Maths Questions

First, equate the equations to determine their points of intersection:

$2x-x^2=-x$

$\Rightarrow 3x-x^2=0$

$\Rightarrow x(3-x)=0$

Therefore, $x=0$ or $x=3$.

For $0\leq x\leq3$, the curve $y=2x-x^2$ lies above the line $y=-x$.

Therefore, the required area is:

$A=\int_0^3\left[(2x-x^2)-(-x)\right],dx$

$A=\int_0^3(3x-x^2),dx$

$A=\left[\frac{3x^2}{2}-\frac{x^3}{3}\right]_0^3$

$A=\left(\frac{3(3)^2}{2}-\frac{(3)^3}{3}\right)-0$

$A=\frac{27}{2}-9$

$A=\frac{27-18}{2}$

$A=\frac{9}{2}$

Hence, the area bounded by the given curve and line is $\frac{9}{2}$ square units.

Let the roots be $\alpha$ and $\beta$. By Vieta’s formulas, $\alpha+\beta=a-2$ and $\alpha\beta=-a-1$.

Now, $\alpha^2+\beta^2=(\alpha+\beta)^2-2\alpha\beta$.

Thus, $\alpha^2+\beta^2=(a-2)^2-2(-a-1)=a^2-2a+6=(a-1)^2+5$.

Since $(a-1)^2\geq0$, the sum is minimum when $(a-1)^2=0$. Hence, $a=1$.

The minimum value of $a\cos x+b\sin x$ is $-\sqrt{a^2+b^2}$.

Therefore, $\min(3\cos x+4\sin x)=-\sqrt{3^2+4^2}=-5$.

Hence, $\min(3\cos x+4\sin x+8)=-5+8=3$.

Using the standard identities:

$\sinh^{-1}x=\log\left(x+\sqrt{x^2+1}\right)$, therefore $a-\mathrm{IV}$.

$\cosh^{-1}x=\log\left(x+\sqrt{x^2-1}\right)$, therefore $b-\mathrm{III}$.

$\tanh^{-1}x=\frac{1}{2}\log\left(\frac{1+x}{1-x}\right)$, therefore $c-\mathrm{II}$.

$\coth^{-1}x=\frac{1}{2}\log\left(\frac{x+1}{x-1}\right)$, therefore $d-\mathrm{I}$.

Let $T(x,y,z)=(u,v,w)$. Then $u=3x$, $v=x-y$ and $w=2x+y+z$.

From $u=3x$, we get $x=\frac{u}{3}$.

From $v=x-y$, we get $y=\frac{u}{3}-v$.

From $w=2x+y+z$, we get $z=w-u+v$.

Therefore, $T^{-1}(u,v,w)=\left(\frac{u}{3},\frac{u}{3}-v,w-u+v\right)$.

By the forward difference definition, the derivative of $f(x)$ at $x=a$ is

$f'(a)=\displaystyle\lim_{h\to 0}\frac{f(a+h)-f(a)}{h}$.

Here, $f(a+h)-f(a)$ represents the forward change in the function, and $h$ represents the corresponding change in $x$.

Hence, the correct answer is A.

The element $2$ is located in the first row and first column. Therefore, its cofactor is

$C_{11}=(-1)^{1+1}M_{11}$

After deleting the first row and first column, the remaining minor is

$M_{11}=5$

Therefore,

$C_{11}=(-1)^2(5)=5$

Hence, the correct answer is 5.

The absolute value of the scalar triple product of three vectors gives the volume of the parallelepiped formed by them.

$V=\left|\vec{a}\cdot(\vec{b}\times\vec{c})\right|$

Therefore, $\vec{a}\cdot(\vec{b}\times\vec{c})$ represents the signed volume, while its absolute value represents the actual volume.

Hence, the correct answer is $\boxed{\text{(C) Volume of a parallelepiped}}$.

The given vector is $\vec{a}=\hat{i}+\hat{j}+\hat{k}$.

Its magnitude is

$|\vec{a}|=\sqrt{1^2+1^2+1^2}=\sqrt{3}$.

The direction cosines are

$\left(\dfrac{1}{|\vec{a}|},\dfrac{1}{|\vec{a}|},\dfrac{1}{|\vec{a}|}\right)
=\left(\dfrac{1}{\sqrt{3}},\dfrac{1}{\sqrt{3}},\dfrac{1}{\sqrt{3}}\right)$.

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