answer» Correct Answer - A::B::C
Applying `R_(3) to R_(3) -xR_(2) " and " R_(2) to R_(2)-xR_(1)` we get
`f(x)= |{:(a,,-1,,0),(0,,a+x,,-1),(0,,0,,a+x):}|=a(a+x)^(2)`
hence
`f(2x)- f(x) =a [(a+2x)^(2) -(a-x)^(2)]`
`=a(a+2x-a-x) (a+2x+a+x)`
`=ax (2a+3x)`
Discussion
464.

In a `triangleABC`, if `|[1,1,1][1+sinA,1+sinB,1+sinC],[sinA+sin^2A,sinB+sin^2B,sinC+sin^2C]|=0`, then prove that `triangleABC` is an isosceles triangle.

Answer» `|(1,1,1),(1+sinA , 1+ sinB, 1+sinC),(sinA +sin^2 A , sin B + sin^2 B , sin C + sin^2 C)|= 0`
`R_1 -> R_2 - R_3`
`|(1,1,1),(sinA , sinB, sinC),(sinA +sin^2 A , sin B + sin^2 B , sin C + sin^2 C)|= 0`
`R_3 -> R_3 -R_2`
`|(1,1,1),(sinA , sinB, sinC),(sin^2 A ,sin^2 B , sin^2 C)|= 0`
`C_2 -> C_2 - C_1 & C_3 -> C_3 - C_1`
`|(1,0,0),(sinA , sinB- sinA, sinC- sinB),(sin^2 A ,(sinB-sinA)(sinB+ sinA) , (sinC-sinB)(sin C + sinB))|= 0`
`(sinB-sinA)(sinC-sinA)|(11,0,0),(sinA,1,1),(sin^2 A , sinB+ sinA, sinC+sinA)|= 0`
`(sin B - sinA)(sin C-sinA)[1(sinC + sinA - sinB - sinA)] = 0`
`sinB = sinA or sinC = sinA or sinC= sinB`
`/_B =/_A ; /_C=/_A ; /_C= /_B`
so, `AC=BC , AB=BC,AB=AC`
`:./_ ABC` is isosceles . hence proved
Discussion
465.

`Delta= |{:(1,,1+ac,,1+bc),(1,,1+ad,,1+bd),(1,,1+ae,,1+be):}|` is independent ofA. aB. bC. c,d,eD. none of these

Answer» Correct Answer - A::B::C
Operating `C_(2) to C_(2)-C_(1),C_(3) to C_(3)-C_(1)` we get
`Delta =|{:(1,,ac,,bc),(1,,ad,,bd),(1,,ae,,be):}|=ab|{:(1,,c,,c),(1,,d,,d),(1,,e,,e):}|= ab (0) =0`
Discussion
466.

If `a, b, and c` are the sides of a triangle and `A, B and C` are the angles opposite to `a, b,and c,` respectively, then `Delta=|(a^2,b sinA,csin A),(b sinA,1,cosA),(c sinA,cos A,1)|`A. `sin A sin B sin C`B. abcC. 1D. 0

Answer» Correct Answer - D
Discussion
467.

If A, B and C denote the angles of a triangle, then `Delta = |(-1,cos C,cos B),(cos C,-1,cos A),(cos B,cos A,-2)|` is independent ofA. `cos A cos B cos C`B. `sin A sin B sin C`C. 0D. none of these

Answer» Correct Answer - C
Discussion
468.

One root of the equation `|{:(3-x,-6,3),(-6,3-x,3),(3,3,-6-x):}|=0" "is:`A. 6B. 3C. 0D. None of these

Answer» Correct Answer - C
Discussion
469.

if `Delta (x) = |{:(tan x,,tan (x+h),,tan(x+2h)),(tan(x+2h),,tan x,,tan(x+h)),(tan(x+h),,tan(x+2h),,tanx):}|, " then "` The value of `lim_(h to 0) .(Delta (pi//3))/(sqrt(3)h^(2)) " is "`A. 144B. 81C. 64D. 36

Answer» Correct Answer - A
`(Delta)/(h^(2)) = |{:(tan x,,(tan (x+h)-tanx)/(h),,(tan (x+2h)-tanx)/(h)),(tan(x+2h),,(tan x-tan(x+2h))/(h),,(tan(x+h)-tan(x+2h))/(h)),(tan(x+h),,(tan(x+2h)-tan(x+h))/(h),,(tanx-tan (x+h))/(h)):}|`
`rArr underset( h to 0)("lim") .(Delta)/(h^(2)) =|{:(tan x,,sec^(2) x,,2sec^(2) x),(tan x,,-2sec^(2) x,,-sec^(2) x),(tan x,,sec^(2)x,,-sec^(2) x):}|`
`= |{:(0,,0,,3sec^(2)x),(0,,-3sec^(2)x,,0),(tan x,,sec^(2)x,,-sec^(2)x):}|`
`= tan x sec^(4) x`
`rArr underset( h to 0)(" lim ") .(Delta (pi//3))/(sqrt(3)h^(2)) =144`
Discussion
470.

If A, B and C are the angles of a triangle and `|(1,1,1),(1 + sin A,1 + sin B,1 + sin C),(sin A + sin^(2) A,sin B + sin^(2)B,sin C + sin^(2) C)|= 0`, then the triangle ABC isA. equilateralB. isoscelesC. any triangleD. right angled

Answer» Correct Answer - B
Discussion
471.

`|(sin^2x,cosx^2x,1), (cos^2x, sin^2x,1),(-10,12,2)|=0`

Answer» Correct Answer - A
Discussion
472.

If `f(x) = |(sin x,cos x,tan x),(x^(3),x^(2),x),(2x,1,1)|, " then " lim_(x rarr 0) (f (x))/(x^(2))`, isA. 3B. `-1`C. 0D. 1

Answer» Correct Answer - D
Discussion
473.

If `0lt0ltpi/2" and "|{:(1+sin^(2)theta,cos^(2)theta,4sin4theta),(sin^(2)theta,1+cos^(2)theta,4sintheta),(sin^(2)theta,cos^(2)theta,1+4sintheta):}|=,"then "theta=`A. `pi/24,(5pi)/(24)`B. `(5pi)/24,(7pi)/24`C. `(7pi)/24,(11pi)/(24)`D. None of these

Answer» Correct Answer - C
Discussion
474.

The roots of the equation `|(1,4,20),(1,-2,5),(1,2x,5x^(2))| = 0` areA. `-1, -2`B. `-1, 2`C. `1, -2`D. `1, 2`

Answer» Correct Answer - B
Discussion
475.

The roots of the equation `|(3x^(2),x^(2) + x cos theta + cos^(2) theta ,x^(2) + x sin theta + sin^(2) theta),(x^(2) + x cos theta + cos^(2) theta,3 cos^(2) theta,1 + (sin 2 theta)/(2)),(x^(2) + x sin theta + sin^(2) theta,1 + (sin 2 theta)/(2),3 sin^(2) theta)| = 0`A. `sin theta, cos theta`B. `sin^(2) theta, cos^(2) theta`C. `sin theta, cos^(2) theta`D. `sin^(2) theta, cos theta`

Answer» Correct Answer - A
Discussion
476.

The roots of the equation `|(x -1,1,1),(1,x -1,1),(1,1,x -1)| =0`, areA. 1, 2B. `-1, 2`C. `1, -2`D. `-1, -2`

Answer» Correct Answer - B
Discussion
477.

From the matrix equation AB = AC we can conclude B = C provided thatA. A is singularB. A is non-singularC. A is symmetricD. A is square

Answer» Correct Answer - B
Discussion
478.

For any `triangleABC`, the value of determinant `|[sin^2A,cotA,1],[sin^2B,cotB,1],[sin^2C,cotC,1]|` is:

Answer» `Delta = |[sin^2A,cotA,1],[sin^2B,cotB,1],[sin^2C,cotC,1]|`
`=>Delta = [sin^2A(cotB-cotC) -sin^2B(cotA - cotC)+ sin^2C( cotA - cotB)]`
Converting, `cot` into `cos/sin` form, we get,
`=>Delta = [sin^2A(sin(C-B)/(sinB*SinC)) +sin^2B(sin(A-C)/(sinC*SinA))+ sin^2C( sin(B-A)/(sinB*SinA))]`
`=>Delta = [(sin^3A(sin(C-B))+sin^3B(sin(A-C))+ sin^3C( sin(B-A)))/(sinAsinBsinC)]`
Now, `sin^3A = sinAsin^2A = sin(pi-(B+C))sin^2A = sin(B+C)sin^2A)`
`sin^3B = sinBsin^2B = sin(pi-(A+C))sin^2B = sin(A+C)sin^2B)`
`sin^3C = sinCsin^2C = sin(pi-(B+A))sin^2C = sin(B+A)sin^2C)`
So, our equation becomes,
`Delta = [ (sin^2A((sinCcosB)^2-(cosCsinB)^2)+sin^2B((sinAcosC)^2-(cosAsinC)^2)+ sin^2C( (sinBcosA)^2-(cosBsinA)^2))/(sinAsinBsinC)]`
`Delta = [0/(sinAsinBsinC)] =0`
Discussion
479.

if `a_(r) = (cos 2r pi + I sin 2 r pi)^(1//9)` then prove that `|{:(a_(1),,a_(2),,a_(3)),(a^(4) ,,a^(5),,a_(6)),( a_(7),, a_(8),,a_(9)):}|=0`

Answer» `a_(r )=(cos 2r pi + I sin 2 rpi )^(1/9) =e^(i(2rpi)/(9))`
`:. Delta = |{:(a_(1),,a_(2),,a_(3)),(a_(4),,a_(5),,c_(6)),(a_(7),,a_(8),,a_(9)):}|=|{:(e^(i(2pi)/(9)),,e^(i(4pi)/(9)),,e^(i(6pi)/(9))),(e^(i(8pi)/(9)),,e^(i(10pi)/(9)),,e^(i(18pi)/(9))),(e^(i(14pi)/(9)),,e^(i(16pi)/(9)),,e^(i(18pi)/(9))):}|`
Now taking `e^(i(6pi)/(9))` common from `R_(2)` we get
`Delta =e^(i(6pi)/(9)) |{:(e^(i(2pi)/(9)),,e^(i(4pi)/(9)),,e^(i(6pi)/(9))),(e^(i(2pi)/(9)),,e^(i(4pi)/(9)),,e^(i(6pi)/(9))),(e^(i(14pi)/(9)),,e^(i(16pi)/(9)),,e^(i(18pi)/(9))):}|`
`(As R_(1) " and " R_(2) " are identical ")`
Discussion
480.

Given `A=|a b2c d e2fl m2n|,B=|f2d e2n4l2m c2a b|`, then the value of `B//A`is ___________.

Answer» Correct Answer - 5
`B =2xx2 |{:(f,,d,,e),(n,,l,,m),(c,,a,,b):}|`
`=2 |{:(2f,,d,,e),(2n,,l,,m),(2c,,a,,b):}|`
`=2|{:(2f,,a,,b),(2n,,d,,e),(2c,,l,,m):}|" "[R_(3) hArr R_(2) " and then "R_(2) hArr R_(2)]`
`=2 |{:(a,,b,,2c),(d,,e,,2f),(l,,m,,2n):}|" "[C_(1) hArr C_(2) " and then " C_(2) hArr C_(3)]`
`=2A`
`:. B//A =2`
Discussion
481.

If `|(1,1,1),(a,b,c),(a^(3),b^(3),c^(3))|= (a -b) (b -c) (c -a) (a + b+c)` where a, b, c are all different, then the determinant `|(1,1,1),((x-a)^(2),(x -b)^(2),(x -c)^(2)),((x -b) (x -c),(x -c) (x -a),(x -a) (x -b))|` vanishes whenA. `a + b + c = 0`B. `x = (1)/(3) (a + b +c)`C. `x = (1)/(2) (a + b +c)`D. `x = a + b + c`

Answer» Correct Answer - B
Discussion
482.

Show that `a x+b y+r=0,b y+c z+p=0a n dc z+a x+q=0`are perpendicular to `x-y ,y-za n dz-x`planes, respectively.A. `-1`B. 0C. 1D. 2

Answer» Correct Answer - A
Discussion
483.

The determinant `Delta = |(b,c,b alpha +c),(c,d,c alpha + d),(b alpha + c,c alpha + d,a a^(3) - c alpha)|` is equal to zero, ifA. b, c, d are in A.PB. b, c, d are in G.PC. b, c, d are in H.PD. `alpha` is a root of `ax^(3) - cx - d = 0`

Answer» Correct Answer - B
Discussion
484.

The value of the determinant `Delta = |((1 - a_(1)^(3) b_(1)^(3))/(1 - a_(1) b_(1)),(1 - a_(1)^(3) b_(2)^(3))/(1 - a_(1) b_(2)),(1 - a_(1)^(3) b_(3)^(3))/(1 - a_(1) b_(3))),((1 - a_(2)^(3) b_(1)^(3))/(1 - a_(2) b_(1)),(1 - a_(2)^(3) b_(2)^(3))/(1 - a_(2) b_(2)),(1 - a_(2)^(3) b_(3)^(3))/(1 - a_(2) b_(3))),((1 - a_(3)^(3) b_(1)^(3))/(1 - a_(3) b_(1)),(1 - a_(3)^(3) b_(2)^(3))/(1 - a_(3) b_(2)),(1 - a_(3)^(3) b_(3)^(3))/(1 - a_(3) b_(3)))|`, is

Answer» Correct Answer - D
Discussion
485.

Find the number of real root of the equation`|0x-a x-b x+a0x-c x+b x+c0|=0,a!=b!=ca n db(a+c)> a c`

Answer» Correct Answer - three
Expanding Using Sarrus Rule
`Delta =(x-a)(x+b)(x-c)+(x-b)(x+a)(x+c)`
`=2x(x^(2)+ac-ab-bc)`
Now `Delta=0` gives x=0
or `x^(2) =b(a+c)-ac`
if b(a+c) `gt ac` we have three roots `0+- sqrt({b(a+c)-ac})`
Discussion
486.

If `C = 2 cos theta`, then the value of the determinant `Delta = |(C,1,0),(1,C,1),(6,1,c)|`, isA. `(sin 4 theta)/(sin theta)`B. `(2 sin^(2) 2 theta)/(sin theta)`C. `4 cos^(2) theta (2 cos theta -1)`D. none of these

Answer» Correct Answer - D
We have,
`Delta = |(C,1,0),(1,C,1),(6,1,c)|`
`rArr Delta = |(0,1,0),(1 -C^(2),C,1),(6-C,1,C)| " " ["Applying " C_(1) rarr C_(1) - C C_(2)]`
`rArr Delta = -|(1 -C^(2),1,),(6 -C,C,)|` [Expanding along `R_(1)`]
`rArr Delta = - (C -C^(3) - 6 + C)`
`rArr Delta = C^(3) - 2C + 6`
`rArr Delta = 8 cos^(3) theta - 4 cos theta + 6`
Hence, option (d) is correct
Discussion
487.

If `a gt 0` and discriminant of `ax^(2) + 2bx + c` is negative, then `Delta = |(a,b,ax +b),(b,c,bx +c),(ax +b,bx +c,0)|`, isA. positiveB. `(ac -b^(2)) (ax^(2) + 2bx +c)`C. negativeD. 0

Answer» Correct Answer - C
It is given that the discriminant of `ax^(2) + 2bx + c` is negative.
`:. 4b^(2) - 4ac lt 0 rArr b^(2) - ac lt 0 rArr ac -b^(2) gt 0`
Also, `a gt 0` and discriminant is negative.
`:. Ax^(2) + 2bx + c gt 0` for all `x in R`
Now,
`Delta = |(a,b,ax + b),(b,c,bx + c),(ax +b,bx+c,0)|`
`rArr Delta = |(a,b,0),(b,c,0),(ax +b,bx +c,- x (ax +b) - (bx +c))|`
`rArr Delta = |(a,b,0),(b,c,0),(ax +b,bx +c,-(ax^(2) + 2bx +c))|`
`rArr Delta = -(ax^(2) + 2bx + c) (ac -b^(2)) lt 0` [Using (i) and (ii)]
Discussion
488.

The equation `|(x-a,x-b,x-c),(x-b,x-a,x-c),(x-c,x-b,x-a)|=0` (a,b,c are different) is satisfied by (A) `x=(a+b+c0` (B) `x= 1/3 (a+b+c)` (C) `x=0` (D) none of theseA. `x = 0`B. `x = a`C. `x = (1)/(3) (a + b +c)`D. `x = a + b + c`

Answer» Correct Answer - C
Discussion
489.

Evaluate :\(\begin{bmatrix} 1^2& 2^2 & 3^2 \\[0.3em] 2^2 & 2^2 & 4^2 \\[0.3em] 3^2 &4^2 & 5^2 \end{bmatrix}\)[(1^2,2^2,3^2)(2^2,2^2,4^2)(3^2,4^2,5^2)]

Answer»

\(\begin{bmatrix}1^2& 2^2 & 3^2 \\[0.3em]2^2 & 2^2 & 4^2 \\[0.3em]3^2 &4^2 & 5^2\end{bmatrix}\) = \(\begin{bmatrix}1 &4 & 9 \\[0.3em]4 & 9 & 16 \\[0.3em]9 & 16 & 25\end{bmatrix}\)

Expanding by first row, we get, 

1(9 × 25 - 16 × 16) + 4(16 × 9 - 4 × 25) + 9(4 × 16 - 9 × 9) = - 31 + 176 - 153 =- 8
Discussion
490.

If `a >0`and discriminant of `a x^2+2b x+c`is negative, then `=a b a x+bb c b x+c a x+bb x+c0`is`+v e`b. `(a c-b)^2(a x^2+2b x+c)`c. `-v e`d. `0`A. `+ve`B. `(ac-b)^(2))(ax^(2)+2bx+c)`C. `-ve`D. `0`

Answer» Correct Answer - C
here `a gt 0 " and " 4b^(2) -4ac lt 0 , i.e., ac -b^(2) gt 0`
`:. .ax^(2) +2bx +c gt 0 ,AA x in R `
Now
`Delta = |{:(a,,b,,ax+b),(b,,c,,bx+c),(0,,0,,-(ax^(2)+2bx+c)):}|`
`"[Operating " R_(3) to R_(3) -xR_(1)-R_(2)"]"`
`=-(ax^(2)+2bx+c)(ac-b^(2))`
` =- (+ve) (+ve) =-ve`
Discussion
491.

The determinant `|(cos C,tan A,0),(sin B,0,-tan A),(0,sin B,cos C)|` has the value, where A, B, C are angled of a triangle

Answer» Correct Answer - A
Discussion
492.

The value of `|(a,a^(2) - bc,1),(b,b^(2) - ca,1),(c,c^(2) - ab,1)|`, isA. 1B. `-1`C. 0D. `-abc`

Answer» Correct Answer - C
Discussion
493.

Evaluate :\(\begin{bmatrix} 102& 18 & 36 \\[0.3em] 1 & 3 & 4 \\[0.3em] 17 & 3 & 6 \end{bmatrix}\) [(102,18,36)(1,3,4)(17,3,6)]

Answer»

\(\begin{bmatrix} 102& 18 & 36 \\[0.3em] 1 & 3 & 4 \\[0.3em] 17 & 3 & 6 \end{bmatrix}\) = 6 x \(\begin{bmatrix} 17& 18 & 6 \\[0.3em] 1 & 6 & 4 \\[0.3em] 17 & 3 & 6 \end{bmatrix}\)[R1’ = R1/6]

Now, for any determinant, if at least two rows are identical, then the value of the determinant becomes zero. 

Here, the first and third rows are identical. 

So, the value of the above determinant evaluated = 0
Discussion
494.

Evaluate:\(\begin{bmatrix} 1^2&2^2 & 3^2 \\[0.3em] 2^2 & 2^2 & 4^2 \\[0.3em] 3^2 & 4^2 & 5^2 \end{bmatrix}\) |(1^2,2^2,3^2)(2^2,2^2,4^2)(3^2,4^2,5^2)|

Answer»

\(\begin{bmatrix} 1^2&2^2 & 3^2 \\[0.3em] 2^2 & 2^2 & 4^2 \\[0.3em] 3^2 & 4^2 & 5^2 \end{bmatrix}\) = \(\begin{vmatrix}1 & 4& 9 \\[0.3em]4 & 9 & 16 \\[0.3em]9 & 16& 25\end{vmatrix}\)

Expanding by first row, we get, 

1(9 × 25 - 16 × 16) + 4(16 × 9 - 4 × 25) + 9(4 × 16 - 9 × 9) 

= - 31 + 176 - 153 

=- 8
Discussion
495.

Evaluate the determinants in(i) `|(costheta,-sintheta),(sintheta,costheta)|` (ii) `|(x^2-x+1,x-1),(x+1,x+1)|`

Answer» (i) `|[costheta,-sintheta],[sintheta,costheta]|`
`=[cos^2theta - (-sin^2theta)]`
`=[cos^2theta + sin^2theta)]`
`=1`
`:. |[costheta,-sintheta],[sintheta,costheta]| = 1`
(ii) `|[x^2-x+1,x-1],[x+1,x+1]|`
`=[(x+1)(x^2-x+1)-(x+1)(x-1)]`
`=[(x+1)(x^2-x+1-x+1)]`
`=[(x+1)(x^2-2x+2)]`
`=x^3-2x^2+2x+x^2-2x+2`
`=x^3-x^2+2`
`:.|[x^2-x+1,x-1],[x+1,x+1]| =x^3-x^2+2`
`
Discussion
496.

Choose the correct answer from the following : `|{:(x,2),(8,x):}|=|{:(6,8),(9,12):}|,"then x:"`A. `+-6`B. `+-4`C. `+-2`D. None of these

Answer» Correct Answer - B
Discussion
497.

The number of distinct real roots of `|(sinx, cosx, cosx),(cos x,sin x,cos x),(cos x,cos x,sin x)|=0` in the interval `-(pi)/4 le x le t (pi)/4` is

Answer» Correct Answer - C
We have, `|(sin x,cos x,cos x),(cos x,sin x,cos x),(cos x,cos x,sin x)| = 0`
`rArr |(sin x + 2 cos x,sin x + 2 cos x,sin x + 2 cos x),(cos x,sin x,cos x),(cos x,cos x,sin x)| =0 " " ["Applying " R_(1) rarr R_(1) + R_(2) + R_(3)]`
`rArr (sinx + 2 cos x) |(1,1,1),(cos x,sin x,cos x),(cos x,cos x,sin x)| = 0`
`rArr (sinx + 2 cos x) |(1,0,0),(cos x,sin x - cos x,0),(cos x,0,sin x - cos x)| = 0 " " ["Applying " C_(3) rarr C_(3) - C_(1)"," C_(2) rarr C_(2) - C_(1)]`
`rArr (sin x + 2 cos x) (sin x - cos x)^(2) = 0`
`rArr tan x = -2 or, tan x = 1`
`rArr x = (pi)/(4) " " [ :. x in [pi//4, pi//4] rArr -1 le tan x le 1]`
Discussion
498.

If a, b and c are real numbers, and `Delta=|b+cc+a a+b c+a a+bb+c a+bb+cc+a|=0`.Show that either `a" "+" "b" "+" "c" "=" "0" "or" "a" "=" "b" "=" "c`.

Answer» ` Delta=|{:(b+c,,c+a,,a+b),(c+a,,a+b,,b+c),(a+b,,b+c,,c+a):}|`
Applying `R_(1) to R_(1)+R_(2)+R_(3)` we have
` Delta=|{:(2(a+b+c),,2(a+b+c),,2(a+b+c)),(c+a,,a+b,,b+c),(a+b,,b+c,,c+a):}|`
`=2 (a+b+c) |{:(1,,1,,1),(c+a,,b-c,,b-a),(a+b,,c-a,,c-b):}|`

Applying `C_(2) to C_(2)-C_(1) " and "C_(3) to C_(3)-C_(1)` we have
`Delta =2(a+b+c) |{:(1,,0,,0),(c+a,,b-c,,b-a),(a+b,,c-a,,c-b),(,,,,):}|`
Expanding along `R_(1)` we have
`Delta =2 (a+b+c) (1) [(b-c)(c-b)-(b-a)(c-a)]`
`=2 (a+b+c) [-b^(2)-c^(2)+2bc-bc+ba+ac-a^(2)]`
`=2 (a+b+c) [ab+bc+ca-a^(2)-b^(2)-c^(2)]`
It is given that `Delta =0` Therefore,
`(a+b+c) [ab+bc+ca-a^(2)-b^(2)-c^(2)]=0`
`" or " (1//2)(a+b+c) [(a-b)^(2)+(b-c)^(2)+(c-a)^(2)=0`
`" or Either " a+b+c =0`
` " or " (a-b)^(2) +(b-c)^(2) +(c-a)^(2)=0`
` rArr " Either " a+b+c =0 " or " a=b=c`
Discussion
499.

Using the factor theorem it is found that `a+b , b+ca n dc+a`are three factors of the determinant`|-2a a+b a+c b+a-2bb+cc+a c+b-2c|dot`The other factor in the value of the determinant is(a) 4 (b)2 (c_ `a+b+c`(d) none of theseA. 4B. 2C. `a + b + c`D. none of these

Answer» Correct Answer - A
Discussion
500.

Evalute the determinants in queations 1 and 2 : Find the values of x, if (i) `|{:(2,4),(5,1):}|=|{:(2x,4),(6,x):}|` (ii)`|{:(2,3),(4,5):}|=|{:(x,3),(2x,5):}|`

Answer» (i) `|{:(2,4),(5,1):}|=|{:(2x,4),(6,x):}|`
`rArr" "2-20=2x^(2)-24`
`rArr" "2x^(2)=2-20+24=6`
`rArr" "x^(2)=3" "rArr" "x=+-sqrt3`
(ii)`|{:(2,3),(4,5):}|=|{:(x,3),(2x,5):}|`
`rArr" "10-12=5x-6x`
`rArr" "-2=-x`
`rArr" "x=2`
Discussion