Modulus Problem with a Classic Solution 😎

[@mr_sam8461]

No one is speaking about sir's unique spectacles collection 😂😎🔥

[@pradeepsrivastava2376]

Yaa

[@shaileshs.3177]

😂😂

[@alien3200]

Sir ka papa lenskart mai kaam karta tha

[@lakshyadhimanld027]

Because we are serious jee aspirants

[@A.liisanitØ]

​@@alien3200karta tha 🤨

[@tanyajaiswal1268]

Question about which sir is talking about is of [JEE Advanced 2013 Paper 2 Question 44] ( i.e; Question 4 of Mathematics section) Q) f(x) = 2|x| + |x+2| - |lx+2l - 2lxl| has a local minimum or local maximum at x= (a)-2 (b)+2 (c)-2/3 (d)+2/3

[@Oldboy99999]

Code 0 Q42 Code 3 Q48

[@apurvajha1917]

+2 aayga answer graph se bnaaye toh pta chala ki 2|x|+|x+2| aaur ||x+2| -|2x|| ka graph coincide krega at all real values of x .

[@groovygames4313]

@@apurvajha1917 mera graph se minimum -2 aur maximum -2/3 aa rha

[@apurvajha1917]

My bad

[@arpitchauhan2147]

a b and c are the answers confirmed by the desmos

[@BORAGALALIKHITYADAV]

Just for all JEE aspirants reference, I used to wonder how does that trick works in my JEE days for ploting such functions, just by marking values at critical points. It is because no matter how the mod opens for different mods the nature of the whole function will either remain linear or constant ( as with only sign difference we cant produce x power terms ) so simply we mark values at critical point and make the graph linear. PS: even tho the LHS is linear nature and RHS is linear nature but we still get 2 solutions which is like mod eq can store multiple solutions in a linear nature, unlike polynomials(eg.quadratic) they store 2 solutions but their nature of equation is different. In short mod equations are collection of solutions of different/indifferent linear equations. I hope this gave a new perspective to your view of mod equations

[@tusharnaagar2260]

GOOD COLLECTION OF BOTH SPECTACLES AND QUESTION

[@PrajalRoyzada-qn6mi]

Aman Sir, what are your opinions about solving the Black Book for maths.Many say that, as it is written by Indian author(Vikas Gupta sir), and apt for jee advance. I am currently in class 11th.

[@UnstoppableAmit]

Bro solve only 44 years jee advanced pyq with coaching material two to three times is much better than opting for any other book 📚....

[@shoumiksadhukhan4621]

Sir how can we solve this by making case

[@QueenDivya43]

Bro am also in class 11 and I am also solving black book it is very good book when I completed 1 chapter from it I Sol jee ad pyq I solved many of them

[@QueenDivya43]

It nice book for jee ad all topers say that

[@SwayamKrishnaartsandcrafts]

My interest in maths to know more depth is increasing day by day and my addiction for maths is also increasing

[@Sanskar99-xyz]

Me too bruh! 🤝

[@NayiSubhaMathAcademy]

Sir you are great mathematician. Your concept and your approach is very beautiful ❤️ sir

[@Vipyograj123]

Sir really I solve question without watching video within 15 -35 min.and find the answer But sir when I watch video then I learn graphical method that is too much. Love you and loving your methods.❤❤❤❤

[@anshu_Dwivedi0008]

At the Beginning, what's the background music name ??

[@upendrakumarmaurya1516]

Sir please make a video for classification of numbers 🙏🙏🙏

[@MOHDRAFIQ-b1r]

Sir please 🥺 aik video chahiye about strategy to solve math questions by self .sir bohot dikkat aati hai .who wants this video from Aman sir please like 👍

[@lya2-xv6wc]

maan gaye sir app ho actually BHANNAT 🙏🙏🙏🙏🙏

[@SagarHotta-xu2hc]

Very nice method sir ❤️

[@_.TUSHAR._]

Sir mujhe bhi Bansal sir ke problems and solutions dekhne hai Mai kaha se dekh sakta hu

[@Qwerty25414]

On 22 July ,you uploaded a vedio in which you solved bansal sir question but you devide the equation by 'x' but x may be 0 . You can't Devide by x until it is confirmed that x can't be 0 , Sir my question is that how do you know that x can't be zero And I am watching your vedios since last year , your questions are nice love you

[@duryodhan903]

Videolink?

[@Qwerty25414]

@@duryodhan903 kzbin.info/www/bejne/oJ65iX97m9l_hK8

[@Im_inevitable07]

Great sir

[@vaisim3450]

Pranaam Sir❤️🙏

[@gyani_studys]

Sir please solve. Sinx+Cosx=tanx find the all possible value of 'x'

[@PwiansOfficially]

Just came from the movie and sir uploaded this video 😂😂

[@babadham3466]

Mai yaha thumbnail dekh kar aaya 😂😂😂😂😂😂GADAR

[@gauravaggarwal]

wese yeh lamba method hai, because left hand side jo points hain wo hain -1,0,1,2 equidistant so right most if it is leaving line at x=2 no need to check it will cross again at x=-2 (if last flip occurs after -2) so need not to plot others

[@saikatde9879]

how to solve it algebrically?

[@FutureWaitHim]

Lots of love ❣️ sir , now I am in IIT ISM Dhanbad ❤

[@aashishkumar9658]

Bhaiya jalao mtt😢

[@amoghyashgupta]

Indirectly cases toh bna rhe?

[@Nostalagic]

Sir agr hmne is question ko cases se solve kiya to bhi to solve hota hai vo glt method hai?

[@GautamVaishnav-b1s]

Koi advanced wala question batayega??

[@Ajeetkumar-zp6uy]

Tell me sir how to join your paid course

[@ajal3025]

JEE ADVANCED 2013 multiple. Modulus question also same approach se ho jayega

[@neilvincent]

Sir aapne point functions ko continuous bola hai lekin ncert unko continuous nahi manti toh jee ke liye use continuous maane ya nahi????????

[@An_ony_mous]

Discontinuous. Always prefer NCERT because even IIT strongly recommends it. I know it's not good for the prep but follow it strictly.

[@An_ony_mous]

It is actually continuous because the graph dosent break.

[@arjunmishra8159]

Sir peheli baar me hogaya dono triko se 😀

[@anjalikumari2649]

Samajne ki koshis kar rahi hu❤

[@swarnabhamitra7233]

Sir plz batado konse year ka hai wo wala iit ka , solve karta phir

[@jagan5398]

But here we need exact sketching know sir but which don't do in general?

[@AjitChaudhary43]

Pranam guru ji❣️

[@gauravaggarwal]

which board are you using??

[@anushripal6005]

What if there is an enequality in this problem?

[@111_aaditya]

Nice

[@solokiller8472]

I will rise soon

[@advaymayank1410]

😬😬😬😬

[@nj_149]

Me too And best of luck

[@aniasprelltinerosdez333]

"I will rise soon"🤓🤡

[@alien3200]

​@@advaymayank1410👍🏼👍🏼

[@nitenuc2341]

karle rise bhai

[@parthnagpal6082]

Just a normal method for modulus question, so much hyped he has done

[@utkarshgoyal7420]

yeah either by cases or graph

[@ankitkumawat-wh5ni]

Love you sir

[@prateekpoonia17]

I did same thing . Jee aspirants but took 2 min

[@c24live46]

sir aap trigonometric functions ke questions aur concepts ka video bana sakte ho

[@gauravaggarwal]

kzbin.info/www/bejne/rJKcaaytrMSLrsUsi=pM8uosm82OEMi9n_

[@omkumarsingh7]

7:40 sir Waha (-3 ) = | (0) - 1 | × 3 , Hoga answer aapne 3 leke 1 k sath jod diya

[@VihaanAgarwal-ey6fn]

legend ho sir

[@project24255]

Graph se jyada jaldi solution aa aata hai sir ji i understand

[@shreyanshdixit9327]

Good evening, sir, I am currently a student of grade 10. I just thought of this way to solve the question. Kindly check this and let me know if this solution has any problem. Take 2 cases 1) x greater than or equal to 2. For this |x+a| = x+a where a belongs to real numbers and |x| > |a| -----{Since x is positive} {Here |x| can also be equal to |a|} => x + 1 - x + 3x - 3 -2x + 4 = x+2 => 2=2 Hence for all values of x greater than or equal to 2, the equation is satisfied --------- {We have not taken below 2 because we need |x| > |a| or are equal} 2) x less than -2 We have taken x less than -2 because, |x+a| can be written as |x| - |a| for all |x| > |a| and 'a' being positive. ------{Hence |x| must have been greater than 2} {Since we have taken all positive values, x must be less than 2} Similarly, |x-a| can be written as |x| + |a| for all |x| > |a| and 'a' being positive. Using these, |x+1| - |x| + 3|x-1| - 2|x-2| = x+2 => |x| -1 -|x| + 3|x| + 3 -2|x| -4 = x+2 => |x| - x = 4 => Either x-x = 4 OR -x-x = 4 Discarding 0 = 4, -2x = 4 => x= (-2) HENCE ONLY ONE POSSIBILITY FOR x < 0. That is (-2) Therefore, x belongs to {-2} U [2, infinity) Kindly confirm this approach to the solution. Thank you!

[@ayushrana124]

Correct ❤❤

[@mathematicalideas7201]

Awesome sir

[@sankalpsharma594]

Hello sir I HAVE A DOUBT 🙌🙌 sir agar hum log 0 with base 2 rakhe to output kya aana chayiye?🙋‍♂️🙋‍♂️🙋‍♂️🙋‍♂️🙋‍♂️🙋‍♂️

[@keshavchoudhary3976]

-(♾️)

[@sankalpsharma594]

@@keshavchoudhary3976 🤝🤝

[@prabuddhabose4033]

Sir madhyamik mai maths hai aache marks kaise laayn pls Batao

[@bollyfan1330]

Bahut badhiya solution hain aapka. Aapke solution koh optimize kiya jaa sakta hain, x+2 koh bhi LHS mein lagaa loh, RHS koh bana doh zero. Phir graph draw karna, aur roots (X-axis crossings) honge solutions.

[@hariomsoni7477]

My jee journey is over (is in iiitA ) but still watching your content ❤️

[@बोलो_राधे_राधे]

Aap kaise h sir

[@PiyushRatnu-hp4rp]

Wao sir " U are great " ❤

[@Rapid_GK]

Just because of you sir I am able to develop the critical thinking in maths.....thank you sir so much..... Coming to this video....I solved this in 1st attempt

[@VishavpreetSingh-i4m]

It hits me really hard that why RUSSIA was written at the top of question

[@Tryha4d]

I feel like this problem is from prelipko aka problems in mathematics by v govorov

[@prasadpatil2118]

Really great method...You are great sir jee ❤

[@utkarshanushka2577]

Ye bahut majedaar tha sir👍👍

[@abinashjena1205]

Sir plz solve this problem X^5+x+1=0 Then x^3-x^2=?

[@gauravaggarwal]

if x^a + x^b + x^c given such that out of a,b,c one is of the form 3k, one is 3k+1, one is 3k-1 then that polynomial is divisible by (x^2+x+1) so 5 = 3*2-1, and other exponents are alraedy 0,1 so perfct triplet so just divide it by x^2+x+1 you will get x^3-x^2+1 so answer is -1

[@Gurvinder__kaur.]

I love the way you teach sir🙇🙇💕

[@als2cents679]

Jabh yeh plot banaya, toh x + 2 koh bhi LHS le jaasaktay hain aur directly roots hee read kar saktay hain, no?

[@nikitasharma7209]

mathsmerizing also give the sam solution of this type of questions

[@adityajha2889]

Sir maine aise hi kiya tha

[@mathsolympiad8808]

sameer sir taught this in class

[@unknown69__]

bhaisahab next level thinking sir jii

[@Aman_nitb_cse]

My favourite teacher is ashish agrawal ❤❤❤ and you too

[@ShubhamSingh-we8hv]

Ashish Aggarwal sir ❤❤❤ Ne bhi aise method sikhaya hai

[@tseries0002]

Sir #karan_Sangwan sir ke bare mein kuchh boliye

[@sahilverma8640]

-2 without seen solution

[@atharvkumar2564]

sir graph me aapne origin alag alag liya hai,jaise f(x) = x + 2 ke liye aapne graph ko (2,4) se start kiya hai,aise hi f(x) = -(x+2) ke liye aapne graph (-1,-1) se start kiya hai,sir ye to galat hai kyuki sabhi graph ka starting point or refernce point aapne ek rule ke hisab se nhi liya.Sahi to tab hoga jab f(x) = x+2 ko aap (0,2)se shuru kroge.Kindly clear my doubt

[@swatishukla5899]

Sir give some of this type

[@ashish78343]

Sir S L loney k coordiinate geometry book review pls

[@physicsandmaths_Op]

My favorite mathematics teacher ---- *Aman sir*

[@knowledgeoffissh0109]

Sir aapko kitne mile hai jee advanced mei phir jab aap try karte ho

[@RishavBansal-m6r]

Mod ke andar linear and quadratic donon a gai to

[@RishavBansal-m6r]

FIR kya karenge sir

[@SaifKhan-mf4ho]

Quadratic ke factors karke use linear me convert aur phir wavy curve apply karna

[@ashutoshkumar9876]

First like and first comment ❤

[@icosahedron27]

😮😮😮

[@abeersharma9487]

My exponential graph has crossed x=1 now it's about time

[@rudranilmukherjee3798]

What

[@abeersharma9487]

@@rudranilmukherjee3798 value of exponential graph increases rapidly after x=1 and also value is >1

[@rudranilmukherjee3798]

@@abeersharma9487 but what do you mean by " my exponential graph has crossed x=1 now it's about time. "

[@abeersharma9487]

@@rudranilmukherjee3798 that means I have improved enough to score enough to clear cutoff for jee advance now I am improving so that i can clear iit as well

[@lovelyaman4779]

Hi sir

[@Ajeetkumar-zp6uy]

Sir I want to join your paid course

[@Ajeetkumar-zp6uy]

How to join your paid course sir ?

[@trulysatyam8231]

Sir in the case 0

[@ashgreninja5699]

X 0 se 1 liye ho toh ans x =-1 aa rh Tumne toh x ko (0,1) liya n toh usme -1 ni aata toh nhi hoga woh

[@duryodhan903]

Bro u are contradicting yourself. First you are saying case 0

[@trulysatyam8231]

Thanks bro😄👍

[@YuezhiTribe]

Bhai vo sol (x= -1)tumhari uper mentioned condition (0

[@dangerzone3200]

Imaginary

[@amay96]

We indians always love Russians 🌚

[@MixbOOsted]

stop it.... atleast here🤡

[@YuezhiTribe]

Irodov🗿

[@Enchantedquillx2]

6k

[@BelieveConfident]

​@@YuezhiTribe haha

[@AvikGupta-ch7oh]

🤡🙂

[@ManishKumar-fy4dz]

Acha laga khud krke

[@ArtcticKnightX]

Not first but fast

[@sr7tech433]

First view sir ji

[@AjayPratap-td7pp]

Hii 🥰🥰🥰🥰

[@बोलो_राधे_राधे]

Sir parnam

[@SnehaMukherjee-q1l]

Bari hea

[@roar-and-rattle]

Sir mera solution to x = -1 and x ≥ 2 aa rha hai.... Maine galat kiya hai kya?

[@advarbindkumar5791]

Obviously

[@roar-and-rattle]

@@advarbindkumar5791 galat hai?

[@advarbindkumar5791]

@@roar-and-rattle ha galat hai

[@roar-and-rattle]

​@@advarbindkumar5791aapne Kiya?

[@ParulSrivastava]

I will bounce back

[@Disabbyhere]

I solved it I don't believe

[@dhruvsharma8256]

Sir ap JEE Advanced 2023 ke maths ke Solutions ki Video bna dijiye . Apke Solutions bohot acche hote ha ❤❤🙏🙏

[@on.football350]

Sir this method cannot be applicable for all questions See this question |x-3| +|2x-3| = 3|x-2| try this question u cannot solve this by graph method

[@bijayanandamahata464]

I bet you can't factor this : x^2-2(p+q)x-pq(p-2)(q-2)

[@gauravaggarwal]

bet accepted (x-2p -2q -pq)(x+pq)=4pq

[@gauravaggarwal]

How i factorized suppose p,q are roots of t^2 -at+b =0 then (p-2),(q-2) will be roots of z^2 - (a-4)z + (2a+b+4) = 0 so (p-2)(q-2) = 2a+b+4 your equation can be rewritten as x^2 -2ax -b(2a+b+4) = (x-a)^2 - (a+b)^2 -4b =(x-2a-b)(x+b) -4b = (x-2p-2q-pq)(x+pq)-4pq

[@bijayanandamahata464]

Yeah thanks

[@bijayanandamahata464]

But the answer is : (x-pq-2p)(x+pq-2q)

[@gauravaggarwal]

@@bijayanandamahata464 hey buddy answer nahi batana tha , could have denied only

[@Mohitsaini-fb5dw]

Sir i have more better approach

[@Mohitsaini-fb5dw]

In less than 50s i have done

[@Mohitsaini-fb5dw]

Plz reply

[@Jimi_hendrix1]

Can you tell?

[@neilvincent]

Sir aapne point functions ko continuous bola hai lekin ncert unko continuous nahi manti toh jee ke liye use continuous maane ya nahi???????

[@neilvincent]

Sir aapne point functions ko continuous bola hai lekin ncert unko continuous nahi manti toh jee ke liye use continuous maane ya nahi??????

[@BHANNATMATHS]

Khaan Bola ??

[@neilvincent]

@@BHANNATMATHS sir aapki continuity wali playlist me ek video hai continuity of point functions usme bola hai aapne.

[@neilvincent]

​@@BHANNATMATHSsir aapne continuity of point function me bola tha please reply