IIT JEE 1982 Maths - 'Adapted' - Subjective to Multi-Correct Q8

Calculus Level 3

Let f f be a twice differentiable function and g , h g, h be some functions, such that f ′ ′ ( x ) = − f ( x ) , f ′ ( x ) = g ( x ) , h ( x ) = [ f ( x ) ] 2 + [ g ( x ) ] 2 , h ( 5 ) = 11 f''(x)=-f(x), f'(x)=g(x), h(x)=[f(x)]^2+[g(x)]^2, h(5)=11 . Then which of the following is/are not incorrect ?

  • (A) h(10)=11
  • (B) h'(10)=0
  • (C) h(0)=5
  • (D) h'(0)=-2

Enter your answer as a 4 digit string of 1s and 9s - 1 for correct option, 9 for wrong. Eg. 1199 indicates A and B are correct, C and D are incorrect. None, one or all may also be correct.

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The answer is 1199.

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Shubhamkar Ayare by Shubhamkar Ayare , 22, India

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1 solution

Tom Engelsman
Dec 19, 2016

Taking the first DE, we obtain:

f ′ ′ ( x ) + f ( x ) = 0 ⇒ f''(x) + f(x) = 0 \Rightarrow which has the characteristic equation r 2 + 1 = 0 ⇒ r = ± i ⇒ f ( x ) = A c o s ( x ) + B s i n ( x ) r^2 + 1 = 0 \Rightarrow r = \pm i \Rightarrow f(x) = Acos(x) + Bsin(x)

and g ( x ) = f ′ ( x ) = − A s i n ( x ) + B c o s ( x ) g(x) = f'(x) = -Asin(x) + Bcos(x) . We finally obtain the function h ( x ) h(x) :

h ( x ) = f ( x ) 2 + g ( x ) 2 = ( A 2 c o s 2 ( x ) + 2 A B s i n ( x ) c o s ( x ) + B 2 s i n 2 ( x ) ) + ( A 2 s i n 2 ( x ) − 2 A B s i n ( x ) c o s ( x ) + B 2 c o s 2 ( x ) ) ; h(x) = f(x)^2 + g(x)^2 = (A^2 cos^2(x) + 2ABsin(x)cos(x) + B^2 sin^2(x)) + (A^2 sin^2(x) - 2ABsin(x)cos(x) + B^2 cos^2(x));

or h ( x ) = A 2 ⋅ ( c o s 2 ( x ) + s i n 2 ( x ) ) + B 2 ⋅ ( c o s 2 ( x ) + s i n 2 ( x ) ) = A 2 + B 2 = C o n s t a n t . h(x) = A^2 \cdot (cos^2(x) + sin^2(x)) + B^2 \cdot (cos^2(x) + sin^2(x)) = A^2 + B^2 = \boxed{Constant}.

Hence h ( x ) = 11 , x ∈ R h(x) = 11, x \in \mathbb{R} , which only makes choices A and B true.

2 Helpful 0 Interesting 0 Brilliant 0 Confused

The third line should be g ( x ) = f ′ ( x ) g(x) = f'(x) instead of − f ( x ) - f(x) .

Calvin Lin Staff - 4 years, 5 months ago

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