Maths-
General
Easy

Question

Let f left parenthesis t with not stretchy bar on top right parenthesis equals left square bracket t right square bracket i with not stretchy bar on top minus left parenthesis t minus left square bracket t right square bracket right parenthesis j with not stretchy bar on top plus left square bracket t plus 1 right square bracket k with not stretchy bar on top comma left square bracket. right square bracket is greatest integer function. If f open parentheses 5 over 4 close parentheses and i with not stretchy bar on top plus lambda j with not stretchy bar on top plus mu k with not stretchy bar on top are parallel vectors thenleft parenthesis lambda comma mu right parenthesis =

  1. (1,1)
  2. open parentheses negative 1 fourth comma 2 close parentheses
  3. open parentheses 1 half comma 2 close parentheses
  4. open parentheses 1 fourth comma 4 close parentheses

hintHint:

We are given a function. We have to find it’s value at particular value. We are given another vector which is parallel to the given function. The second vector has two variables. We have to find the values of the variable.

The correct answer is: open parentheses negative 1 fourth comma 2 close parentheses


    The given function is f with rightwards arrow on top open parentheses t close parentheses space equals space left square bracket t right square bracket i with hat on top space minus space left parenthesis t space minus space left square bracket t right square bracket right parenthesis j with hat on top space plus space left square bracket t space plus space 1 right square bracket k with hat on top
    We have to find it’s value at t = 5/4
    Let’s denote the vector parallel to this vector by P.
    So, P = P space equals i with hat on top plus lambda j with hat on top plus mu k with hat on top
    We have to find the values of (λ, μ)
    Greatest integer means we have to write the integer which is closest and smaller than the value in the square bracket.
    We will find the value of f open parentheses 5 over 4 close parentheses
    stack f left parenthesis t right parenthesis with rightwards arrow on top space equals space open square brackets t close square brackets i with hat on top space minus space left parenthesis t space minus space open square brackets t close square brackets right parenthesis j with hat on top space plus space open square brackets t space plus space 1 close square brackets k with hat on top
stack f open parentheses 5 over 4 close parentheses with rightwards arrow on top equals open square brackets 5 over 4 close square brackets i with hat on top space minus open parentheses 5 over 4 minus open square brackets 5 over 4 close square brackets close parentheses j with hat on top space plus open square brackets 5 over 4 plus 1 close square brackets k with hat on top
space space space space space space space space space space space space equals open square brackets 1.25 close square brackets i with hat on top space minus open parentheses 5 over 4 minus left square bracket 1.25 right square bracket close parentheses j with hat on top space plus open square brackets 9 over 4 close square brackets k with hat on top
space space space space space space space space space space space space equals space 1 i with hat on top space minus open parentheses 5 over 4 minus 1 close parentheses j with hat on top space plus space left square bracket 2.25 right square bracket k with hat on top
space f with rightwards arrow on top open parentheses 5 over 4 close parentheses equals i with hat on top space minus 1 fourth j with hat on top space plus 2 k with hat on top
L e t apostrophe s space d e n o t e space i t space b y space B with rightwards arrow on top.
B with rightwards arrow on top space equals i with hat on top minus 1 fourth i with hat on top space plus 2 k with hat on top
    Now, as the vectors B with rightwards arrow on top and p with rightwards arrow on top are parallel we can express one of the vector as a scalar multiple of the other vector. Let the scalar multiple be a.
    We can write
    B with rightwards arrow on top equals a P with rightwards arrow on top
left parenthesis i with hat on top space minus 1 fourth j with hat on top space plus space 2 k with hat on top right parenthesis space equals a left parenthesis i with hat on top plus lambda j with hat on top plus mu k with hat on top right parenthesis
space space space space space space space space space space space space space space space space space space space space space space space space space space equals a i with hat on top space plus a lambda j with hat on top plus a mu k with hat on top
C o m p a r i n g space b o t h space t h e space s i d e s.
a space equals space 1
a lambda space equals negative 1 fourth
left parenthesis 1 right parenthesis lambda space equals negative 1 fourth
lambda space equals negative 1 fourth
a mu space equals space 2
left parenthesis 1 right parenthesis mu space equals 2
mu space equals space 2
    The required values are open parentheses lambda comma u close parentheses equals open parentheses negative 1 fourth comma 2 close parentheses

    For such questions, we should know the concept of greatest integer number. We should also know the properties of parallel vectors.

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