Question
If f'(x) = x + b, f'(1) = 5, f'(2) = 13, find f'(x).
✓
Answer
$\text{f}'\text{(x)}=\text{x + b},\text{f}'(1)=5,\text{f}'(2)=13$
$\text{f}'\text{(x)}=\text{x + b}$
$\int\text{f}'\text{(x) dx}=\int(\text{x + b})\text{dx}$
$\text{f}'\text{(x)}=\frac{\text{x}^2}{2}+\text{bx}+\text{C}\ \dots(1)$
$\text{f}'(1)=5,\text{f}'(2)=13$ (Given)
putting x = 1 in (1)
$\text{f}'(1)=\frac{1^2}{2}+\text{b}_1+\text{C}$
$5=\frac{1}{2}+\text{b + C}\ \dots(2)$
Putting x = 2 in (1)
$\text{f}'(2)=\frac{2^2}{2}+\text{b}_2+\text{C}$
$13=\frac{4}{2}+2\text{b + C}$
$13=2+2\text{b + C}\ \dots(3)$
Solving (2) and (3) we get,
$\text{b}=\frac{13}{2}\text{ and }\text{C}=-2$
Thus, $\text{f}'\text{(x)}=\frac{\text{x}^2}{2}+\frac{13}{2}\text{x}-2$
$\text{f}'\text{(x)}=\text{x + b}$
$\int\text{f}'\text{(x) dx}=\int(\text{x + b})\text{dx}$
$\text{f}'\text{(x)}=\frac{\text{x}^2}{2}+\text{bx}+\text{C}\ \dots(1)$
$\text{f}'(1)=5,\text{f}'(2)=13$ (Given)
putting x = 1 in (1)
$\text{f}'(1)=\frac{1^2}{2}+\text{b}_1+\text{C}$
$5=\frac{1}{2}+\text{b + C}\ \dots(2)$
Putting x = 2 in (1)
$\text{f}'(2)=\frac{2^2}{2}+\text{b}_2+\text{C}$
$13=\frac{4}{2}+2\text{b + C}$
$13=2+2\text{b + C}\ \dots(3)$
Solving (2) and (3) we get,
$\text{b}=\frac{13}{2}\text{ and }\text{C}=-2$
Thus, $\text{f}'\text{(x)}=\frac{\text{x}^2}{2}+\frac{13}{2}\text{x}-2$
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
Evaluate the following integrals:
$\int\limits_{0}^{{\pi}}\frac{1}{5+3\cos\text{x}}\text{ dx}$
→$\int\limits_{0}^{{\pi}}\frac{1}{5+3\cos\text{x}}\text{ dx}$
Find the area of the bounded by $\text{y}=\sqrt{\text{x}}$ and $y^2 = x.$
→Solve the following differential equation:
$(1+\text{y}^2)+(\text{x}-\text{e}^{\tan^{-1}\text{y}})\frac{\text{dy}}{\text{dx}}=0$
→$(1+\text{y}^2)+(\text{x}-\text{e}^{\tan^{-1}\text{y}})\frac{\text{dy}}{\text{dx}}=0$
Evaluate the following integrals:
$\int\frac{(\text{x}\tan^{-1}\text{x})}{(1+\text{x}^2)^{\frac{3}{2}}}\text{dx}$
→$\int\frac{(\text{x}\tan^{-1}\text{x})}{(1+\text{x}^2)^{\frac{3}{2}}}\text{dx}$
Differentiate $\sin^{-1}\sqrt{1-\text{x}^2}$ with respect to $\cos^{-1}\text{x},$ if
$\text{x}\in(0, 1)$
→$\text{x}\in(0, 1)$
Solve the following systems of homogeneous linear equations by matrix method: $x + y - z = 0 , x - 2y + z = 0a , 3x + 6y - 5z = 0$
→Using properties of determinants, prove that
; background:rgba(220,220,220,0.5))
If $\text{y}\sqrt{1-\text{x}^2}+\text{x}\sqrt{1-\text{y}^2}=1,$ prove that $\frac{\text{dy}}{\text{dx}}=-\sqrt{\frac{1-\text{y}^2}{1-\text{x}^2}}$
→If $\text{x}=2\cos\theta-\cot2\theta$ and $\text{y}=2\sin\theta-\sin2\theta,$ prove that $\frac{\text{dy}}{\text{dx}}=\tan\big(\frac{3\theta}{2}\big)$
→Solve the following linear programming problem graphically.Minimise $\text{z = 3x + 5y}$
subject to the constraints
$\text{x + 2y}\geq 10$
$\text{x + y}\geq 6$
$\text{3x + y}\geq 8$
$\text{x, y}\geq 0.$
→subject to the constraints
$\text{x + 2y}\geq 10$
$\text{x + y}\geq 6$
$\text{3x + y}\geq 8$
$\text{x, y}\geq 0.$