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251.

A filter paper soaked in salt X tums brown when exposed to `HNO_(3)` capor. The salt X isA. KClB. KBrC. KlD. `K_(2)SO_(4)`

Answer» Correct Answer - C
`2KI+4HNO_(3)toI_(2)+2NO_(2)+2KNO_(3)+2H_(2)O`
Iodide ion is a strong reducing agent and reduces `HNO_(3)" vapours to " NO_(2)` (Brown gas)
Discussion
252.

A filter paper soaked in salt X turns brown when exposed to HNO3 vapor. The salt X is –  (A) KCl (B) KBr (C) KI (D) K2SO4 

Answer»

Correct option (C) KI

Explanation: 

Filter paper soaked with KI turns brown when exposed to HNO3 vapour due to liberation of iodine. The reaction is as follows : 

6KI + 8HNO3 → 6KNO3 + 4H2O + 2NO + 3I2
Discussion
253.

The formal oxidation of Cr and Cl in the ions `Cr_(2)O_(7)^(2-)" and "ClO_(3)^(-),` respectively, areA. `+6 and +7`B. `+7 and +5`C. `+6 and +5`D. `+8 and +7`

Answer» Correct Answer - C
Oxidation number of Cr in `Cr_(2)O_(7^(2-))=+2x-14=-2rArrx=+6`
Oxidation number of Cl in `ClO_(3)^(-)=x-6=-1rArrx=+5`
Discussion
254.

The weight of calcium oxide formed by burning 20 g of calcium in excess oxygen is-A. 36gB. 56gC. 28gD. 72g

Answer» Correct Answer - A::B::C::D
CaO formed `=(1)/(2)` moles
`w=(1)/(2)xx56=28gm`
Discussion
255.

The amount (in mol) of bromoform `(CHBr_(3))` produced when 1.0 mol of acetone reacts completely with 1.0 mol of bromie in the presence of aqueous NaOH isA. `1/3`B. `2/3`C. `1`D. `2`

Answer» Correct Answer - A
`underset( 1 "mole")(CH_(3)-overset(O) overset(||)C-CH_(3))+3Br^(2)+underset(1 "mole")(4NaOH)toCH_(3)-overset(O)overset(||)C-overset(-)(O)Na^(+)+3NaBr+3H_(2)O+underset(1//3 " mol" )(CHBr_(3))`
Moles of bromoform produced `=1//3` mole
Discussion
256.

X and Y areA. enantiomersB. diastereomersC. constitutional isomersD. conforms

Answer» Correct Answer - D
Both x and y represent meso-2,3-dichlorobutane. They are conformers, where X is eclipsed and Y is anti form
Discussion
257.

A graph of species richness vs area on log-log axes isA. linearB. sigmoidalC. oscillatoryD. parabolic

Answer» Correct Answer - A
log S= log c +z log A
Discussion
258.

Filariasis is caused by (A) Entamoeba histolytica (B) Plasmodium falciparum (C) Trypanosoma brucei (D) Wuchereria bancrofti 

Answer»

Correct Option :- (D) Wuchereria bancrofti

Explanation :-

Wuchereria bancrofti lives in lymphatic vessels and causes swelling of lower limps and scrotum
Discussion
259.

Which ONE of the following molecules is dervied from pantothenic acid ?A. Thiamine pyrophosphateB. Nicotinamide adenine dinucleotide phosphateC. Flavin adenine dinucleotide phosphateD. Acetyl-CoA

Answer» Correct Answer - D
Vitamin `B_(5)` is pantothenic acid, that synthesize Co-enzyme A (CoA)
Discussion
260.

Which ONE of the following conversions does NOT happen under anaerobic conditions ?(A) Glucose to ethanol by Saccharomyces. (B) Lactose to lactic acid by Lactobacillus. (C) Glucose to CO2 and H2O by Saccharomyces. (D) Cellulose to glucose by Cellulomonas.

Answer»

Correct Option :- (C) Glucose to CO2 and H2O by Saccharomyces.

Explanation :-

Glucose to CO2 and H2O is formed during aerobic respiration.
Discussion
261.

Which ONE of the following conversions does NOT happen under anaerobic conditions ?A. Glucose to ethanol by Saccharomyces.B. Lactose to lactic acid by Lactobacillus.C. Glucose to `CO_(2)" and "H_(2)O` by Saccharomyces.D. Cellulose to glucose by Cellulomonas.

Answer» Correct Answer - C
Glucose to `CO_(2)" and "H_(2)O` is formed during aerobic respiration.
Discussion
262.

The number of electrons required to reduce one molecule of oxygen to water during mitochondrial oxidation isA. 4B. 3C. 2D. 1

Answer» Correct Answer - A
`O_(2)+4e^(-)+4H^(+)rarr2H_(2)O`
Discussion
263.

An amont of 18 g glucose corresponds toA. 1.8 moleB. 1 moleC. 0.18 moleD. 0.1 mole

Answer» Correct Answer - D
`"Mole "=("mass in gram")/("molecular weight")=(18)/(180)=0.1`
Discussion
264.

The gas produced on heating `MnO_(2)` with cone. HCl is -A. `Cl_(2)`B. `H_(2)`C. `O_(2)`D. `O_(3)`

Answer» Correct Answer - A
`MnO_(2)` Oxidises `Cl^(-)` of HCl to `Cl_(2)`
Discussion
265.

The number of real number `lambda` for which the equality `(sin(lamdaalpha))/(sinalpha)-(cos(lamdaalpha))/(cosalpha)=lamda-1`, holds for all real `alpha` which are not integral multiples of `pi//2` is-A. 1B. 2C. 3D. Infinite

Answer» Correct Answer - B
`(sin(lambdaalpha))/(sinalpha)-(cos(lambdaalpha))/(cosalpha)=lambda-1`
By observation
`sin(lambdaalpha)cosalpha-cos(lambdaalpha)sinalpha=(lambda-1)sinalphacosalpha`
`sin(alpha-1)alpha=(lambda-1)sinalphacosalpha`
clearly `lambda=1, lambda=3` is solution
Discussion
266.

The value of the limit `lim_(x to 0) ((x)/(sinx))^(6//x^(2))` isA. eB. `e^(-1)`C. `e^(-1//6)`D. `e^(6)`

Answer» Correct Answer - A
`underset(xto0)lim((x)/sinx)^(6//x^(2))" "(1)^(oo)`
`underset(e)underset(" "xto0)(" "lim)(6)/(x^(2))((x)/(sinx)-1)`
`underset(e)underset(" "xto0)(" "lim)(6)/(x^(2))((x-sinx)/(sinx))`
`underset(e)underset(" "xto0)(" "lim)(6)/(x^(2)){(x-(x-x^(3)/(3!)+x^(5)/(5!) . . . . . .))/{x-(x-x^(3)/(3!)+x^(5)/(5!) . . . . . .)}}`
`underset(e)underset(" "xto0)(" "lim)(6)/(x^(2)){{(x^(3))/(3!)-x^(5)/(5!) . . . . . .}}/((x-x^(3)/(3!)+x^(5)/(5!) . . . . . .))`
`underset(e)underset(" "xto0)(" "lim)(6x^(3))/(x^(2)){{(1)/(3!)-x^(5)/(5!) . . . . . .}}/{{1-x^(2)/(3!) . . . . . .}}=e^(1)`
Discussion
267.

The diagram below shows two circle loops of wire (A and B) centred on and perpendicular to the x-axis, and oriented with their planes parallel to each other. The y-axis passes vertically through loop A (dashed line). There is a current IB in loop B as shown. Possible actions which we might perform on loop A are: (i) Move A to the right along x axis closer to B (ii) Move A to the left along x axis away from B (iii) As viewed from above, rotate A clockwise about y axis (iv) As viewed from above, rotate A anticlockwise about y axis Which of these actions will induce a current in A only in the direction shown. (A) Only (i) (B) Only (ii) (C) Only (i) and (iv) (D) Only (ii) and (iii)

Answer»

Correct Option :- A) Only (i)

Explanation :-

According to Lenz's Law
Discussion
268.

A parachutist with total weight 75 kg drops vertically onto a sandy ground with a speed of `2 ms^(-1)` and comes to a halt over a distance of `0.25 m`. The average force from the ground on her is close to -A. 600 NB. 1200 NC. 1350 ND. 1950 N

Answer» Correct Answer - C
Discussion
269.

Force `vecF` applied on a body is written as `vecF=(hatn.hatF)" "hatn+vecG`, where `hatn` is a unit vector. The vector `vecG` is equal toA. `hatnxxvecF`B. `hatnxx(hatnxxvecF)`C. `(hatnxxvecF)xxvecF//|vecF|`D. `(hatnxxvecF)xxhatn`

Answer» Correct Answer - D
Discussion
270.

Consider the R-L-C circuit given below. The circuit is driven by a 50 Hz AC source with peak voltage 220 V. If R = 400 `Omega`, C = 200`mu`F and L = 6 H, the maximum current in the circuit is closest to A. `0.120` AB. `0.55` AC. `1.2` AD. `5.5` A

Answer» Correct Answer - A
Discussion
271.

A certain 12 - hour digital clock displays the hour and the minute of a day. Due to a defect in the clock whenever the digit 1 is supposed to be displayed it displays 7. What fraction of the day will the clock show the correct time ?A. `1/2`B. `5/8`C. `3/4`D. `5/6`

Answer» Correct Answer - A
The clock will show the incorrect time (between 1 - 2, 10 - 11, 11- 12, 12 - 1 day and night night both)
`:." in correct time "8 xx 60 = 480` (each minute it will display 1)
Remaining 20 hours it will show the incorrect time ` 16 xx 15 =240`
Total incorrect time = `240 + 480 =720`
correct time = 1 - incorrect time
` = 1-(720)/(24 xx 60)=1//2`
Discussion
272.

There are 30 questions in a multiple –choice test. A student gets 1 mark for each unattempted question, 0 mark for each wrong answer and 4 marks for each corrent answer. A student answered x question correctly and scored 60. Then the number of possible value of x is  (A) 15 (B) 10 (C) 6 (D) 5 

Answer»

Correct option (C) 6 

Explanation:

Right (4 marks) Wrong (0 mark) Unattempted (1 marks)
15150
144
138
1212
1116
1020

6 cases. 
Discussion
273.

Let f `(x) = ax^(2)+bx+c` where a, b, c are integers. Suppose f (1)=0,`40 lt f (6) lt 50, 60 lt f (7) lt 70, and 1000t lt f(50) lt 1000(t+1)` for some integer t. Then the value fo t isA. 2B. 3C. 4D. 5 or more

Answer» Correct Answer - C
f(x)=`ax^(2)+bx+c`
given f (1) = 0
`rArr a+b+c = 0`
and `40 lt f(6) lt 50`
`rArr 40 lt 36a+6b+c lt 50`
`rArr 40 lt 35a+5b lt 50`
`rArr 8 lt 7a+b lt 10`
7a+b = integer = 9 …..(1)
and `60 lt f(7) lt 70`
`rArr 60 lt 49a+7b +c lt 70`
`rArr 60 lt 48a+6b lt 70`
`rArr 10 lt 8a+b lt 11.6`
8a+b = integer = 11 ....(2)
Solving (1)&(2)
a =2, b = -5, c = 3
`:. f (x) 2x^(2)-5x+3`
f(50) = 4753
`1000 t lt f(50) lt 1000 (t+1)`
`(1000 xx 4) lt 4753 lt 1000 (4+1)`
`:. t = 4`
Discussion
274.

A person looks at the image of two parallel finite length lines PQ and RS in a convex mirror (see figure). Which of the following represents schematically the image correctly ? (Note : Letters P, Q, R and S are used only to denote the endpoints of the lines.) A. AB. BC. CD. D

Answer» Correct Answer - B
Discussion
275.

The number of roots of equation `cos^(7)theta - sin^(4)theta = 1` that lie in the interval `[0,2pi]` is-A. 2B. 3C. 4D. 8

Answer» Correct Answer - A
`cos^(7)theta = 1 + sin^(4)theta ge 1` but `cos^(7) le 1`
so `cos theta = 1 , sin theta = 0`
`theta = 0, 2pi`
Discussion
276.

Let `(1+x+x^(2))^(2014)=a_(0)+a_(1)x+a_(2)x^(2)+a_(3)x^(3)+ . . . +a_(4028)x^(4028),` and let `A=a_(0)-a^(3)+a_(6)- . . . +a_(4026)` `B=a_(1)-a_(4)+a_(7)- . . . . . . . -a_(4027)`, `C=a_(2)-a_(5)+a_(8)- . . . .+a_(4028)`, Then-A. `|A|=|B|gt|C|`B. `|A|=|B|lt|C|`C. `|A|=|C|gt|B|`D. `|A|=|C|lt|B|`

Answer» Correct Answer - D
`(1+x+x^(2))^(2014)=a_(0)+a_(1)x+a_(2)x^(2)+a_(3)x^(3)+ . . . +a_(4028)x^(4028),`
put x=-1
`1=1=a_(0)-a_(2)+a_(2)-a_(3)+a_(4)-a_(5)+a_(6)-a_(7). . . .` . . . (1)
`"put"x=-omega`
`(2omega)^(2014)=(1-omega+omega^(2))^(2014)=a_(0)-a_(1omega)+a_(2omega^(2))+a_(3)+a_(4)omega^(2)+a_(6). . . . . . . ` . . .(2)
`"Put"x=-omega^(2)`
`(2omega^(2))^(2014)=(a-omega^(2)+omega)^(2014)=a_(0)-a_(3)+a_(4)omega^(2)+a_(2)omega-a_(3)+a_(4)omega^(2)-a_(5)omega . . . . . . .` (3)
Now, (1)+(2)+(3)
`rArr1+(2omega)^(2014)+(2omega^(2))^(2014)=3(a_(0)-a_(3)+a_(6) . . . . . . . . .`
`rArra_(0)-a_(3)+a_(6). . . . . . . .=(1+2^(2014)omega+2^(2014)omega^(2))/(3)`
`A=(1-2^(2014))/(3)`
`|A|=(2^(2014)-1)/(3)`
`and (1)+(2)xxomega+(3)omega^(2)`
`rArr(1+2^(2014).omega^(2014).omega^(2015)+2^(2014).omega^(4030))/(3)=a_(2)-a_(5)+a_(8). . . . .`
`rArr(1+2^(2014)+omega^(2015)+2^(2014).omega^(4030))/(3)=C`
`rArrC=(1-2^(2014))/(3)rArr|C|=(2^(2014)-1)/(3)`
and similarly `(1)+(2)xxomega^(2)+(3)xxomega`
`B=(1+2^(2014).omega^(2014).omega^(2)+2^(2014).(omega^(2))^(2014).omega)/(3)`
`=(1+2^(2014).omega^(2016)+2^(2014).omega^(4029))/(3)`
`|B|=(1+2^(2015))/(3)`
`:.|B|gt|A|=|C|`
Discussion
277.

The number of solution pairs (x , y) of the simultaneous equations `log_(1//3) (x + y) + "log"_(3) (x-y) = 2` and `2^(y^(2)) = 512^(x + 1)` is

Answer» Correct Answer - B
`"log"_(3^(-1)) (x + y) + "log"_(3)( x - y) = 2`
`-"log"_(3) (x +y) + "log"_(3) (x - y) = 2`
`"log"_(3) ((x - y)/(x + y)) = 2 `
`( x - y)/(x + y) = 9`
`2^(y^(2)) = (2^(9))^(x + 1)`
`2^(y^(2)) = 2^(9^(x + 1))`
`y^(2) = 9(x + 1)`
Solve eliminate y
`16x^(2) - 225 x -225 = 0 `
x = 15 , `(-15)/(16)`
At x = 15 , y = `-12`
x = `(-15)/(16) , y = (3)/(4)` ( not possible )
only sol . x = 15, y = `-12` only one sol.
Discussion
278.

Consider an incomplete pyramid of balls on a square base having 18 layers; and having 13 balls on each side of the top layer. Then the total number N of balls in that pyramid satisfiesA. `9000 lt N lt 10000`B. `8000 lt N lt 9000`C. `7000 lt N lt 8000`D. `10000 lt N lt 12000`

Answer» Correct Answer - B
Top layer has `(13 xx 13)` balls similarly are layer below top layer will have `(14 xx 14)` balls
We have 18 layer
So total number of balls
`N=(13)^(2)+(14)^(2)+……..+(30)^(2)`
`N = (30xx31xx61)/6-(12xx13xx25)/6`
`N= 8805`.
Discussion
279.

The number of distinct prime divisors of the number ` 512^(3) - 253^(3) - 259^(3)` isA. 4B. 5C. 6D. 7

Answer» Correct Answer - C
`(512)^(3)-253^(3)-259^(3)`
`=(512)^(3)-[(253^(3)+(259)^(3)]`
=`(512)^(3)-(253+259)(253^(2)+259^(2)-(253)(259))`
`=(512)^(3)-(512)[253+259)^(2)-2(253)(259)-(253)(259)]`
`= 512[(512)^(2)-{(512)^(2)-3(253)(259)}]`
`=(512)[3(253)(259)]`
`=2^(9).3. (253)(259)`
`=2^(9).3(11)(23)(7)(37)`
6 prime divisors.
Discussion
280.

Suppose `z` is any root of `11z^(8)+20 i z^(7)+10 iz-22=0`, where `i=sqrt(-1)`. Then `s=|z|^(2)+|z|+1` satisfiesA. `S le 3`B. `3 lt S lt 7`C. `7 le S lt 13`D. `S ge 13`

Answer» Correct Answer - B
Discussion
281.

The maximum possible area bounded by the parabola `y= x^(2)+x+10` and a chord of the parabola of length 1 isA. `(1)/(12)`B. `1/6`C. `1/3`D. `1/2`

Answer» Correct Answer - B
Discussion
282.

Let `A={ theta in R|cos^(2) (sin theta)+sin^(2) (cos theta)=1}` and `B={ theta in R| cos (sin theta) sin (cos theta)=0}`. Then `A nn B`A. is the empty setB. has exactly one elementC. has more than one but finitely many elementsD. has infinitely many elements

Answer» Correct Answer - A
for `A nn B`
`cos (sin theta)=1` or `-1` & `sin (cos theta)=0`
Which is not possible
or `cos (sin theta)=0` & `sin (cos theta)=1` or `-1`
also not possible
so `A nn B` is an empty set
Discussion
283.

For each positive interger n, define `f_(n)(x)=` minimum `(x^(n)/(n!), ((1-x)^(n))/(n!))`, for `0 le x le 1`. Let `I_(n)= int_(0)^(1) f_(n) (x) dx, n ge 1`. Then `I_(n)=sum_(n=1)^(oo) I_(n)` is equal to -A. `2sqrt(e)-3`B. `2sqrt(e)-2`C. `2sqrt(e)-1`D. `2sqrt(e)`

Answer» Correct Answer - A
`I_(n)=underset(0)overset(1//2)(int) x^(n)/(n!) dx +underset(1//2)overset(1)(int) ((1-x)^(n))/(n !) dx=1/((n+1)!) ((1/2)^(n+1)+(1/2)^(n+1))=((1/2)^(n))/((n+1)!)`
`sum_(n=1)^(oo) I_(n)=((1//2)/(2!)+((1//2)^(2))/(3!)+....)=2 sqrt(e)-3`
Discussion
284.

the least postitive interger n from which `root(3) (n+1) -root(3)(n)lt (1)/(12)` is -A. 6B. 7C. 8D. 9

Answer» Correct Answer - C
`(n+1)^(1//3)-n^(1//3) lt (1)/(12) `
` (n+1)-n-3(n+1)^(1//3)n^(1//3)-n^(1//3)lt ((1)/(12))^(3)`
` 1-3n^(1//3) (n+1)^(1//3)xx(1)/(12)lt (1)/((12)^(3))`
`(12)^(3) -3.(12)^(2) n^(1//3)(n+1)^(1//3) lt 1`
` (12)^(3) -1lt 3.(12)^(12)n^(1//3)(n+1)^(1//3)`
`(1727)/(3xx144)lt n^(1//3)(n+1)^(1//3)`
`n(n+1)gt((1727)/(3xx144))^(3)`
`n(n+1)gt 63.88`
`n=8`
Discussion
285.

IF rice is cooked in a pressure cooker on the siachen glacirer , at sea beach , and on Deccan plain , which of the following Is correct about the time taken for cooking rice ?A. Get cooked faster on the siachen glacierB. Gets cooked faster at sea breachC. Gets cooked faster on Deccon plainD. Gets cooked at the same time at all three places

Answer» Correct Answer - D
Pressure cooker is used .
Discussion
286.

Light enters an isosceles right triangular prism at normal incidence through face AB and undergoes total internal reflection at face BC as shown below. The minimum value of the refractive index of the prism is close to A. `1.10`B. `1.55`C. `1.42`D. `1.72`

Answer» Correct Answer - C
Discussion
287.

Which of the following intervals is a possible domain of the function f (x) = log{x} [x] + log[x] {x}, where [x] is the greatest integer not exceeding x and {x} = x – [x] ? (A) (0, 1) (B) (1, 2) (C) (2, 3) (D) (3, 5) 

Answer»

Correct Option :- (C) (2, 3)

Explanation :-

x∉I & [x] > 1 

⇒ x∈(2, 3) only option satisfy. 
Discussion
288.

Let `"f(x)"=(sin(x-a)+sin(x+a))/(cos(x-a)-cos(x+a)),` then-A. `f(x+2pi)=f(x)"but f"(x+alpha)nef(x)"for any"0ltalphalt2pi`B. f is strictly increasing functionC. f is strictly decreasing functionD. f is constant function

Answer» Correct Answer - D
`f(x)=(sin(x-a)+sin(x+a))/(cos(x-a)-cos(x+a))=(2sin(x).cosa)/(2sinx.sina)=cota`
Discussion
289.

Let [x] and {x} be the integer part and fractional part of a real number x respectively. The value of the integral `int_(0)^(5)[x]{x}dx` is -A. `5//2`B. 5C. `34.5`D. `35.5`

Answer» Correct Answer - B
`int_(0)^(5)[x]{x}dx=int_(0)^(5)[x](x-[x])dx=int_(0)^(1)0.dx+int_(1)^(2)1.(x-1)dx+int_(2)^(3)2(x-2)dx+int_(3)^(4)3(x-3)dx+int_(4)^(5)4(x-4)dx`
`(((x-1)^(2))/(2))^(2)+2(((x-2)^(2))/(2))_(2)^(3)+3(((x-3)^(2))/(2))_(3)^(4)+4(((x-4)^(2))/(2))_(4)^(5)`
`(1)/(2)+(2)/(2)+(3)/(2)+(4)/(2)`
= 5
Discussion
290.

the value of `tan9^@-tan2 7^@-tan6 3^@+tan8 1^@` is equal to

Answer» Correct Answer - D
`tan81^(@)-tan63^(@)-tan27^(@)+tan9^(@)`
`tan(90^(@)-9^(@))-tan(90^(@)-27^(@))-tan27^(@)+tan9^(@)`
`cot9^(@)-cot27^(@)-tan27^(@)+Tan9^(@)`
By solving we get
= 4
Discussion
291.

The value of the integral `int_(0)^(pi)(1-|sin 8x|)dx` is

Answer» Correct Answer - C
`I=int_(0)^(pi)1-|sin 8x|dx`
`= pi-int_(0)^(pi)(x|sin 8x|)dx`
`= pi-int_(0)^(8.(8)/(pi))(x|sin 8x|)dx`
`= pi-int_(0)^((8)/(pi))xsin 8xdx`
`=pi-8 [(- cos 8x)/(8)]^((pi)/(8)) `
`P+(-1-phi)=pi-2`
Discussion
292.

Let `a=cos 1^(@)` and `b= sin 1^(@)`. We say that a real number is algebraic if is a root of a polynomial with integer coefficients. ThenA. a is algebraic but b is not algebraicB. b is algebraic but a is not algebraicC. both a and b are algebraicD. neither a nor b is algebraic

Answer» Correct Answer - C
`cos x =(e^(ix)+e^(-ix))/(2)`
`cos 60x =(e^(60ix)+e^(-60ix))/(2)`
`cos 60x =((cos s + i sin x)^(60)+(cos x-i sin x)^(60))/(2)`
`(1)/(2)=((cos^(60)+1^(@)+60c_(2) cos^(58) i sin^(2)1+.......oo)^(60)c_(60)sin^(60)1^(@))/(2)`
Change all `sin 1^(@)` to `cos 1^(@)` using the identity `sin^(2)1^(@)= 1 cos^(2) 1^(@)` equation with root `cos 1^(@)` so it is algebraic . Similarly for `b = sin 1` also algebric .
let a poly. `-x^(2)+1 cos^(2) 1^(@)=0`
`-x^(2)+1= - sin^(2) 1^(@)=0`
Discussion
293.

Let X be a set of 5 elements. The number d of ordered pairs (A,B) of subsets of X such that `Anephi,Bnephi,AnnB=phi` satisfiesA. `50le dle100`B. `101le d le150`C. `151le d le200`D. `200 led`

Answer» Correct Answer - C
`""^(5)C_(2)*2!+"^(5)C_(2)((3!)/(1!2!)xx2!)+.^(5)C_(4)[(4!)/(1!3!)xx2!+(4!)/(2!)xx(2!)/(2!)]+.^(5)C_(5)[(5!)/(1!4!)xx2!+(5!)/(2!3!)xx2!]`
`10(2)+10(6)+5(8+6)+(10+20)`
=20+60+70+30=180
Discussion
294.

Among all sectors of fixed perimeter, choose the one with maximum area. Then the angle at the center of this sector (i.e.,the angle between the boundibg radii) isA. `(pi)/(3)`B. `(3)/(2)`C. `sqrt(3)`D. 2

Answer» Correct Answer - D
Given that `2r_rtheta=P" "r=(p)/(2+theta)`
area `=(1)/(2)r^(2)theta=(1)/(2)theta((P)/(2+theta))^(2)=(1)/(2)(P^(2)theta)/((2+theta)^(2))`
`(dA)/(d theta)=0" "theta=2^(C)`
Discussion
295.

If P(x) is a cubic polynomial with P(1)=3,P(0)=2 and P(-1)=4, then `underset(-1)overset(1)fP(x)dx` isA. 2B. 3C. 4D. 5

Answer» Correct Answer - D
`Let P(x)-ax^(3)+bx^(2)+cx+d`
a+b+c+d=3
d=2
-a+b-c+d4
2b+2d=7
2b+4=7
2b=3
`b=(3)/(2)`
`underset(-1)overset(1)(f)(ax^(3)+bx^(2)+cx+d)dx`
`2underset(0)overset(1)(f)(bx^(2)+d)dx`
`=2((b)/(3)+d)`
`=2((1)/(2)+2)=5`
Discussion
296.

Consider the following statements. For any integer n, I. `n^(2)+3` is never divisible by 17. II. `n^(2)+4` is never divisible by 17. ThenA. both I and II are trueB. both I and II are falseC. I is false and II is trueD. I is true and II is false

Answer» Correct Answer - D
`n^(2)+4` is divisible by 17 check at n=9
`because(n^(2)+3)/(17)=(n^(2)+4)/(17)-(1)/(17)" not divisible by 17 "`
Discussion
297.

Let `nge4` be a positive integer and let `l_(1),l_(2),.....,l_(n)` be the lengths of the sides of arbitrary n-sided non-degenerate polygon P. Suppose `l_(1)/(l_(2))+l_(2)/(l_(3))+....l_(n-1)/(l_(n))+l_(n)/(l_(1))=n.` Consider the following statements: I. The lengths of the sides of P are equal. II. The angles of P are equal. III. P is a regular polygon if it is cyclic. ThenA. I is true and I implies IIB. II is trueC. III is falseD. I and III are true

Answer» Correct Answer - D
`"given ":l_(1)/(l_(2))+l_(2)/(l_(3))......+l_(n)/(l_(1))=n.....(i)`
`therefore"Use A.M"geG.M`
We get
`((l_(1)/(l_(2))+l_(2)/(l_(3)).....+l_(n)/(l_(1))))/(n)geroot(n)(l_(1)/(l_(2)).l_(2)/(l_(3))....l_(n)/(l_(1)))`
`therefore(n)/(n)ge1`
`rArrn=n`
So A.M=G.M
`"Hence "l_(1)/(l_(2))=l_(2)/(l_(3))......=l_(n)/(l_(1))=k`
`rArrk=(l_(1)+l_(2).....+l_(n))/(l_(2)+l_(3).....+l_(n)+l_(1))=1`
`l_(1)=l_(2).....=l_(n)`
Discussion
298.

Let f : R → R be the function f(x) = (x – a1)(x–a2)+(x–a2)(x–a3) + (x– a3) (x–a1) with a1, a2, a3 ∈ R. Then f(x) > 0 if and only if – (A) At least two of a1, a2, a3 are equal (B) a1 = a2 = a3 (C) a1, a2, a3 are all distinct (D) a1, a2, a3 , are all positive and distinct  

Answer»

Correct :- (B) a1 = a2 = a3 

Explanation :-

Only when a1 = a2 = a3 In other cases f(x) will take both positive and negative values 
Discussion
299.

Let r be a real number and `n in N` be such that the polynomial `2x^(2) + 2x + 1` divides the polynomial `(x + 1)^(n) - r`. Then `(n,r)` can be-(A) `(4000, 4^(1000))` (B) `(4000, 1/4^(1000))` (C) `(4^(1000),1/4^1000)` (D) `(4000,1/4000)`A. `(4000, 4^(1000))`B. `(4000, (1)/(4^(1000)))`C. `(4^(1000),(1)/(4^(1000)))`D. `(4000, (1)/(4000))`

Answer» Correct Answer - B
`2x^(2) + 2x + 1 = 0`
`x = (-1+i)/(2), (-1-i)/(2)`
x satisfies `(x + 1)^(n) - r = 0`
`((-1+- i)/(2) + 1)^(n) - r = 0`
`((1+-i)/(2))^(n) -r = 0`
`((1)/(sqrt(2)))^(n) ((1+i)/(sqrt(2)))^(n) = r`
`((1)/(sqrt(2)))^(n) (e^(+-( ipi)/(4)))^(n)= r`
RHS = real
LHS = real only when n = multiply of 4
`n = 4000`
`r = ((1)/(sqrt(2)))^(4000) = 1/(4^(1000))`
Discussion
300.

Let `a_(0) = 0` and ` a_(n) = 3a_(n-1) + 1` for `n ge 1`. Then the remainder obtained dividing `a_(2010)` by 11 is -

Answer» Correct Answer - C
`a_(n) = 3a_(n-1) +1`
`a_(2010) = 3a_(2009) + 1`
`= 3(3a_(2008)+1) + 1 = 3^(2)a_(2008) + 3 + 1`
`{:(3^(3)a_(2007),+3+3+1),(.,),(.,),(.,),(.,):}`
`3^(2010)a_(0) + ubrace((3+3+"....."3))_("2009 times")+1`
` = 0 + 6027 + 1 = 6028`
Remainder `((6028)/(11)) = 0`
Discussion