IIT JEE 1982 Maths - MCQ Question 5

Calculus Level 3

The area bounded by the curves y = f ( x ) y=f(x) , the x x -axis and the ordinates x = 1 x=1 and x = b x=b is ( b − 1 ) sin ⁡ ( 3 b + 4 ) (b-1) \sin(3b+4) . Then what is f ( x ) f(x) ?

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s i n ( 3 x + 4 ) sin(3x+4) ( x − 1 ) c o s ( 3 x + 4 ) (x-1)cos(3x+4) None of the others s i n ( 3 x + 4 ) + 3 ( x − 1 ) c o s ( 3 x + 4 ) sin(3x+4)+3(x-1)cos(3x+4)

Shubhamkar Ayare by Shubhamkar Ayare , 22, India

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1 solution

Kushal Bose
Nov 30, 2016

As stated area under f ( x ) f(x) form x = 1 → x = b x=1 \to x=b then

∫ 1 b f ( x ) = ( b − 1 ) sin ⁡ ( 3 b + 4 ) \int_{1}^{b} f(x)=(b-1) \sin (3 b+4)

Consider it as a function of b b .So,by using Newtob-Leibnitz rule both side.

f ( b ) = sin ⁡ ( 3 b + 4 ) + 3 ( b − 1 ) cos ⁡ ( 3 b + 4 ) f(b)=\sin (3 b+4) + 3(b-1) \cos (3 b+4)

Now put b = x b=x then our function is

f ( x ) = sin ⁡ ( 3 x + 4 ) + 3 ( x − 1 ) cos ⁡ ( 3 x + 4 ) f(x)=\sin (3 x+4) + 3(x-1) \cos (3 x+4)

2 Helpful 0 Interesting 0 Brilliant 0 Confused

As area is said to follow the given condition, wouldn't (-1)*the answer also be a solution? would there be finitely many such continuous functions possible?

Ajinkya Shivashankar - 4 years, 6 months ago

yes you are correct if f(x) bounds area A then -f(x) also bounds same area.

Kushal Bose - 4 years, 6 months ago

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