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  • How to solve this problem- - Limit , continuity and differentiability - JEE Main-9

let f(x) and g(x) are continous \forall x\epsilon R . if f(1)=5 , f(2) =2 , g(1)= 7 , g(2) = 1
 then there will be at least one solution lieing in (1,2) for the equation 

  • Option 1)

    f(x)+g(x)=2 

  • Option 2)

    f(x)+g(x)=3

  • Option 3)

    f(x)+g(x)=1

  • Option 4)

    f(x)+g(x)=5

 

Answers (1)

best_answer

 As we have learned

Properties of Continuous function -

If  f  is continuous on [a, b] the f assumes at least once every value between its minimum and maximum values then there exists at least one solution of the equation f(x) = K in the open interval (a, b)

-

 

Let h(x)= f(x)+g(x) ,   f and g are continous so h will also be 

h(1)= 5+7=12  and h(2)= 2+1=3

\therefore h(x) wil take all values between 3 and 12 atleast once in (1,2)

 

 

 

 

 

 


Option 1)

f(x)+g(x)=2 

Option 2)

f(x)+g(x)=3

Option 3)

f(x)+g(x)=1

Option 4)

f(x)+g(x)=5

Posted by

Himanshu

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