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
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
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.
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
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.❤❤❤❤
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 👍
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
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
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!
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
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
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
@@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
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
No one is speaking about sir's unique spectacles collection 😂😎🔥
Yaa
😂😂
Sir ka papa lenskart mai kaam karta tha
Because we are serious jee aspirants
@@alien3200karta tha 🤨
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
Code 0 Q42
Code 3 Q48
+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 .
@@apurvajha1917 mera graph se minimum -2 aur maximum -2/3 aa rha
My bad
a b and c are the answers confirmed by the desmos
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
GOOD COLLECTION OF BOTH SPECTACLES AND QUESTION
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.
Bro solve only 44 years jee advanced pyq with coaching material two to three times is much better than opting for any other book 📚....
Sir how can we solve this by making case
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
It nice book for jee ad all topers say that
My interest in maths to know more depth is increasing day by day and my addiction for maths is also increasing
Me too bruh! 🤝
Sir you are great mathematician. Your concept and your approach is very beautiful ❤️ sir
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.❤❤❤❤
At the Beginning, what's the background music name ??
Sir please make a video for classification of numbers 🙏🙏🙏
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 👍
maan gaye sir app ho actually BHANNAT 🙏🙏🙏🙏🙏
Very nice method sir ❤️
Sir mujhe bhi Bansal sir ke problems and solutions dekhne hai Mai kaha se dekh sakta hu
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
Videolink?
@@duryodhan903 th-cam.com/video/jmVXLEftLNM/w-d-xo.html
Great sir
Pranaam Sir❤️🙏
Sir please solve. Sinx+Cosx=tanx find the all possible value of 'x'
Just came from the movie and sir uploaded this video 😂😂
Mai yaha thumbnail dekh kar aaya 😂😂😂😂😂😂GADAR
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
how to solve it algebrically?
Lots of love ❣️ sir , now I am in IIT ISM Dhanbad ❤
Bhaiya jalao mtt😢
Indirectly cases toh bna rhe?
Sir agr hmne is question ko cases se solve kiya to bhi to solve hota hai vo glt method hai?
Koi advanced wala question batayega??
Tell me sir how to join your paid course
JEE ADVANCED 2013 multiple. Modulus question also same approach se ho jayega
Sir aapne point functions ko continuous bola hai lekin ncert unko continuous nahi manti toh jee ke liye use continuous maane ya nahi????????
Discontinuous. Always prefer NCERT because even IIT strongly recommends it. I know it's not good for the prep but follow it strictly.
It is actually continuous because the graph dosent break.
Sir peheli baar me hogaya dono triko se 😀
Samajne ki koshis kar rahi hu❤
Sir plz batado konse year ka hai wo wala iit ka , solve karta phir
But here we need exact sketching know sir but which don't do in general?
Pranam guru ji❣️
which board are you using??
What if there is an enequality in this problem?
Nice
I will rise soon
😬😬😬😬
Me too
And best of luck
"I will rise soon"🤓🤡
@@advaymayank1410👍🏼👍🏼
karle rise bhai
Just a normal method for modulus question, so much hyped he has done
yeah either by cases or graph
Love you sir
I did same thing . Jee aspirants but took 2 min
sir aap trigonometric functions ke questions aur concepts ka video bana sakte ho
th-cam.com/video/va98yww_Xxc/w-d-xo.htmlsi=pM8uosm82OEMi9n_
7:40 sir Waha (-3 ) = | (0) - 1 | × 3 ,
Hoga answer aapne 3 leke 1 k sath jod diya
legend ho sir
Graph se jyada jaldi solution aa aata hai sir ji i understand
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!
Correct ❤❤
Awesome sir
Hello sir I HAVE A DOUBT 🙌🙌
sir agar hum log 0 with base 2 rakhe to output kya aana chayiye?🙋♂️🙋♂️🙋♂️🙋♂️🙋♂️🙋♂️
-(♾️)
@@keshavchoudhary3976 🤝🤝
Sir madhyamik mai maths hai aache marks kaise laayn pls Batao
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.
My jee journey is over (is in iiitA ) but still watching your content ❤️
Aap kaise h sir
Wao sir " U are great " ❤
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
It hits me really hard that why RUSSIA was written at the top of question
I feel like this problem is from prelipko aka problems in mathematics by v govorov
Really great method...You are great sir jee ❤
Ye bahut majedaar tha sir👍👍
Sir plz solve this problem
X^5+x+1=0
Then x^3-x^2=?
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
I love the way you teach sir🙇🙇💕
Jabh yeh plot banaya, toh x + 2 koh bhi LHS le jaasaktay hain aur directly roots hee read kar saktay hain, no?
mathsmerizing also give the sam solution of this type of questions
Sir maine aise hi kiya tha
sameer sir taught this in class
bhaisahab next level thinking sir jii
My favourite teacher is ashish agrawal ❤❤❤ and you too
Ashish Aggarwal sir ❤❤❤
Ne bhi aise method sikhaya hai
Sir #karan_Sangwan sir ke bare mein kuchh boliye
-2 without seen solution
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
Sir give some of this type
Sir
S L loney k coordiinate geometry book review pls
My favorite mathematics teacher ---- *Aman sir*
Sir aapko kitne mile hai jee advanced mei phir jab aap try karte ho
Mod ke andar linear and quadratic donon a gai to
FIR kya karenge sir
Quadratic ke factors karke use linear me convert aur phir wavy curve apply karna
First like and first comment ❤
😮😮😮
My exponential graph has crossed x=1 now it's about time
What
@@rudranilmukherjee3798 value of exponential graph increases rapidly after x=1 and also value is >1
@@abeersharma9487 but what do you mean by " my exponential graph has crossed x=1 now it's about time. "
@@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
Hi sir
Sir I want to join your paid course
How to join your paid course sir ?
Sir in the case 0
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
Bro u are contradicting yourself. First you are saying case 0
Thanks bro😄👍
Bhai vo sol (x= -1)tumhari uper mentioned condition (0
Imaginary
We indians always love Russians 🌚
stop it.... atleast here🤡
Irodov🗿
6k
@@YuezhiTribe haha
🤡🙂
Acha laga khud krke
Not first but fast
First view sir ji
Hii 🥰🥰🥰🥰
Sir parnam
Bari hea
Sir mera solution to x = -1 and x ≥ 2 aa rha hai.... Maine galat kiya hai kya?
Obviously
@@advarbindkumar5791 galat hai?
@@roar-and-rattle ha galat hai
@@advarbindkumar5791aapne Kiya?
I will bounce back
I solved it I don't believe
Sir ap JEE Advanced 2023 ke maths ke Solutions ki Video bna dijiye . Apke Solutions bohot acche hote ha ❤❤🙏🙏
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
I bet you can't factor this :
x^2-2(p+q)x-pq(p-2)(q-2)
bet accepted
(x-2p -2q -pq)(x+pq)=4pq
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
Yeah thanks
But the answer is :
(x-pq-2p)(x+pq-2q)
@@bijayanandamahata464 hey buddy answer nahi batana tha , could have denied only
Sir i have more better approach
In less than 50s i have done
Plz reply
Can you tell?
Sir aapne point functions ko continuous bola hai lekin ncert unko continuous nahi manti toh jee ke liye use continuous maane ya nahi???????
Sir aapne point functions ko continuous bola hai lekin ncert unko continuous nahi manti toh jee ke liye use continuous maane ya nahi??????
Khaan Bola ??
@@BHANNATMATHS sir aapki continuity wali playlist me ek video hai continuity of point functions usme bola hai aapne.
@@BHANNATMATHSsir aapne continuity of point function me bola tha please reply