Question

If $f(x)=(ax+b)sin~x+(cx+d)cos~x$, then the values of a, b, c and d such that $f^\prime(x)=x~cos~x$ for all x, are

• A. $b=c=0, a=d=1$
• B. $b=d=0, a=c=1$
• C. $c=d=0, a=b=1$
• D. None of the above

Hint:

If $f(x)=(ax+b)sin~x+(cx+d)cos~x$, then the values of a, b, c and d such that $f

Answer 1

The Correct Option is A

Step 1: Concept
Differentiate $f(x)$ and equate the resulting expression to $x~cos~x$.
Step 2: Analysis
$f'(x) = a~sin~x + (ax+b)cos~x + c~cos~x - (cx+d)sin~x$. Rearrange: $f'(x) = (a-cx-d)sin~x + (ax+b+c)cos~x$.
Step 3: Evaluation
Equate coefficients with $x~cos~x$: Coefficient of $x~cos~x \Rightarrow a=1$. Coefficient of $x~sin~x \Rightarrow c=0$. Constant for $cos~x \Rightarrow b+c=0 \Rightarrow b=0$. Constant for $sin~x \Rightarrow a-d=0 \Rightarrow d=1$.
Step 4: Conclusion
Thus, $a=1, b=0, c=0, d=1$, which corresponds to $b=c=0$ and $a=d=1$.
Final Answer: (a)