Let f: R rightarrow R be defined as f(x)=a e^ | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Given $f(x)=a e^{2 x}+b e^x+c x$.$f(0)=-1 \Rightarrow a+b=-1 \quad (1)$$f^{\prime}(x)=2 a e^{2 x}+b e^x+c$.$f^{\prime}(\ln 2)=2 a(4)+b(2)+c=8 a+2

Vedclass · Accepted Answer

$8$

Vedclass · Answer

(D) Given $f(x)=a e^{2 x}+b e^x+c x$.$f(0)=-1 \Rightarrow a+b=-1 \quad (1)$$f^{\prime}(x)=2 a e^{2 x}+b e^x+c$.$f^{\prime}(\ln 2)=2 a(4)+b(2)+c=8 a+2 b+c=21 \quad (2)$Given $\int_0^{\ln 4}(f(x)-c x) d x=\frac{39}{2}$.$\int_0^{\ln 4}(a e^{2 x}+b e^x) d x=\left[\frac{a e^{2 x}}{2}+b e^x\right]_0^{\ln 4}=\frac{39}{2}$.$\left(\frac{a(16)}{2}+b(4)\right)-\left(\frac{a}{2}+b\right)=\frac{39}{2}$.$8 a+4 b-\frac{a}{2}-b=\frac{39}{2} \Rightarrow \frac{15 a}{2}+3 b=\frac{39}{2} \Rightarrow 15 a+6 b=39 \Rightarrow 5 a+2 b=13 \quad (3)$.From $(1)$,$b=-1-a$. Substitute into $(3)$:$5 a+2(-1-a)=13 \Rightarrow 3 a-2=13 \Rightarrow 3 a=15 \Rightarrow a=5$.Then $b=-1-5=-6$.From $(2)$,$8(5)+2(-6)+c=21 \Rightarrow 40-12+c=21 \Rightarrow 28+c=21 \Rightarrow c=-7$.Thus,$a+b+c=5-6-7=-8$.Therefore,$|a+b+c|=|-8|=8$.

Let $f: R \rightarrow R$ be defined as $f(x)=a e^{2 x}+b e^x+c x$. If $f(0)=-1$,$f^{\prime}(\log _e 2)=21$,and $\int_0^{\log _e 4}(f(x)-c x) d x=\frac{39}{2}$,then the value of $|a+b+c|$ equals:

Similar Questions

The function $f(x) = \int\limits_0^x \sqrt{1 - t^4} \, dt$ is such that

If $\alpha \in (2, 3)$,then the number of solutions of the equation $\int_{0}^{\alpha} \cos(x + \alpha^2) \, dx = \sin \alpha$ is:

$\int_{0}^{2} ( |2x^2 - 3x| + [x - \frac{1}{2}] ) dx$,where $[ \cdot ]$ is the greatest integer function,is equal to:

Value of the $\int_0^9 \sqrt{x} \,dx + \int_0^{\pi/2} (\cos x + \sin x) \,dx$ is:

Assertion $(A)$: $\int_0^{\frac{\pi}{2}} (\sin^6 x + \cos^6 x) dx$ lies in the interval $(\frac{\pi}{8}, \frac{\pi}{2})$.
Reason $(R)$: $\sin^6 x + \cos^6 x$ is a periodic function with period $\frac{\pi}{2}$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module