Q: Let R denote the set of all real numbers. Let f: R R and g: R (0,4) be functions defined by f(x)= _e (x^2+2 x+4 ), and g(x)= 4 1+e^ -2 x Define the composite function f g^ -1 by (f g^ -1 )(x)= (g^ -1

aligned & Let h ( x )= f ( g ^ -1 ( x ) ) and g (0)=2 \\ & aligned h ^ ( x ) & = f ^ ( g ^ -1 ( x ) ) ( g ^ -1 ( x ) )^ \\ h ^ (2) & = f ^ ( g ^ -1 (2) ( g ^ -1 )^ (2) . \\ & = f ^ (0) ( g ^ -1 )^ (2) aligned aligned Now aligned & f(x)= _0 (x^2+2 x+4 ) \\ & f^ (x)= 2 x+2 x^2+2 x+4 \\ & f^ (0)= 1 2 \\ & g(x)= 4 1+e^ -2 x , g(0)=2 \\ & g^ -1 (g(x))=x aligned array l|l ( ( g ^ -1 )^ ( g ( x )) ) g ^ ( x )=1 & ~g ( x )= 4 1+ e ^ -2 x \\ ( g ^ -1 )^ (2)= 1 ~g ^ (0) & g ^ ( x )= 8 e ^ -2 x (1+ e ^ -2

Question 1

Let R denote the set of all real numbers. Consider the polynomial function f: R R defined by f(x) = d^ 10 dx^ 10 ((x^2 - 1)^ 10 ), for all x R . Here d^ 10 dx^ 10 ((x^2 - 1)^ 10 ) is the 10th order de

Accepted Answer

Given f(x) = d^ 10 dx^ 10 ((x^2 - 1)^ 10 ). Using the binomial expansion, we have: (x^2 - 1)^ 10 = _ k=0 ^ 10 (-1)^k \, ^ 10 C_ k x^ 20-2k Differentiating 10 times with respect to x, we get: f(x) = _ k=0 ^ 5 (-1)^k \, ^ 10 C_ k (20-2k)! (10-2k)! x^ 10-2k For the coefficient of x^8, we set 10 - 2k = 8 k = 1. The coefficient is (-1)^1 \, ^ 10 C_ 1 18! 8! = -10 ( 18! 8! ). Thus, statement (A) is true. The highest power of x in f(x) corresponds to k = 0, which gives x^ 10 with a non-zero coefficient

Question 2

Let R denote the set of all real numbers. Let f: R R and g: R (0,4) be functions defined by f(x)= _e (x^2+2 x+4 ), and g(x)= 4 1+e^ -2 x Define the composite function f g^ -1 by (f g^ -1 )(x)= (g^ -1

Accepted Answer

aligned & Let h ( x )= f ( g ^ -1 ( x ) ) and g (0)=2 \ & aligned h ^ ( x ) & = f ^ ( g ^ -1 ( x ) ) ( g ^ -1 ( x ) )^ \ h ^ (2) & = f ^ ( g ^ -1 (2) ( g ^ -1 )^ (2) . \ & = f ^ (0) ( g ^ -1 )^ (2) aligned aligned Now aligned & f(x)= _0 (x^2+2 x+4 ) \ & f^ (x)= 2 x+2 x^2+2 x+4 \ & f^ (0)= 1 2 \ & g(x)= 4 1+e^ -2 x , g(0)=2 \ & g^ -1 (g(x))=x aligned array l|l ( ( g ^ -1 )^ ( g ( x )) ) g ^ ( x )=1 & ~g ( x )= 4 1+ e ^ -2 x \ ( g ^ -1 )^ (2)= 1 ~g ^ (0) & g ^ ( x )= 8 e ^ -2 x (1+ e ^ -2

Differentiation — Previous Year Questions & Solutions