﻿

Ask questions which are clear, concise and easy to understand.

can you please solve the important questions part 3 in introduction of trignometry

Posted by Aditya Talwar (Jul 23, 2017 6:36 p.m.) (Question ID: 6877)

• which book....if possible write question

Answered by Swami Jee (Jul 23, 2017 8:07 p.m.)
Thanks (0)

If SECa =X+1/4x, prove that SECa+ TANa =2x+1/2x

Posted by Mayank Mishra (Jul 22, 2017 9:55 p.m.) (Question ID: 6851)

• Given: {tex}\sec {\rm{A}} = x + {1 \over {4x}}{/tex}

Then, {tex}\tan {\rm{A}} = \sqrt {{{\sec }^2}{\rm{A}} - 1} {/tex}

=>          {tex}\tan {\rm{A}} = \sqrt {{x^2} + {1 \over 2} + {1 \over {16{x^2}}} - 1} {/tex}

=>          {tex}\tan {\rm{A}} = \sqrt {{{16{x^4} + 8{x^2} + 1 - 16{x^2}} \over {16{x^2}}}} {/tex}

=>          {tex}\tan {\rm{A}} = \sqrt {{{16{x^4} - 8{x^2} + 1} \over {16{x^2}}}} {/tex}

=>          {tex}\tan {\rm{A}} = \sqrt {{x^2} - {1 \over 2} + {1 \over {16{x^2}}}} {/tex}

=>          {tex}\tan {\rm{A}} = \sqrt {{{\left( {x - {1 \over {4x}}} \right)}^2}} {/tex}

=>          {tex}\tan {\rm{A}} = \pm \left( {x - {1 \over {4x}}} \right){/tex}

Therefore, {tex}\sec {\rm{A}} + \tan {\rm{A}} = x + {1 \over {4x}} + x - {1 \over {4x}} = 2x{/tex}

Or {tex}\sec {\rm{A}} + \tan {\rm{A}} = x + {1 \over {4x}} - x + {1 \over {4x}} = {1 \over {2x}}{/tex}

Answered by Rashmi Bajpayee (Jul 23, 2017 4:48 p.m.)
Thanks (0)

In an AP, the sum of first 10 terms is -150 and the sum of its next 10 terms is  -550.Find the arithmetic progression.

Posted by Anubhab Majumder (Jul 22, 2017 4:20 p.m.) (Question ID: 6834)

• in continuation from above

so AP is  3 , -1 , -5, -9, -13, -17 , -21, -25, -29, -33

sum of 1st 10 terms = -150

-37 , -41, -45, -49, -53, - 57, -61, -65, -69, -73

sum of next 10 terms = - 550

Answered by Hans Raj (Jul 22, 2017 8:06 p.m.)
Thanks (0)
• sorry due to power supply failure i could not complete

from equation i and ii above we get

2a + 9d = - 30

2a + 19d = - 70

- 10d = + 40

d = - 4

2a - 36 = - 30

2a = 6

a = 3

Answered by Hans Raj (Jul 22, 2017 7:40 p.m.)
Thanks (0)
• Let first term = a

common difference = d

According to question,

{tex}S_{10}= -150{/tex}

{tex}=> {10\over 2}[2a+9d]=-150{/tex}

{tex}=> 2a+9d= -30 \ ........(1){/tex}

Also

{tex}=> S_{20}-S_{10}=-550{/tex}

{tex}=> S_{20}+150=-550{/tex}

{tex}=> S_{20} = -700{/tex}

{tex}=> {20\over 2}[2a+19d]=-700{/tex}

{tex}2a+19d= -70\ .......(2){/tex}

Solving (1) and (2)

10 d = -40

=> d = -4

=> a = 3

{tex}So AP = 3,-1,-5,-9,-13,.. {/tex}

Answered by Sahdev Sharma (Jul 22, 2017 7:12 p.m.)
Thanks (0)
• Let a and d be the first term and common difference of A.P.

{tex}S_{10}= -150{/tex}

{tex}=> {10\over 2}[2a+9d]=-150{/tex}

{tex}=> 2a+9d= -30 \ ........(1){/tex}

Again

{tex}a_{11}+a_{12}+......+a_{20}= -550{/tex}

{tex}=> S_{20}-S_{10}=-550{/tex}

{tex}=> S_{20}+150=-550{/tex}

{tex}=> S_{20} = -700{/tex}

{tex}=> {20\over 2}[2a+19d]=-700{/tex}

{tex}2a+19d= -70\ .......(2){/tex}

Subtract (1) from (2), we get

10 d = -40

=> d = -4

Putting value of d in (1) we get

a = 3

So AP = 3,-1,-5,-9,....

Answered by Payal Singh (Jul 22, 2017 7:06 p.m.)
Thanks (0)
• sn = n/2 (2a + (n - 1)d)

-150 = 5(2a + 9d)      - 30 = 2a + 9d ....... i

- 550 + -150 = 10(2a +19d)

- 700 = 10 (2a + 19d.....ii

Answered by Hans Raj (Jul 22, 2017 7:05 p.m.)
Thanks (0)

The sum of first q terms of an A.P. is 162.The ratio of its 6th term to its 13th term is 1:2.Find the first and 15th term of the A.P.

Posted by Anubhab Majumder (Jul 22, 2017 4:14 p.m.) (Question ID: 6833)

• The question is wrong. It  should be 9 in place of q

<hr>

Let first term of AP is a And Common difference is d.

Given:

{tex}a_6 = a+5d\\a_{13}=a+12d{/tex}

{tex}{a_6\over a_{13}}={1\over 2}{/tex}

{tex} =>{ a+5d\over a+12d}={1\over 2}{/tex}

{tex}=> 2a+ 10 d = a+12d {/tex}

{tex}=> a = 2d{/tex}

Also,

{tex}S_9= {9\over 2}[2a+(9-1)d]{/tex}

{tex}=> 162 = {9\over 2}(2a+8d){/tex}

{tex}=> 36=[4d+8d]{/tex}

{tex}=> d = 3 {/tex}

Then a = 6

{tex}a_{15}= 6+14\times 3 = 48{/tex}

{tex}{/tex}

Answered by Payal Singh (Jul 22, 2017 6:53 p.m.)
Thanks (0)
• The data is insufficient. The q should be replaced with 9.

Let first term of AP = a

Common difference = d

{tex}a_6 = a+5d\\a_{13}=a+12d{/tex}

According to question,

{tex}{a_6\over a_{13}}={1\over 2}{/tex}

{tex} =>{ a+5d\over a+12d}={1\over 2}{/tex}

=> 2a+ 10 d = a+12d

=> a = 2d ... (1)

Also,

{tex}S_9=> 162= {9\over 2}[2a+8d]{/tex}

{tex}=> 36=[4d+8d]{/tex}

=> d = 3

So, a = 6

{tex}a_{15}= 6+14\times 3 = 48{/tex}

{tex}{/tex}

Answered by Sahdev Sharma (Jul 22, 2017 6:49 p.m.)
Thanks (0)

If one zero of the polynomial, (a⁴+4)x² + 13x + 4a, is reciprocal of the othe,Find ' a'

Posted by Gagan Sai (Jul 22, 2017 2:28 p.m.) (Question ID: 6831)

• Let one root = {tex}\alpha {/tex}

Then second root = {tex}1\over \alpha{/tex}

We know

Product of roots = {tex}c\over a{/tex}

{tex}=> \alpha\times {1\over \alpha}= {4a\over a^2+4}{/tex}

=> {tex}a^2+4=4a{/tex}

{tex}=> a^2-4a+4= 0{/tex}

{tex}=> (a-2)^2= 0{/tex}

=> a-2= 0

=> a = 2

Answered by Sahdev Sharma (Jul 22, 2017 2:50 p.m.)
Thanks (0)

if x and y are two irrational nubers then tell whether x-y is always irrational or not?

Posted by Sona Shree (Jul 22, 2017 10:51 a.m.) (Question ID: 6826)

• It may be rational or irrational both.

Answered by Sahdev Sharma (Jul 22, 2017 11:41 a.m.)
Thanks (0)

what is the difference between  CaO and CaOH.

Posted by Gauri Misra (Jul 21, 2017 10:08 p.m.) (Question ID: 6823)

• CaO is Calcium Oxide, otherwise known as Quick Lime.

CaOH is Calcium Hydroxide otherwise known as slaked lime which is obtained when calcium oxide is mixed or slaked with water.

Answered by Sona Shree (Jul 22, 2017 10:54 a.m.)
Thanks (0)
• CaO is known as quick lime while Ca(OH)2     is slaked lime , aqueous form of slaked lime is known as lime water which is used to show the presence of CO2.

Answered by Shridha Sharma (Jul 21, 2017 10:19 p.m.)
Thanks (1)

2/x^2-5/x+2=0

Bye using the method of compeleting the square.

Posted by Ananya Chaurasia (Jul 20, 2017 1:32 p.m.) (Question ID: 6788)

• {tex}{2\over x^2}{/tex}-{tex}{5\over x}{/tex}+2=0

{tex}{2-5x+2x^2\over x^2}{/tex}=0

2-5x+2x2=0

2x2-5x+2=0 ------(1)

Now first of all we will make coefficient of x2 to 1

So divide eq.(1) by 2 we get

x2-{tex}{5\over 2}{/tex}x+1=0

Now add and subtract the square of half of the coefficient of x we get

x2-{tex}{5\over2}{/tex}x+{tex}({5\over 4})^2 -({5\over 4})^2{/tex}+1=0

(x-{tex}{5\over 4})^2{/tex}-{tex}{25\over 16}{/tex}+1=0       [ by using identity (a-b)2=a2-2ab+b2]

(x-{tex}{5\over 4})^2{/tex}-{tex}{25+16\over 16}{/tex}=0

(x-{tex}{5\over 4})^2{/tex}-{tex}{9\over 16}{/tex}=0

(x-{tex}{5\over 4})^2{/tex}={tex}{9\over 16}{/tex}

(x-{tex}{5\over 4}{/tex})={tex}{\pm \sqrt{9 \over 16}}{/tex}

(x-{tex}{5\over 4}){/tex}={tex}{\pm 3\over 4}{/tex}

So x-{tex}{5\over 4}{/tex}={tex}{3\over 4}{/tex} or x-{tex}{5\over 4}{/tex}={tex}{-3\over 4}{/tex}

x={tex}{3\over 4} +{5\over 4}{/tex}  or x={tex}{-3\over 4}+{5\over 4}{/tex}

x={tex}{3+5\over 4}{/tex} or x={tex}{-3+5\over 4}{/tex}

x={tex}{8\over 4}{/tex} or  x={tex}{2\over 4}{/tex}

x=2 or x= {tex}{1\over 2}{/tex}

Answered by Dharmendra Kumar (Jul 20, 2017 5:49 p.m.)
Thanks (0)
• {tex}2x^2-5x+2=0{/tex}

{tex}=> (\sqrt 2x)^2-2\times \sqrt 2 \times {5\over 2\sqrt2}x+2 + ({5\over 2\sqrt 2})^2-({5\over 2\sqrt 2})^2=0{/tex}

{tex}=> (\sqrt 2x)^2-2\times \sqrt 2 \times {5\over 2\sqrt2}x + ({5\over 2\sqrt 2})^2+2 -{25\over 8}=0{/tex}

{tex}=> (\sqrt 2x)^2-2\times \sqrt 2 \times {5\over 2\sqrt2}x + ({5\over 2\sqrt 2})^2-{9\over 8}=0{/tex}

{tex}=> (\sqrt 2x-{5\over 2\sqrt 2})^2-({3\over 2\sqrt 2})^2=0{/tex}

{tex}=> (\sqrt 2x-{5\over 2\sqrt 2} +{3\over 2\sqrt 2})(\sqrt 2x-{5\over 2\sqrt 2} -{3\over 2\sqrt 2}){/tex}

{tex}=> (\sqrt 2x-{1\over \sqrt 2} )(\sqrt 2x-{\sqrt 2}){/tex}

Answered by Naveen Sharma (Jul 20, 2017 3:07 p.m.)
Thanks (0)

If (x-6) is a factor of x​​​​​3+ ax​​​​​​2​​​​+bx-b =0 and a-b = 7 find the value of a and b

Posted by Abhishek Raj (Jul 19, 2017 11:11 p.m.) (Question ID: 6782)

• Since (x-6) is a factor of x3+ax2+bx-b=0------(1)

Therefore x=6 is one of the roots of the given cubic polynomial.

So put x=6 in eq.(1) we get,

63+a×62+b×6-b=0

216+36a+6b-b=0

36a+5b=-216   -------(2)

given that  a-b=7

a=7+b  -----(3)

Substituting the value of a from eq.(3) into eq.(2) we get

36(7+b)+5b=-216

252+36b+5b=-216

41b=-216-252

41b=-468

{tex}b = {-468 \over 41}{/tex}

Substituting this value in eq.(3) we get

a=7+({tex}{-468\over 41}){/tex}={tex}{41×7-468\over 41}{/tex}={tex}{287-468 \over 41}{/tex}={tex}{-181\over 41}{/tex}

Answered by Dharmendra Kumar (Jul 19, 2017 11:43 p.m.)
Thanks (1)

Prove that a positive integer n is a prime number, if no Prime 'p' less than or equal to √n divides n

Posted by Lipika Bisht (Jul 19, 2017 10:29 p.m.) (Question ID: 6780)

Find the smallest number which leaves remainder 8 and 12 when divided by 28 and 32 respectively.

Posted by Lipika Bisht (Jul 19, 2017 10:25 p.m.) (Question ID: 6779)

• Let the required number be z.

z = 28x + 8 and z = 32y+12

=> 28x+8 = 32y+12

=>  7x = 8y+1  ..... (1)

Here, x =8n -1, y =7n -1 satisfies the equation (1).

Putting n =1,

we get x = 8-1 = 7 and y =7-1 =6 ?

=> z = 28 × 7 + 8 = 204

Thus, the smallest number which leaves remainder 8 and 12 when divided by 28 and 32 respectively is 204

Answered by Sahdev Sharma (Jul 20, 2017 6:28 p.m.)
Thanks (0)

p(x) is a polynomial of degree 4.g(x) is anotger polynomial of degree one. when p(x) is divided by g(x) the quotient q(x) &the remainder r(x) [r(x) not equal to 0] . find the degree of q(x)&r(x)

Posted by Mujeeb Korambayil (Jul 19, 2017 8:12 p.m.) (Question ID: 6771)

• Let p(x) is a polynomial of degree 4 and g(x) is a polynomial of degree 1.

Now when p(x) is divided by g(x) the quotient is q(x) and remainder is r(x)

So by Division algorithm for polynomials we get

p(x)=q(x)×g(x)+r(x)  where either r(x)=0 or degree of r(x)<degree of q(x) -----(1)

Now since p(x) is a polynomial of degree 4 and we divide it by polynomial g(x) of degree 1 ,therefore the degree of quotient q(x) must be 3.

Since r(x) cannot be 0 (given)

Therefore degree of r(x) < degree of q(x) =3     (by eq.1)

degree of r(x) < 3 so degree of r(x) is either 1 or 2.

Answered by Dharmendra Kumar (Jul 20, 2017 12:12 a.m.)
Thanks (0)

Comment upon the nature of roots of the following quadratic equation.    (x-2a)(x-2b)=4ab

Posted by Ayush Sharma (Jul 19, 2017 6:38 a.m.) (Question ID: 6753)

• (x-2a)(x-2b)=4ab

x(x-2b)-2a(x-2b)=4ab

x2-2bx-2ax+4ab=4ab

x2-(2b+2a)x+4ab-4ab=0

x2-(2a+2b)x+0=0

So on comparing with quadratic eqaution Ax2+Bx+C=0

We get A=1 B=-(2a+2b)  C=0

Discriminent D=B2-4AC

=(-(2a+2b))2-4×1×0

=(2a+2b)2-0

=(2a+2b)2 =(2(a+b))2=4(a+b)2

Since square of any number cannot be negative, therefore D cannot be negative.

So either D>0 or D=0

Case1:- D=4(a+b)2>0 when a+b>0 , in this case roots are real and distinct.

Case 2:- D=4(a+b)2=0 when a+b=0, in this case roots are real and repeated.

Answered by Dharmendra Kumar (Jul 19, 2017 11:17 a.m.)
Thanks (0)

If two water taps together can fill all tank in 78/9 hours. If one water tap can take 3 hour less than the other tap to fill tank separately, then find the time taken by both taps to fill water seperately.

Posted by Kumar Vivek (Jul 17, 2017 7:35 p.m.) (Question ID: 6717)

16.4x+2-16.2x+1+1=0

Posted by Deepa Jain (Jul 16, 2017 12:15 p.m.) (Question ID: 6684)

• {tex}16.4\ ^ {x+2}-16.2\ ^ {x+1}+1=0{/tex}

{tex}16(4\ ^ x\times4\ ^2)-16(2\ ^ x\times2\ ^1)+1=0{/tex}

{tex}16\times16\times4\ ^x-16\times2\times2\ ^x+1=0{/tex}

{tex}256\times4\ ^ x-32\times2\ ^x+1=0{/tex}

{tex}256\times(2\times2)\ ^ x-32\times2\ ^ x+1=0{/tex}

Let,{tex}2\ ^x=p{/tex}

{tex}256\times p\times p-32\times p+1=0{/tex}

{tex}256p\ ^2-32p+1=0{/tex}

{tex}256p\ ^2-16p-16p+1=0{/tex}

{tex}16p(16p-1)-1(16p-1)=0{/tex}

{tex}(16p-1)(16p-1)=0{/tex}

Either {tex}16p-1=0{/tex},or,{tex}16p-1=0{/tex}

{tex}p={1\over16},{1\over16}{/tex}

Now, {tex}p=2\ ^ x{/tex}

{tex}{1\over 16}={ 2\ ^ x}{/tex},

{tex}({1\over2})\ ^ 4=2\ ^ x{/tex}

{tex}(2)\ ^ {-4}=2\ ^ x{/tex}

{tex}x= -4{/tex}

Answered by Poulami Dasgupta (Jul 16, 2017 4:14 p.m.)
Thanks (0)

If sinA +sin (2)^A=1 Prove that cos (2)^A+cos (4)^A=1

Posted by Aditya Narayan Panda (Jul 16, 2017 10:08 a.m.) (Question ID: 6677)

• {tex}SinA + Sin\ ^ 2A = 1{/tex}

{tex}SinA = 1-Sin\ ^ 2A{/tex}

{tex}SinA = Cos\ ^ 2 A{/tex}( Using identity)

Squaring both sides,

{tex}(SinA)\ ^ 2 = ( Cos\ ^2 A )\ ^ 2{/tex}

{tex}Sin\ ^ 2 A=Cos\ ^ 4 A{/tex}

{tex}1- Cos\ ^ 2 A= Cos \ ^ 4 A{/tex}( Using identity)

{tex}1 = Cos \ ^ 2A + Cos \ ^ 4 A{/tex}

{tex}Cos \ ^ 2A + Cos \ ^ 4 A = 1{/tex}

Answered by Poulami Dasgupta (Jul 16, 2017 10:19 a.m.)
Thanks (0)

If two of the roots of f (x) = x^3-5x^2 - 16x +80 are equal in magnitude but opposite in sign then find all the. Zeroes

Posted by Vansh Kataria (Jul 15, 2017 8:12 p.m.) (Question ID: 6663)

• let the zeroes be a , -a , b

then (x-a),(x+a),(x-b) are the factors of f(x)

therefore f(x)=(x-a)(x+a)(x-b)=x3-bx2-a2x+a2b

therefore x3-5x2-16x+80=x3-bx2-a2x+a2b

equating the coefficients we get a=+or-4 and b=5

Answered by Joshina Learnel (Jul 20, 2017 10:08 p.m.)
Thanks (0)

find the valid of a and b for which (2a-1)x-3y=5 and 3x+(b-2)y=3 has unique solution  which comes out to be y= px+q.  then find p and q

Posted by Harsha Bansal (Jul 15, 2017 8:08 p.m.) (Question ID: 6662)

(SinA +1-cosA)/(cosA-1+sinA ) =1+sinA/cosA

Posted by Aditya Narayan Panda (Jul 15, 2017 7:22 p.m.) (Question ID: 6658)

• divide numerator and denomintor of LHS by cos A

{tex}LHS=\frac{tan A+sec A-1}{1-sec A+tanA} but 1=sec^2A-tan^2A LHS=\frac{tan A+sec A-(sec^2A-tan^2A)}{1-sec A+tanA} =\frac{tan A+sec A-(secA-tanA)(secA+tanA)}{1-sec A+tanA} from the numerator take (sec A+tan A) =\frac{(tan A+sec A)(1-(secA+tanA)}{1-sec A+tanA} cancel (1-(secA+tanA) =tanA+secA =\frac{sinA}{cosA}+\fract{1}{cosA} =\fract{sinA+1}{cosA {/tex}

Answered by Joshina Learnel (Jul 20, 2017 10:26 p.m.)
Thanks (0)

Find the value of 7/cot2A - 7/cos2A

Posted by Angdoopt Pt (Jul 15, 2017 11:35 a.m.) (Question ID: 6648)

Answered by Rasmi Rv (Jul 15, 2017 12:53 p.m.)
Thanks (0)

If the zeros of the polynomial f(x)=2xcube -15xsquare +37x -30 are in A.P.,find them.

Posted by Anubhab Majumder (Jul 13, 2017 1:55 p.m.) (Question ID: 6612)

• Given {tex}f(x)=2{ x }^{ 3 }-15{ x }^{ 2 }+37x-30{/tex}

Let the zeroes of the polynomial in A.P  be a-d,a,a+d

Using  the relationship between the zeroes and  coefficients of a cubic polynomial ,we get

Sum of the roots = {tex}a-d+a+a+d=\frac { -\left( -15 \right) }{ 2 } {/tex}

{tex}\Rightarrow 3a=\frac { 15 }{ 2 } \\ \Rightarrow a=\frac { 5 }{ 2 } {/tex}

Now the product of the roots= {tex}\left( a-d \right) a\left( a+d \right) =\frac { -\left( -30 \right) }{ 2 } {/tex}

{tex}\Rightarrow a\left( { a }^{ 2 }-{ d }^{ 2 } \right) =\frac { 30 }{ 2 } \\ \Rightarrow \frac { 5 }{ 2 } \left( \frac { 25 }{ 4 } -{ d }^{ 2 } \right) =15\\ \Rightarrow \left( \frac { 25 }{ 4 } -{ d }^{ 2 } \right) =15\times \frac { 2 }{ 5 } =6\\ \Rightarrow { d }^{ 2 }=\frac { 25 }{ 4 } -6=\frac { 25-24 }{ 4 } =\frac { 1 }{ 4 } \\ \Rightarrow { d }=\pm \frac { 1 }{ 2 } {/tex}

Hence the roots are {tex}\frac { 5 }{ 2 } -\frac { 1 }{ 2 } ,\frac { 5 }{ 2 } ,\frac { 5 }{ 2 } +\frac { 1 }{ 2 } =2,2.5,3{/tex}

Answered by Rasmi Rv (Jul 13, 2017 10:30 p.m.)
Thanks (0)

The 6th and 17th terms of an A.P. are 19 & 41,find the 40th term?

Posted by Anubhab Majumder (Jul 13, 2017 1:50 p.m.) (Question ID: 6611)

• We have nth term of an A.P is  {tex}{ T }_{ n }=a+\left( n-1 \right) d{/tex}

Now {tex}{ T }_{ 6 }=19\Rightarrow a+5d=19.............(i)\\ { T }_{ 17 }=41\Rightarrow a+16d=41..........(ii){/tex}

Subtracting  equation(i) from(ii), we get

{tex}11d=22\Rightarrow d=\frac { 22 }{ 11 } =2{/tex}

Now from equation (i) we get

{tex}a+10=19\Rightarrow a=9{/tex}

Hence we get {tex}{ T }_{ 40 }=a+39d=9+39\left( 2 \right) =9+78=87{/tex}

So the 40th term is 87.

Answered by Rasmi Rv (Jul 13, 2017 10:42 p.m.)
Thanks (0)

Is 302 a term of the A.P. 3,8,13,...?

Posted by Anubhab Majumder (Jul 13, 2017 1:44 p.m.) (Question ID: 6610)

• Let 302 be the nth term of the A. P.

an=a+(n-1)d

302=3+(n-1)5

299/5=n-1

As n is not a whole number, so 302 is not a term for the given A. P.

Answered by Shweta Gulati (Jul 14, 2017 12:24 a.m.)
Thanks (0)

Prove that no matter what the real numbers and b are,the sequence with nth term a+nb is always an A.P. What is the common difference?

Posted by Anubhab Majumder (Jul 13, 2017 1:42 p.m.) (Question ID: 6609)

• Let a=a+nb where a and b are real numbers.

Then put n=1we get a1=a+b

Put n=2  a2=a+2b

Put n=3  a3=a+3b

Put n=4  a4=a+4b......and so on

Now a2-a1= a+2b-(a+b)=a+2b-a-b=b

a3-a2=a+3b-(a+2b)=a+3b-a-2b=b

a4-a3=a+4b-(a+3b)=a+4b-a-3b=b ........

So we get  a2-a1=a3-a2=a4-a3=..........=b

Clearly the difference between two consecutive terms are same.

Hence the given sequence of real numbers is in an A.P. with common difference d=a2-a1=a3-a2=a4-a3=..........=b

Answered by Dharmendra Kumar (Jul 18, 2017 11:52 a.m.)
Thanks (0)

If the zeros of the polynomial f(x)=2xcube +15xsquare 37x-30 are in A.P.,Find them.

Posted by Anubhab Majumder (Jul 12, 2017 10:01 p.m.) (Question ID: 6601)

• The  question is incomplete

Posted by Rasmi Rv (Jul 13, 2017 11:49 a.m.)

Cos(x)(tan(x)+1)(2tan(x)+1)= 2sec(x)+5sin(x)

Posted by Rahul Kumar (Jul 12, 2017 9:32 p.m.) (Question ID: 6599)

1/(x-1)(x-2)+1/(x-2)(x-3)+1/(x-3)(x-4)=1/6

Posted by Anubhab Majumder (Jul 10, 2017 1:43 p.m.) (Question ID: 6527)

• [{tex}1\over(x-1)(x-2){/tex}+{tex}1\over(x-2)(x-3){/tex}]+{tex}1\over(x-3)(x-4){/tex}={tex}1\over6{/tex}

=>{tex}(x-3)+(x-1)\over(x-1)(x-2)(x-3){/tex}+{tex}1\over (x-3)(x-4){/tex}={tex}1\over6{/tex}

=>{tex}2x-4\over(x-1)(x-2)(x-3){/tex}+{tex}1\over(x-3)(x-4){/tex}={tex}1\over6{/tex}

=>{tex}2(x-2)\over(x-1)(x-2)(x-3){/tex}+{tex}1\over(x-3)(x-4){/tex}={tex}1\over6{/tex}

=>{tex}2\over(x-1)(x-3){/tex}+{tex}1\over(x-3)(x-4){/tex}={tex}1\over6{/tex}

=>{tex}2(x-4)+(x-1)\over(x-1)(x-3)(x-4){/tex}={tex}1\over6{/tex}

=>{tex}2x-8+x-1\over(x-1)(x-3)(x-4){/tex}={tex}1\over6{/tex}

=>{tex}3x-9\over(x-1)(x-3)(x-4){/tex}={tex}1\over6{/tex}

=>{tex}3(x-3)\over(x-1)(x-3)(x-4){/tex}={tex}1\over6{/tex}

=>{tex}3\over(x-1)(x-4){/tex}={tex}1\over6{/tex}

=>{tex}(x-1)(x-4)=18{/tex}

=>{tex}x\ ^2-4x-x+4=18{/tex}

=>{tex}x\ ^2-5x-14=0{/tex}

=>{tex}x\ ^2-7x+2x-14=0{/tex}

=>{tex}x(x-7)+2(x-7)=0{/tex}

=>{tex}(x+2)(x-7)=0{/tex}

=>{tex}either ,x+2=0 or, x-7=0{/tex}

{tex}x=-2 or7{/tex}

Answered by Poulami Dasgupta (Jul 10, 2017 5:28 p.m.)
Thanks (0)

A dealer sells an article for rs. 24 and gains as much % as the C.P. of the article.Find the cost price of the article.

Posted by Anubhab Majumder (Jul 10, 2017 1:25 p.m.) (Question ID: 6526)

• Let the C.P. of the article be Rs.{tex}x {/tex}

Then according to question, gain = {tex}x {/tex}%

Therefore,

C.P. + Gain = S.P.

=>     {tex}x + x\% {\rm\\ {\ of }}\ x = 24{/tex}

=>     {tex}x + {x \over {100}} \times x = 24{/tex}

=>     {tex}x + {{{x^2}} \over {100}} = 24{/tex}

=>     {tex}{x^2} + 100x - 2400 = 0{/tex}

=>     {tex}{x^2} + 120x - 20x - 2400 = 0{/tex}

=>     {tex}x\left( {x + 120} \right) - 20\left( {x + 120} \right) = 0{/tex}

=>     {tex}\left( {x - 20} \right)\left( {x + 120} \right) = 0{/tex}

=>     {tex}x -20=0{/tex} or {tex}x +120=0{/tex}

=>     {tex}x = 20{/tex} or {tex}x = -120{/tex} [Neglected]

Therefore, the C.P. of the article is Rs.20

Answered by Rashmi Bajpayee (Jul 10, 2017 3:20 p.m.)
Thanks (0)

Prove that tanA +sinA/tanA-sinA = secA +1/secA-1=1+cosA/1- cosA

Posted by Aditya Narayan Panda (Jul 10, 2017 6:35 a.m.) (Question ID: 6522)

• {tex}{{\tan {\rm{A}} + \sin {\rm{A}}} \over {\tan {\rm{A}} - \sin {\rm{A}}}}{/tex}

= {tex}{{{{\sin {\rm{A}}} \over {\cos {\rm{A}}}} + \sin {\rm{A}}} \over {{{\sin {\rm{A}}} \over {\cos {\rm{A}}}} - \sin {\rm{A}}}}{/tex}

= {tex}{{\sin {\rm{A}} + \cos {\rm{A}}.\sin {\rm{A}}} \over {\sin {\rm{A}} - \cos {\rm{A}}.\sin {\rm{A}}}}{/tex}

= {tex}{{\sin {\rm{A}}\left( {1 + \cos {\rm{A}}} \right)} \over {\sin {\rm{A}}\left( {1 - \cos {\rm{A}}} \right)}}{/tex}

= {tex}{{1 + \cos {\rm{A}}} \over {1 - \cos {\rm{A}}}}{/tex}

Proved.

Also, {tex}{{1 + \cos {\rm{A}}} \over {1 - \cos {\rm{A}}}}{/tex}

= {tex}{{1 + {1 \over {\sec {\rm{A}}}}} \over {1 - {1 \over {\sec {\rm{A}}}}}}{/tex}

= {tex}{{\sec {\rm{A}} + 1} \over {\sec {\rm{A}} - 1}}{/tex}

Proved.

Answered by Rashmi Bajpayee (Jul 11, 2017 12:16 p.m.)
Thanks (0)

To fill a swimming pool 2 pipes are used. If the pipe of larger diameter used for 4 hours and the pipe of smaller diameter for 9 hours,only half of the pool can be filled. Find, how long it would take for each pipe to fill the seperately, if the pipe of smaller diameter takes 10 hours more than the pipe of larger diameter to fill the pool.

Posted by Anubhab Majumder (Jul 09, 2017 10:02 p.m.) (Question ID: 6521)

• Let the time taken by the pipe of larger diameter = x hours

The time taken by the pipe of smaller diameter = x + 10 hours

In 1 hour pipe of larger diameter fills  {tex}1\over x{/tex}part of the pool , So in 4 hour the pipe of larger diameter fills = {tex}4\over x{/tex}

In 1 hour pipe of smaller diameter fills {tex}1\over x + 10{/tex} part of the pool , So in 9 hour the pipe of larger diameter fills ={tex} 9\over x + 10{/tex}

ATQ

{tex}{4\over x} + {9\over x + 10} = {1\over 2}{/tex}

=> {tex}{4x+40+9x\over x^2+10x}= {1\over 2}{/tex}

=> {tex}x^2+10x = 26x+80{/tex}

=> {tex}x^2-16x-80=0{/tex}

=> {tex}x^2-20x+4x-80=0{/tex}

=> x(x-20)+4(x-20)= 0

=> (x-20)(x+4)=0

x  = - 4  and 20,

But we assume x as hour , So we neglect negative term and get

x  =  20

The time taken by the pipe of larger diameter = 20 hours
And
The time taken by the pipe of smaller diameter = 20 + 10 =  30 hours

Answered by Sahdev Sharma (Jul 10, 2017 6:12 a.m.)
Thanks (0)

#### Subscribe by E-mail

For Latest News and Updates from myCBSEguide.com