Let f: R R be a differentiable function such that its derivative … - Vidyadip

MCQ

Let $f: R \rightarrow R$ be a differentiable function such that its derivative $f^{\prime}$ is continuous and $f(\pi)=-6$. If $F:[0, \pi] \rightarrow R$ is defined by $F(x)=\int_0^{ x } f( t ) dt$, and if $\int_0^\pi\left(f^{\prime}( x )+ F ( x )\right) \cos x dx =2$ then the value of $f(0)$ is. . . . . .

✓
$4$
B
$5$
C
$6$
D
$7$

✓

Answer

Correct option: A.

$4$

a
$F ( x )=\int_0^{ x } f( t ) \cdot dt$

$\Rightarrow F ^{\prime}( x )=f( x )$

$I =\int_0^\pi f^{\prime}( x ) \cdot \cos x d x +\int_0^\pi F ( x ) \cos ( x ) dx =2 .$ $. . . . . . (1)$

$I _1=\int_0^\pi f^{\prime}( x ) \cdot \cos xdx \text { (Let) }$

Using by parts

$I _1=(\cos x \cdot f( x ))_0^\pi+\int_0^\pi \sin x . f( x ) dx$

$I _1=6-f(0)+\int_0^\pi \sin x \cdot F ^{\prime}( x ) dx$

$I _1=6-f(0)+ I _2$ $. . . . . . (2)$

$I _2=\int_0^\pi \sin x \cdot F ^{\prime}( x ) \cdot dx$

Using by part we get

$I_2=(\sin x \cdot F ( x ))_0^\pi-\int_0^\pi \cos x F( x ) dx$

$I _2=-\int_0^\pi \cos x \cdot F ( x ) dx$

$(2)$ $\Rightarrow I _1=6-f(0)-\int_0^\pi \cos x \cdot F ( x ) dx$

$(1) \Rightarrow I =6-f(0)=2 \Rightarrow f(0)=4$

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Explore more

More "M.C.Q (1 Marks)" from STD 12 - 7.2 definite integral→STD 12 - 7.2 definite integral — all question types→Mathematics STD 12 Science question papers→CBSE Board STD 12 Science question papers→CBSE Board question paper generator→

Similar questions

The ratio of the altitude of the cone of greatest volume which can be inscribed in a given sphere to the diameter of the sphere is :-

If $u, v, w$ are non$-$coplanar vector and $p, q$ are real numbers, then the equality $\text{[3u pv pw] - [pv w qw] - [2w qv qu] = 0}$ holds for:

Point $(4, 0)$ lies on $..........?$

If $\left(a_1, b_1, c_1\right)$ and $\left(a_2, b_2, c_2\right)$ be the direction ratios of two parallel lines then

If $A$ is a square matrix of order 3 and $|2 A|=k|A|$, then find the value of $k$.

The area bounded by the curvey $=\sqrt{\text{x}}$ the line 2y + 3 = x and the x - axis in the first quadrant is:

A man 2 metres tall walks away from a lamp post 5 metres height at the rate of 4.8km/hr. The rate of increase of the length of his shadow is:

The value of the definite integral, $\int\limits_0^{\sqrt {\ln \left( {\frac{\pi }{2}} \right)} } {\cos \left( {{e^{{x^2}}}} \right)} {\mkern 1mu} \cdot {\mkern 1mu} 2x{\mkern 1mu} {e^{{x^2}}}dx$ is

$\big(\vec{\text{a}}+\vec{\text{b}}\big).\big(\vec{\text{b}}+\vec{\text{c}}\big)\times\big(\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}\big)=$

If $y=y(x)$ is the solution of the differential equation $x \frac{d y}{d x}+2 y=x e^{x}, y(1)=0$ then the local maximum value of the function $z(x)=x^{2} y(x)-e^{x}$, $x \in R$ is