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Another Method Follows: यहाँ हम f(x) को x^2 की ही form में देख सकते हैं, क्योंकि f(x/2) आदि केवल x^2 के coefficient में ही change करेंगे, जिससे x^2 को equation से cancel ही हो जाना है, तो फिर हमें result में जो constant=constant जैसा मिलेगा उसे ही adjust करना है। मान लो f(x) = kx^2 Equation में रखने पर, kx^2 - 2k(x/2)^2 + k(x/4)^2 = x^2 Or, k - k/2 + k/16 = 1 Or, k = 16/9 Thus, we get f(x) = (16/9)x^2 Hence, 9[f(100)-f(0)] = 160000 The method of this video is called Telescopic Method wherein we mutually cancel out the terms between the initial and final terms of a series. This is also a thought provoking one. There are also the standard methods for these types of equations, which are similar to the Recurrence Relations.
Sir is using resemble recursive pattern from algebra and number theory which of next term is depending on its values at other points, or previous points, which is smaller also it creates the dependence and function can defined .....I am also math teacher ....I am very much inspired by sir (sir doing a great job keep going sir)!!!!!
Q. Suppose k > 0. Let c and d be the roots of x² - kax - (k + 1)b = 0 and a, b be the roots of x² - kcx - (k + 1)d = 0, and a + b + c + d = 180, then k is equal to A. 11 B. 9 C. 7 D. 5
I have my own methord if someone wanna hear degre of fx , f(x /2) , f (x /4) will be polynomial of degrrr 2 now f dash 0 is zero hence ax² + bx + c term b will be zero as f dash zero is zero now put the given f = àx² + c in above relation
Sir I have solved this question by other method.. continous function h to Maine quadratic function let kar liya aur fir jo condition diya hua h use same aagya .. answer mja aagya . Shi h ki ni..
Sur we could also do this by assuming f(x) to be a quadratic expression as it must HV atleast degree 2 then we will see that coefficient of x must be 0 and that power higher than x^2 cannot be cancelled and comparing with rha coefficient of x^2 will come 16/9
This seems incorrect .When n tends to infinity , the denominator tends to infinity ,but it is wrong to say f(x/2^(n+1)) is same as f(0) . x is an element of all real numbers which means x can be infinitely large too . To say this is true for all x is mathematically wrong .Infact if x -> infinity the form is indeterminate of infinity/infinity and hence cant be 0 . The answer may be right but the approach is fundamentally wrong .
Fiitjee mein ase question ko koi importance dete bhi nahi sach bolu mein to mein abhi top batch mein hu aur yah Sare hod teacher khudke ke banye hue question karate hai jee advance se bhi khatarnaak.
