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20901.

Statement-1: The (x – p) (x – r) + λ (x – q) (x – s) = 0 where p < q < r < s has non real roots if λ > 0.Statement-2: The equation (p, q, r ∈R) βx2 + qx + r = 0 has non-real roots if q2 – 4pr < 0.(A) Statement – 1 is True, Statement – 2 is True; Statement – 2 is a correct explanation for Statement – 1. (B) Statement – 1 is True, Statement – 2 is True; Statement – 2 is NOT a correct explanation for Statement – 1.(C) Statement – 1 is True, Statement – 2 is False. (D) Statement – 1 is False, Statement – 2 is True

Answer»

 Correct option (A) 

Explanation:

 x2 − x + 1

= (x - 1/2)2 + 3/4
Discussion
20902.

Statement-1:  ax2 + bx + C = 0 is a quadratic equation with real coefficients, if 2 + 3 is one root then other root can be any other real number.Statement-2: If P + √q + is a real root of a quadratic equation, then P - √q is other root only when the coefficients of equation are rational(A) Statement – 1 is True, Statement – 2 is True; Statement – 2 is a correct explanation for Statement – 1. (B) Statement – 1 is True, Statement – 2 is True; Statement – 2 is NOT a correct explanation for Statement – 1.(C) Statement – 1 is True, Statement – 2 is False. (D) Statement – 1 is False, Statement – 2 is True

Answer»

 Correct option (D) Statement – 1 is False, Statement – 2 is True

Explanation:

R is obviously true. So test the statement let f(x) = (x – p) (x – r) + λ (x – q) (x – s) = 0

 Then f(p) = λ (p – q) (p – s) f(r) = λ (r – q) (r – s) If λ > 0 then f(p) > 0, f(r) < 0

⇒ There is a root between p & r Thus statement-1 is false.
Discussion
20903.

Statement-1:  One is always one root of the equation (l – m)x2 + (m – n) x + (n – l ) = 0, where l, m, n∈R.Statement-2:  If a + b + c = 0 in the equation ax2 + bx + c = 0, then 1 is the one root.(A) Statement – 1 is True, Statement – 2 is True; Statement – 2 is a correct explanation for Statement – 1. (B) Statement – 1 is True, Statement – 2 is True; Statement – 2 is NOT a correct explanation for Statement – 1.(C) Statement – 1 is True, Statement – 2 is False. (D) Statement – 1 is False, Statement – 2 is True

Answer»

  Correct option (A) Statement – 1 is True, Statement – 2 is True; Statement – 2 is a correct explanation for Statement – 1. 

Explanation:

The roots of the given equation will be of opposite signs. If they are real and their product is negative D ≥ 0 and product of root is < 0

⇒ (a3 – 8a – 1)2 – 8(a2 – 4a) ≥ 0 and  a2 - 4a/2 < 0

⇒ a2 – 4a < 0

⇒ 0 < a < 4. 
Discussion
20904.

Statement-1 : If 1 + √2 + is a root of x2 – 2x – 1 = 0, then 1 -√2 − will be the other root.Statement-2 : Irrational roots of a quadratic equation with rational coefficients always occur in conjugate pair. (A) Statement – 1 is True, Statement – 2 is True; Statement – 2 is a correct explanation for Statement – 1. (B) Statement – 1 is True, Statement – 2 is True; Statement – 2 is NOT a correct explanation for Statement – 1.(C) Statement – 1 is True, Statement – 2 is False. (D) Statement – 1 is False, Statement – 2 is True

Answer»

(A) Statement – 1 is True, Statement – 2 is True; Statement – 2 is a correct explanation for Statement – 1. 
Discussion
20905.

Let a, b, c be real such that ax2 + bx + c = 0 and x2 + x + 1= 0 have a common rootStatement–1 : a = b = cStatement–2 : Two quadratic equations with real coefficients can not have only one imaginary root common.(A) Statement – 1 is True, Statement – 2 is True; Statement – 2 is a correct explanation for Statement – 1. (B) Statement – 1 is True, Statement – 2 is True; Statement – 2 is NOT a correct explanation for Statement – 1.(C) Statement – 1 is True, Statement – 2 is False. (D) Statement – 1 is False, Statement – 2 is True

Answer»

Correct option  (A) Statement – 1 is True, Statement – 2 is True; Statement – 2 is a correct explanation for Statement – 1. 

Explanation:

x2 + x + 1 = 0 

D = – 3 < 0 

∴ x2 + x + 1 = 0 and ax2 + bx + c = 0 have both the roots common 

⇒ a = b = c. 
Discussion
20906.

Statement-1:  If (a2 – 4) x2 + (a2 – 3a + 2) x + (a2 – 7a + 0) = 0 is an identity, then the value of a is 2.Statement-2:  If a = b = 0 then ax2 + bx + c = 0 is an identity.(A) Statement – 1 is True, Statement – 2 is True; Statement – 2 is a correct explanation for Statement – 1. (B) Statement – 1 is True, Statement – 2 is True; Statement – 2 is NOT a correct explanation for Statement – 1.(C) Statement – 1 is True, Statement – 2 is False. (D) Statement – 1 is False, Statement – 2 is True

Answer»

(C) Statement – 1 is True, Statement – 2 is False. 
Discussion
20907.

Let f(x) = x2 = –x2 + (a + 1) x + 5Statement–1 : f(x) is positive for same α &lt; x &lt; β and for all a∈R Statement–2 : f(x) is always positive for all x∈R and for same real 'a'.(A) Statement – 1 is True, Statement – 2 is True; Statement – 2 is a correct explanation for Statement – 1. (B) Statement – 1 is True, Statement – 2 is True; Statement – 2 is NOT a correct explanation for Statement – 1.(C) Statement – 1 is True, Statement – 2 is False. (D) Statement – 1 is False, Statement – 2 is True

Answer»

 Correct option (C) Statement – 1 is True, Statement – 2 is False. 

Explanation:

Here f(x) is a downward parabola

D = (a + 1)2 + 20 > 0

From the graph clearly st (1) is true but st (2) is false 
Discussion
20908.

Statement-1:  The roots of the equation 2x2 + 3i x + 2 = 0 are always conjugate pair.Statement-2:  Imaginary roots of a quadratic equation with real coefficients always occur in conjugate pair. (A) Statement – 1 is True, Statement – 2 is True; Statement – 2 is a correct explanation for Statement – 1. (B) Statement – 1 is True, Statement – 2 is True; Statement – 2 is NOT a correct explanation for Statement – 1.(C) Statement – 1 is True, Statement – 2 is False. (D) Statement – 1 is False, Statement – 2 is True

Answer»

(D) Statement – 1 is False, Statement – 2 is True
Discussion
20909.

Consider the equation (a + 2)x2 + (a – 3) x = 2a – 1  Statement–1 : Roots of above equation are rational if 'a' is rational and not equal to –2.Statement–2 : Roots of above equation are rational for all rational values of 'a'.A) Statement – 1 is True, Statement – 2 is True; Statement – 2 is a correct explanation for Statement – 1. (B) Statement – 1 is True, Statement – 2 is True; Statement – 2 is NOT a correct explanation for Statement – 1.(C) Statement – 1 is True, Statement – 2 is False. (D) Statement – 1 is False, Statement – 2 is True

Answer»

(C) Statement – 1 is True, Statement – 2 is False.

Obviously x = 1 is one of the root 

∴ Other root = – 2a - 1/a + 2

 = rational for all rational a ≠ –2.

(C)  is correct option. 
Discussion
20910.

Statement–1 : The number of values of a for which (a2 – 3a + 2) x2 + (a2 – 5a + b) x + a2 – 4 = 0 is an identity in x is 1.Statement–2 : If ax2 + bx + c = 0 is an identity in x then a = b = c = 0.(A) Statement – 1 is True, Statement – 2 is True; Statement – 2 is a correct explanation for Statement – 1. (B) Statement – 1 is True, Statement – 2 is True; Statement – 2 is NOT a correct explanation for Statement – 1.(C) Statement – 1 is True, Statement – 2 is False. (D) Statement – 1 is False, Statement – 2 is True  

Answer»

 

Correct option (A) Statement – 1 is True, Statement – 2 is True; Statement – 2 is a correct explanation for Statement – 1. 

Explanation:

 

(a2 – 3a + 2) x2 + (a2 – 5a + 6) x + a2 – 4 = 0

Clearly only for a = 2, it is an identify. 
Discussion
20911.

If `bar(OA)= hat(i) +3hat(J)-2hat(K), "then" bar(OC)` which bisects the angle AOB is given by:A. `hat(i)-hat(J)-hat(K)`B. `hat(i)+hat(J)+hat(K)`C. `hat(-i)+hat(J)-hat(K)`D. `hat(i)+hat(J)-hat(K)`

Answer» Correct Answer - 4
We know that internal bisector of the angle between the vectors bara and barb is
`therefore barOC = lambdabar(OA+barOB)`
`=lambda[hati+3hatj-2hatk)//sqrt(14)+(3hatj+hatj-2hatk)//sqrt(14)]`
`={lambda//sqrt(14)}(4hati + 4 hatj- 4 hat k)`
={4lamba//sqrt(14)}(hati+hatj-hatk)`
`"Taking" lambda = sqrt(14)//4barOC "can be taken as" (hati +hat j-hatk)`
Discussion
20912.

A 100 cm3 solution of sodium carbonate is prepared by dissolving 8.653 g of the salt in water. The density of solution is 1.0816 g per mililitre. What are the molality and molarity of the solution?

Answer»

We have given,

density of solution = 1.0816 g/ml

\(\therefore\) Numbers of moles of Na2CO3 = \(\frac{8.653}{100}\)

= 0.08 mol

\(\because\) 1cm3 = 1ml

\(\therefore\) 100 cm3 = 100 ml

\(\therefore\) molarity of Na2CO3 solution = \(\frac{0.08}{100}\times1000\)

= 0.8 m

weight of solution = 100 ml x 1.816 g/ml

= 108.16 g

weight of solvent = weight of solution - weight of solute

= 108.16 g - 8.653

= 99.507 g

\(\therefore\) Molality of Na2CO3 solution = \(\frac{0.08}{99.507}\times1000\) 

= 0.8039 m

Hence, molality and molality of Na2CO3 solution will be 0.8039 m and 0.8 m respectively.
Discussion
20913.

In `Delta ABC` with usual notation `r_1/(bc)+r_2/(ca)+r_3/(ab)` isA. `(1)/(2R) -(1)/(r )`B. `2R-r`C. `r-2R`D. `(1)/(r )-(1)/(2R)`

Answer» Correct Answer - 4
`(r_(1))/(bc)=(4R sin (A//2)cos(B//2)cos(c//2))/(2R sin B.2R sin c)`
`=(sin(A//2))/(4R(sin(B//2)sin(C//2)`
`=(sin^(2)(A//2))/(R )`
`"So that" (r_(1))/(bc)+(r_(2))/(ca)+(r_(3))/(ab)`
`=(1)/(r )[(sin^(2)(A//2)+sin^(2)(B//2)+sin^(2)(C//2)]`
`=(1)/(2r)[3-(1+4 sin(A//2)sin(B//2)sin(C//2))]`
`=(1)/(2r)[2-(r)/(R)]=1/r-1/R`
Discussion
20914.

The matrix `A={:[((1)/(sqrt(2)),(1)/(sqrt(2))),((-1)/(sqrt(2)),(-1)/(sqrt(2)))]:}` isA. idempotentB. orthogonalC. nilpotentD. involutory

Answer» Correct Answer - 3
`A^(2)=[{:(" "(1)/(2)-(1)/(2)," "(1)/(2)-(1)/(2)),(-(1)/(2)+(1)/(2),-(1)/(2)+(1)/(2)):}]`
`=[(o,o),(o,o)]`
Discussion
20915.

If the number of ways of selecting 3 numbers out of `1, 2, 3, ……., 2n+1` such that they are in arithmetic progression in 441, then they are in arithemtric progression is 441, then the sum of the divisors of n is equal toA. 21B. 22C. 32D. None of these

Answer» Correct Answer - 3
`C_(2) +^(n+1)C_(2)=441 rArr (n(n-1))/(2)+((n+1)n)/(2)=441`
`n^(2)=441 rArr n =21`
`"sum of divisors of" 21 = (1+3)(1+7)`
`=4xx8=32`
Discussion
20916.

If `log_(2)(5.2^(x)+1),log_(4)(2^(1-x)+1)` and 1 are in A.P,then x equalsA. `log^(5)`B. `1-log^(5)`C. `log_(5) 2`D. None of these

Answer» Correct Answer - 2
The given no are in A.P
`rArr 2 log_(4) (2^(1-x)+12)=log_(2)(5.2^(x)+1)+1`
`rarr 2 log_(2)((2)/(2^(x))+1) = log_(2) (5.2^(x)+1)+ log_(2)`
`rArr (2)/(2) log_(2) (2)/(2^(x))+1=10.2^(x)+2`
`rArr2/y+1=10y+2`
`rArr10 Y^(2)+y-2=0`
`rArr y=2/5 or y= -1/2`
`rArr 2^(x)=2//5 or 2^(x) =-1/2`
`rArr x=log_(2)(2//5)`
`rArr x=log_(2)- log_(2)5`
`rArr x= 1-log_(2)5.`
Discussion
20917.

If `alpha and beta "are the roots of the quadratic equation" ax^(2) +bx+c=0`, then `lim_(xrarr(1)/(alpha)sqrt(1-(cos ex^(2) + bx +a)/(2(1-alphax)^(2)))=`A. `|(c )/(2alpha)((1)/(alpha)-(1)/(beta))|`B. `|(c )/(2beta)((1)/(alpha)-(1)/(beta))|`C. `|(c )/(2beta)((1)/(alpha)-(1)/(beta))|`D. None of these

Answer» Correct Answer - 1
`ax^(2)+bx+c rarr roots are alpha,beta`
`cx^(2) + bx + a rarr "roots are" (1)/(alpha), (1)/(beta)`
`underset(xrarrl//alpha)limsqrt(1-cos c(x-1//alpha)(x-1//beta).C^(2)(x-1//beta)^(2))/(2a^(2)(x-1//alpha)a^(2)(x-1//beta)^(2))c^(2)`
`=underset(xrarr//a)limsqrt((1)/(2).(1)/(2alpha^(2))c^(2)(x-1//beta))^(2)`
`1/2.|c/alpha|1/alpha-1/beta|`
Discussion
20918.

If the ratio of working efficiency of X and Y is 6 ∶ 5 and that of Y and Z is 5 ∶ 4. If X can complete the work in 4 days, then in how many days the total work will be completed if X and Y are assisted by Z on every alternate day?1. 21/11 days2. 3 days3. 28/15 days4. 2 days

Answer» Correct Answer - Option 3 : 28/15 days

Given:

Ratio of work efficiency of X and Y is 6 ∶ 5 and that of Y and Z is 5 ∶ 4.

X can complete the work in 4 days. 

Z assists X and Y on every alternate day.

Concepts used:

Time = Work/Efficiency

Efficiency = Work/Time

Work = Efficiency × Time

Calculation:

Ratio of work efficiency of X and Y is 6 ∶ 5 and that of Y and Z is 5 ∶ 4. 

X ∶ Y = 6 ∶ 5

Y ∶ Z = 5 ∶ 4

⇒ X ∶ Y ∶ Z = 6 ∶ 5 ∶ 4

X can complete the work in 4 days. 

Work = Efficiency × Time

Total work done by X = 6 × 4 = 24 unit

Time = Work/Efficiency

First day work of X and Y = 6 + 5 = 11 unit 

Remaining work = 24 – 11 = 13 unit

Second day work of X, Y and Z = 6 + 5 + 4 = 15 unit

15 unit work is done by X, Y and Z in 1 day.

Time taken by X, Y and Z to complete 13 unit work = 13/15 

Total time taken by X, Y and Z = 1 day + 13/15 days = 28/15 days

∴ X, Y and Z took 28/15 days to complete the work together if X and Y are assisted by Z on every alternate day.
Discussion
20919.

If `alpha in(0,1)` and `f:R->R` and `lim_(x->oo)f(x)=0,lim_(x->oo)(f(x)-f(alphax))/x=0,` then `lim_(x->oo)f(x)/x=lambda` where `2lambda+7` is

Answer» Correct Answer - 7
`lim_(x to oo) (f(x)-f(alphax))/x=0`
`implies` For any `epsilongt0` there is `deltagt0` such that `|x|lt delta` and `|f(x)-f(alphax)|lt epsilon |x|`
using triangle inequality `|f(x)-f(alpha^(n)x)le|f(x)-f(alpha x)|+|f(alpha^(2)x)|+..+|f(alpha^(n-1)x)-f(alpha^(n)x)|`
`gt epsilon|x|(1+alpha+alpha^(2)+........+alpha^(n-1))=epsilon (1-alpha^(n))/(1-alpha)|x| le(epsilon|x|)/(1-alpha)`
As, `nto oo`
`|f(x)|le (epsilon)/(1-alpha)|x|`
`implieslim_(x to oo) (f(x))/x=0`
Discussion
20920.

if a `in` R and equation `(a-2)(x-[x])^(2)+2(x-[x])+a^(2)=0` (where [x] represent G.I.F) has no integral solution and has exactly one solution in the interval (12,13) then a lies in `(alpha,beta)` the value of `alpha^(2)+3beta^(2)` is

Answer» Correct Answer - 1
let `x-[x]=t`
`f(t)=(a-2)t^(2)+2t+a^(2)=0` ..(i)
case (i) `ane2` exactly one root of equation (i) lies in (0,1)
`f(0),f(1)lt0` ltbr. `a^(2)(a^(2)+a)lt0`
`a in (-1,0)`
case (ii) if `a=2` the `t=2`
`implies` no root
Hence `a in (-1,0)`
`alpha=-1`
`beta=0`
Discussion
20921.

Consider a sequence `{a_n}` with `a_1 =(a_(n-1)^2)/(a_(n-2))` for all ` n ge 3` terms of the sequence being distinct .Given that `a_2 " and " a_5` are positive integers and `a_5 le 162`, then the possible values (s) of `a_5` can beA. 162B. 64C. 32D. 2

Answer» Correct Answer - A::C::D
`(a_(n))/(a_(n-1))=(a_(n-1))/(a_(n-2))`
Here `a_(1),a_(2),a_(3)`…..are in G.P.
Let `a_(2)=x` then for `n=3`
`(a_(3))/(a_(2))=(a_(2))/(a_(1))impliesa_(2)^(2)=a_(1)a_(3)impliesa_(3)=(x^(2))/2`
`:.` G.P. is `2, x, (x^(2))/2, (x^(3))/4`,………..
Comon ratio `r=x/2`
Given `(x^(4))/8le162impliesx^(4)1296impliesxle6`
Also, `x` & `(x^(4))/8` are integers
so, if `x` is also even then only `(x^(4))/8` will be an integer
Hence, the possible values of `x` are 4 & 6, because `x!=2` as terms are distinct hence possible values of `a_(5)=(x^(4))/8` are `(4^(4))/8` & `(6^(4))/8`
Discussion
20922.

if `cos^(-1)((1)/(sqrt(2))(cos((7pi)/(5)-sin((2pi)/(5)))=(ppi)/(q)` (where p,q are in lowest form), then value of `(q-p)` is

Answer» Correct Answer - 3
`cos^(-1)((1)/(sqrt(2))(cos((7pi)/(5))-sin((2pi)/(5)))`
`=cos^(-1)(cos((2pi)/(5))cos((3pi)/(4))-sin((3pi)/(4))sin((2pi)/(5)))`
`=cos^(-1)(cos((23pi)/(20)))=(2pi-(23pi)/(20))=(17pi)/(20)`
Discussion
20923.

The sum of three numbers forming a geometric progression is equal to 56. if we subtract 1,7,21 from these numbers, respectively, then the newly obtained numbers will form an arithmetic progression. Find the common ratio (integral value) of the G.P.

Answer» Correct Answer - 2
`(a)/(r)+a+ar=56`
`((a)/(r)-1)+(ar-21)=2(a-7)`
`implies(a)/(r)+ar-22=2a-14`
`implies(a)/(r)+ar=2a+8`
`impliesa+(a)/(r)+ar=3a+8=56`
`implies3a=48impliesa=16`
Also `16((1)/(r)+1+r)=56`
`(1+r+r^(2))/(r)=(7)/(2)`
`implies2+2r+2r^(2)=7t`
`implies2r^(2)-5r+2=0`
`implies2r^(2)-4r-r+2=0`
`2r(r-2)-1(r-2)=0`
`impliesr=(1)/(2),2`
`because` numbers are `(16)/(2),16,16xx2`
Discussion
20924.

if the equation `x^(4)-4x^(3)+ax^(2)+bx+1=0` has four positive roots, then find a.

Answer» Correct Answer - 6
`x^(4)-4x^(3)+ax^(2)+bx+1=0`
let root are `alpha,beta,gamma&lamda`
`=alpha+beta+gamma+lamda=4`
`alphabetagammalamda=1`
As all roots are positive
`=a.m.gtg.m`
`=(alpha+beta+gamma+lamda)/(4)gt(alphabetagammalamda)^((1//4))`
all the numbers are equal
`alpha=beta=gamma=lamda`
`alpha+beta+gamma+lamda=4`
`implies4alpha=4impliesalpha=1`
`alpha=beta=gamma=lamda=1`
Discussion
20925.

Consider the equation (a2 – 3a + 2) x2 + (a2 – 5a + 6)x + a2 – 1 = 0Statement – 1: If a = 1, then above equation is true for all real x.Statement – 2: If a = 1, then above equation will have two real and distinct roots.(A) Statement – 1 is True, Statement – 2 is True; Statement – 2 is a correct explanation for Statement – 1. (B) Statement – 1 is True, Statement – 2 is True; Statement – 2 is NOT a correct explanation for Statement – 1.(C) Statement – 1 is True, Statement – 2 is False. (D) Statement – 1 is False, Statement – 2 is True

Answer»

(C) Statement – 1 is True, Statement – 2 is False. 
Discussion
20926.

Find the number of positive integral solution of the equation `tan^(-1)x+cos^(-1)(y)/(sqrt(1+y^(2)))=sin^(-1)((3)/(sqrt(10)))`.

Answer» Correct Answer - 2
`tan^(-1)x+tan^(-1)((1)/(y))=tan^(-1)3impliesy=(1+3x)/(3-x)`
put `x=1,y=2` and `x=2,y=7`
only two possible integral solution.
Discussion
20927.

Find the integral values of a for which the equation `x^4-(a^2-5a+6)x^2-(a^2-3a+2)=0` has only real roots

Answer» Correct Answer - 3
for the equation
`x^(4)-(a^(2)-5a+6)x^(2)-(a^(2)-3a+2)=0`
to have real root only the equation
`t^(2)-(a^(2)-5a+6)t-(a^(2)-3a+2)=0`
must have both roots greater then or equal to zero
`(a^(2)-5a+6)^(2)+4(a^(2)-3a+2)gt0`
`(a^(2)-5a+6)/(2(a))gt0=a^(2)-5a+6gt0`
`(a-2)(a-3)gt0`
`a in (-infty,2]cup[3,infty)`
`(3)`
`-(a^(2)-3a+2)gt0`
`-(a^(2)-3a+3)gt0`
`-(a-1)(a-2)gt0`
`a in [1,2]`
From 2 & 3 we have integral value of a equal to 1 and 2 which also satisfy condition 1. ltbr `a=1,2`
Discussion
20928.

The value of x for which `|{:(x,2,2),(3,x,2),(3,3,x):}|+|{:(1-x,2,4),(2,4-x,8),(4,8,16-x):}|gt33` isA. `0ltxlt1`B. `-(1)/(2)ltxlt(1)/(2)`C. `xlt-(1)/(7)`D. `xgt1`

Answer» Correct Answer - C::D
`|{:(x,2,2),(3,x,2),(3,3,x):}|+|{:(1-x,2,4),(2,4-x,8),(4,8,16-x):}|ge33`
`implies21x^(2)-18x+30ge33`
`implies21x^(2)-18x-3ge0`
`implies7x^(2)-6x-1ge0`
`implies7x^(2)-7x+x-1ge0`
`(7x+1)(x-1)ge0`
`impliesx in (-infty,(-1)/(7))cup(1,infty)`
Discussion
20929.

If `f : IR rarr IR` be defined by `f(x)={(2x,xgt3),(x^(2),1ltxle3."thenf(-1)+f(2)+f(5)=),(3x,xle1):}`A. 10B. 11C. 12D. 14

Answer» Correct Answer - B
If `f: IR rarr IR` be ……….
`f(-1)= -3, f(2)= 4, f(5) =10`
Discussion
20930.

Let `f(x)=(sin^(-1)x)/(cos^(-1)x)+(cos^(-1)x)/(tan^(-1)x)+(tan^(-1)x)/(sec^(-1)x)+(sec^(-1)x)/(cosec^(-1)x)+(cosec^(-1)x)/(cot^(-1)x)+(cot^(-1)x)/(sin^(-1)x)`. Then which of the following statements holds good?A. Minimum value of f(x) is 6.B. `f(x)` is a continous function.C. `f(-1)=(-107)/(12)`D. `f(x)` is non-derivable at `x=-1`

Answer» Correct Answer - B::C::D
`f(x)` is defined for `x=-1` only and `f(-1)=-(107)/(12)`
Discussion
20931.

If `F(alpha)=[(cosalpha, -sinalpha,0),(sinalpha, cosalpha, 0),(0,0,1)] and G(beta)=[(cosbeta, 0, sinbeta),(0, 1, 0),(-sinbeta, 0, cosbeta)], then [F(alpha)G(beta)]^-1` is equal to (A) `F(-alpha)G(-beta)` (B) `G(-beta)F(-alpha0` (C) `F(alpha^-1)G(beta^-1)` (D) `G(beta^-1)F(alpha^-1)`A. `(A(alpha))^(-1)=A(-alpha)`B. `A(alpha)A(beta)=A(alpha+beta)`C. `B(alpha)B(beta)=B(alpha+beta)`D. `(B(beta))^(-1)=B(-beta)`

Answer» Correct Answer - A::B::C::D
`A(alpha).A(-alpha)=[{:(cosalpha,sinalpha,0),(sinalpha,cosalpha,0),(0,0,1):}][{:(cosalpha,sinalpha,0),(-sinalpha,cosalpha,0),(0,0,1):}]`
`=[{:(1,0,0),(0,1,0),(0,0,1):}]=I`
Also `|A(alpha)|ne0`
`because(A(alpha))^(-1)=A(-alpha)`
`B(beta)B(-beta)=[{:(cosbeta,0,sinbeta),(0,1,0),(-sinbeta,0,cosbeta):}][{:(cosbeta,0,-sinbeta),(0,1,0),(sinbeta,0,cosbeta):}]`
`=[{:(1,0,0),(0,1,0),(0,0,1):}]=I`
And `|B(beta)|ne0`
`implies(B(beta))^(-1)` exist
`(B(beta))^(-1)=B(beta)`
`A(alpha)A(beta)=[{:(cosalpha,-sinalpha,0),(sinalpha,cosalpha,0),(0,0,1):}][{:(cosbeta,-sinbeta,0),(sinbeta,cosbeta,0),(0,0,1):}]`
`=[{:(cos(alpha+beta),-sin(alpha+beta),0),(sin(alpha+beta),cos(alpha+beta),0),(0,0,1):}]=A(alpha+beta)`

`B(alpha)B(beta)=[{:(cosalpha,0,sinalpha),(0,1,0),(-sinalpha,0,cosalpha)][{:(cosbeta,0,sinbeta),(0,1,0),(-sinbeta,0,cosbeta):}]`
`=[{:(cos(alpha+beta),0,sin(alpha+beta)),(0,1,0),(-sin(alhpa+beta),0,cos(alpha+beta)):}]=B(alpha+beta)`
Discussion
20932.

If `(1)/(log_(2)a)+(1)/(log_(4)a)+(1)/(log_(8)a)+(1)/(log_(16)a)+….+` `(1)/(log_(2^(n))a) = (n(n+1))/(k)` then k `log_(a)2` is equal toA. 2B. 3C. 1D. 4

Answer» Correct Answer - A
`If (1)/(log_(2)a)+(1)/(log_(4)a)+(1)/(log_(8)a)` …………
Given
`log_(a)2+log_(a)2^(2)+……+log_(a)2^(n) = (n(n+1))/(k)`
`implies (1+2+3+…….+n)log_(a)2 = (n(n+1))/(k)`
`implies (n(n+1))/(2). Log_(a)2 = (n(n+1))/(k)`
`implies k. log_(a)2 = 2`
Discussion
20933.

The value of `lamda` so that the matric `A^(-1)lamdaI` is singular where `A=[{:(6,-2,2),(-2,3,-1),(2,-1,3):}]`A. `(1)/(2),(-1)/(2)`B. `1,(-1)/(2)`C. `1,(1)/(8)`D. `(1)/(2),(1)/(8)`

Answer» Correct Answer - D
`|A^(-1)-lamdal|=0`
`|A||A^(-1)-lamdal|=0`
`|l-lamdaA|=0`
`|(l)/(lamda)-A|=0`
`|A-(1)/(lamda)l|=0`
`|{:(6-(1)/(lamda),-2,2),(-2,3-(1)/(lamda),-1),(2,-1,3-(1)/(lamda)):}|=0`
`lamda=(1)/(2),(1)/(8)`
Discussion
20934.

If `aN = {ax|x in N}` and `bN nn cN = dN`, where `b,c epsilon N`, thenA. `d = bc`B. `c = bd`C. `b = cd`D. `d = LCM (b,c)`

Answer» Correct Answer - D
If `aN = {ax|x in N}` …………….
We have.
`bN = {bx|x in IN}` = the set of the positive integral multiples of b
`cN = {cx|x in IN}` = the set of the positive integral multiples of c
`:. bN nn cN` = the set of positive integral multiple of LCM of b & c
`implies d = LCM (b,c)`
Discussion
20935.

if `b_(1),b_(2),b_(3)(b_(1)gt0)` are three successive terms of a G.P. with common ratio `r`, the value of for which the inequality `b_(3)gt4b_(2)-3b_(1)`, holds is given byA. `rgt3`B. `0ltrlt1`C. `r=3.5`D. `r=5.2`

Answer» Correct Answer - A::B::C::D
`b_(3)ge4b_(2)-3b_(1)`
`impliesb_(1)r^(2)ge4.b_(1)r-3b_(1)`
`r^(2)ge4r-3`
`impliesr^(2)-4r+3ge0`
`(r-1)(r-3)ge0`
`impliesr in (-infty,1)cup(3, infty)`
Discussion
20936.

if `(3sin^(-1)x+pix-pi)^(2)+{sin(cos^(-1)((x)/(5)))}^(2)-2sin(cos^(-1)((x)/(5)))=-1` then which of the following is/are Correct?A. sum of values of x is 1B. product of non-zero solution is `(1)/(2)C. number of solution are two number of solutions are zero.D.

Answer» Correct Answer - D
From given equation
`(3sin^(-1)x+pix-pi)^(2)+{sin(cos^(-1)((x)/(5))-1}^(2)=0`
Which provide
`3sin^(-1)x+pix-pi=0impliesx=(1)/(2)` .(i)
and `sin(cos^(-1)((x)/(5)))=1`
`impliescos^(-1)((x)/(5))=(pi)/(2)impliesx=0` .. .(ii)
From (i) & (ii) given equation has no solution
Discussion
20937.

If `S_(n)` denotes the sum to `n` terms of the series `(1 le n le 9)1 + 22 + 333 + "……" + ("nnn…..n")/(n"times")`, then for `n ge 2`A. `S_(n) - S_(n - 1) = (1)/(9)(10^(n) - n^(2) + n)`B. `S_(n) = (1)/(9)(10^(n) - n^(2) + 2n - 2)`C. `9(S_(n) - S_(n-1)) = n(10^(n) - 1)`D. `S_(3) = 356`

Answer» Correct Answer - C::D
`{:(S_(n)=1+(2)/(9)(10^(2)-1)(3)/(9)(10^(3)-1)+"...."+(n)/(9)(10^(n)-1)),(S_(n-1)=1+(2)/(9)(10^(2)-1)(3)/(9)(10^(3)-1)+"..."+):}/(S_(n)-S_(n-1)=(n)/(9)(10^(n)-1))`
`:.9(S_(n)-s_(n-1))=n(10^(n)-1)`
`:. S_(3) = 356 , :. 3` & `4`
Discussion
20938.

If `sum_(r=1)^(n)r(r+1)(2r+3)=an^(4)+bn^(3)+cn^(2)+dn+e`, then.A. `a+c=b+d`B. `e=0`C. `a,b-2//3,c-1` are in A.P.D. `c//a` is an integer

Answer» Correct Answer - A::B::C::D
`sum_(r=1)^(n)r(r+1)(2r+3)=sum_(r=1)^(n)2r^(3)+5r^(2)+3r`
`=2((n^(2)(n+1)^(2))/(4))+5((n(n+1)(2n+1))/(6))+3(n(n+1))/(2)`
`=(3n^(2)(n+1)^(2)+5n(n+1)(2n+1)+9n(n+1))/(6)impliesa=(3)/(6)=(1)/(2)`
`b=(16)/(6)=(8)/(3)`
`c=(27)/(6)=(9)/(2)`
`d=(14)/(6)=(7)/(3)`
`e=0`
Discussion
20939.

Let `a, x, b` in A.P, `a, y, b` in GP `a, z, b` in HP where `a` and `b` are disnict positive real numbers. If `x = y + 2` and `a = 5z`, then which of the following is/are `COR RECT` ?A. `y^(2) = zx`B. `x gt y gt z`C. `a = 9 , b = 1`D. `a = (1)/(4), b = -(9)/(4)`

Answer» Correct Answer - A::B::C
`{:(a_(1)x_(1)b,"inA.P.",rArr,2x=a+b),(a_(1)y_(1)b,"inG.P.",rArr,y^(2)=ab),(a_(1)z_(1)b,"inA.P.",,):}`
`rArr z = (2ab)/(a + b)`
`:. x gt y gt z (A gt G gt H)`
at `y^(2) = xz`
`:. a = 5z` , (Given)…….(i)
`(a)/(5) = (2ab)/(a + b)` or `a = 9b`
`x = y + 2`
or `(a + b)/(2) = sqrt(ab) + 2`
or `5b = 3b + 2 , (b gt 0)`
`:. (1,2,3)`
Discussion
20940.

If `sum_(r=1)^(n)r(r+1)+sum_(r=1)^(n)(r+1)(r+2)=(n(an^(2)+bn+c))/(3),(a,b,cinN)` thenA. `a + b + c = 24`B. `2a + b - c = 0`C. `a - b + c= 6`D. `ab + bc + ca = 11`

Answer» Correct Answer - A::B::C
`overset(n)underset(r=1)sumr(r+1)+overset(n)underset(r=1)sum(r+1)(r+2)`
`=overset(n)underset(r=1)sum2(r+1)^(2)=(n(2n^(2)+9n+13))/(3)`
`rArr (1)/(3)(n+1)(n+2)(n+n+3)`
`rArr (1)/(3)(n+2)(2n+3)(n+1)`
`rArr a_(1) = 1, a_(2) = 2, a_(3) = 3`
Discussion
20941.

Five digit number divisible by 3 is formed using 0 1 2 3 4 6 and 7 without repetition Total number of such numbers are

Answer» Correct Answer - 6
Since a five digits no. is formed using the digits `(0,1,2,3,4` and `5`)
divisible by `3` i.e. only possible
Case`-1` Using digits `0,1,2,4,5` no. of ways `= 4.4.3.2.1 = 96`
Case`-2` Using digits `1,2,3,4,5` no. of ways `5.4.3.2.1= 120`
Total numbers formed `= 120 + 96 = 216 = 6^(3)`
Discussion
20942.

A number 3257X33 is divisible by 11. Find the value of X + 3.1. 42. 13. 34. 5

Answer» Correct Answer - Option 1 : 4

Given:

3257X33 is divisible by 11

Concept Used:

A number is divisible by 11 if the difference of sum of the alternate terms is 0 or a multiple of 11.

Calculations:

Sum of alternate terms = (3 + 5 + X + 3)

⇒ Sum of alternate terms = (11 + X)      ----(1)

Sum of terms that are left = (2 + 7 + 3)

⇒ Sum of terms that are left = 12      ----(2)

Difference of  sum of the alternate terms is 0.

(2) - (1) = 0

⇒ 12 - (11 + X) = 0

⇒ X = 1

Now, X + 3 = 1 + 3

⇒ X + 3 = 4

∴ The value of (X + 3) is 4.
Discussion
20943.

`a, b, c` are positive integers froming an increasing `G.P`. whose common ratio is rational number, `b - a` is cube of natural number and `log_(6)a + log_(6)b + log_(6)c = 6` then `a + b + c` is divisible byA. `3`B. `5`C. `7`D. `9`

Answer» Correct Answer - A::C::D
`log_(6)a + log_(6)b + log_(6)c = 6`
`:. abc = 6^(6) rArr b^(3) = 6^(6)`
`b - a` is cube of a natural number
`b = 36, a = 9, c = 144`
`a + b + c = 169`
which is divisible by `3` & `7`.
Discussion
20944.

Third term in expression of `(x + x^(log_(10)x))^(5)` is `10^(6)` than possible value of `x` areA. `1`B. `10`C. `10^(-5//2)`D. `10^(6)`

Answer» Correct Answer - B::C
`T_(3) = .^(5)C_(2)x^(3)(x^(log_(10)x))^(2) = 10.x^(3+2log_(10)x)=10^(6)`
`rArr x^(3+2log_(10)x) = 10^(5)`
taking `log` on both sides
`(3 + 2log_(10)x)log_(10)x=5`
`rArr log_(10)x = 1` or `-(5)/(2)`
Discussion
20945.

For any integral value of n, 73n + 14n + 24 when divided by 7 will leave the remainder.1. 42. 33. 54. 2

Answer» Correct Answer - Option 2 : 3

Given:

73n + 14n + 24 divided by 7

Concept used:

Number = Divisor × Dividend + Remainder

Calculation:

73n + 14n + 24 divided by 7

⇒ 73n + 14n + 21 + 3

⇒ 7(73n - 1 + 2n + 3) + 3

Number = Divisor × Dividend + Remainder

So, the remainder is 3

∴ 73n + 14n + 24 when divided by 7 will leave the remainder 3.
Discussion
20946.

What is the difference between \(\frac{7}{16}\) and \(\frac{7}{48}\)?1. \(\frac{7}{48}\)2. \(\frac{7}{24}\)3. \(\frac{7}{12}\)4. \(\frac{7}{32}\)

Answer» Correct Answer - Option 2 : \(\frac{7}{24}\)

Given:

\(\frac{7}{{16}}{\rm{\;}},\frac{7}{{48}}\)

Calculation:

7/16 – 7/48

⇒ (21 – 7)/8

⇒ 14/48

⇒ 7/24

∴ Difference between  \(\frac{7}{16}\) and \(\frac{7}{48}\) is \(\frac{{7}}{{24}}\).
Discussion
20947.

If `(1 + x + x^(2) + x^(3))^(n) = a_(0) + a_(1)x + a_(2)x^(2)+"……….."a_(3n)x^(3n)` then which of following are correctA. `a_(0) + a_(1) + a_(2) +……+a_(3n) = 2^(2n)`B. `a_(0) + a_(2) + a_(4) +"……"= a_(1) + a_(3) + a_(5) +"…….."`C. `a_(0) = a_(3n), a_(1) = a_(3n - 1), a_(2) = a_(3n - 2)`D. `a_(0) + a_(2) + a_(4) + "……." = 2^(2n - 1)`

Answer» Correct Answer - A::B::D
Put `x = 1` to get
`4^(n) = a_(0) + a_(1) + a_(2) + "………" + a_(3n)"………"`
Put `x = -1` to get
`rArr a_(0) + a_(2) + a_(4) "…….." = a_(1) + a_(3) + a_(5) "……"`
Discussion
20948.

Sides of `DeltaABC` are in `A.P.` if a `lt min {b,c}` then `cosA` is equal toA. `(4c - 3b)/(2b)`B. `(4c - 3b)/(2c)`C. `(3c - 4b)/(2c)`D. `(4b - 3c)/(2b)`

Answer» Correct Answer - B::D
Sides are in `A.P.` and be `a lt min{b, c}`
`:.` order of `A.P.` can be `b, c, a` or `c, b, a`
if `2c = a + b` then `cos A = (b^(2) + c^(2) - a^(2))/(2bc)`
`= (b^(2) + c^(2) - (2c - b)^(2))/(2bc) = (4b - 3c)/(2b)`
if `2b = a + c`
then `cos A = (.b^(2) + c^(2) - (2b-c)^(2))/(2bc) = (4c - 3b)/(2c)`
Discussion
20949.

Find the total number of odd divisors of P = 540.1. 242. 213. 204. 23

Answer» Correct Answer - Option 2 : 21

Given:

P = 540

Concept Used:

x = ap × bq

Total number of divisors = (p + 1) × (q + 1)

where, a and b → Prime numbers.

Total number of divisors = Odd divisors + Even divisors

Calculations:

P = 540

⇒ P = 22 × 33 × 51

⇒ Total number of divisors = (2 + 1) × (3 + 1) × (1 + 1)

⇒ Total number of divisors = 3 × 4 × 2

⇒ Total number of divisors = 24

Even divisors = (2 + 1) = 3

Total number of divisors = Odd divisors + Even divisors

⇒ Odd divisors = Total number of divisors - Even divisors

⇒ Odd divisors = 24 - 3

⇒ Odd diviors = 21

∴ The total number of odd divisors is 21.
Discussion
20950.

The number of ways in which `20` identical coins be distributed in `4` persons if each person receive at least `2` coins and atmost `5` coins, areA. `.^(15)C_(12) - 4. .^(11)C_(8) + 6. .^(7)C_(4) - 4`B. `.^(15)C_(3) - 6. . ^(11)C_(8) + .^(7)C_(4)`C. `.^(23)C_(3)`D. `.^(15)C_(0)`

Answer» Correct Answer - A::D
Number of ways `=` coefficient of `x^(20)` in
`(x^(2) + x^(3) + x^(4) + x^(5))^(4) = x^(B) (1 + x + x^(2) + x^(3))^(4)`
`=` coefficient of `x^(12)` in `((1 - x^(4))/(1 - x))^(4) = (1 - 4^(4))^(4)(1 - x)^(-4)`
`=.^(15)C_(12) - 4 . .^(11)C_(8) + 6. .^(7)(C_(4) - 4 = 1 = .^(15)C_(0)`
Discussion