Mathematics - MCQ Practice Questions

Using the standard limits ${\lim }_{u\to 0}\frac{\sin u}{u}=1$ and ${\lim }_{u\to 0}\frac{\tan u}{u}=1$.

Critical points: $x=1$ and $x=3$. \\ $f^{\prime \prime }(x)=6x-12$. \\ At $x=1$: $f^{\prime \prime }(1)=-6<0$ → local maximum. \\ At $x=3$: $f^{\prime \prime }(3)=6>0$ → local minimum. \\ Hence the local minimum is at $\boxed{x=3}$.

Hence the area is $\boxed{\frac{26}{3}}$ square units.

Using the chain rule: $\frac{d}{dx}[\ln (u)]=\frac{u^{\prime }}{u}$ with $u=\sin x$.

Using the standard formula $\int \frac{dx}{x^2+a^2}=\frac{1}{a}\arctan \left(\frac{x}{a}\right)+C$ with $a=2$:

$f(x)=|x|$ satisfies ${\lim }_{x\to 0^{-}}|x|=0={\lim }_{x\to 0^{+}}|x|=f(0)$, so it is continuous. However, the left-hand derivative $=-1$ and the right-hand derivative $=+1$ are unequal, so it is not differentiable at $x=0$. Hence $f$ is $\boxed{\text{continuous but not differentiable at }x=0}$.

Rolle's Theorem requires: (i) continuity on $[a,b]$, (ii) differentiability on $(a,b)$, (iii) $f(a)=f(b)$. \\ For $f(x)=1-x^2$ on $[-1,1]$: it is a polynomial (continuous and differentiable everywhere), and $f(-1)=0=f(1)$. All conditions are met. \\ $\frac{1}{x}$ is undefined at $0$; $|x|$ is not differentiable at $0$; $\tan x$ has a discontinuity inside. Hence $\boxed{f(x)=1-x^2}$.

Each differentiation multiplies by $2$ (chain rule), giving $2^n$ after $n$ differentiations.

Equation of tangent using point-slope form:

Thus $x=\boxed{4}$.

Let $\alpha$ be the common root. Then:

Subtracting: $(b-c)\alpha +(c-b)=0\Rightarrow (b-c)(\alpha -1)=0$. Since $b\ne c$, $\alpha =1$. Substituting back: $1+b+c=0\Rightarrow b+c=\boxed{-1}$.

Let $u=|x|\ge 0$. Then $u^2-3u+2=0\Rightarrow (u-1)(u-2)=0\Rightarrow u=1$ or $u=2$.

So there are $\boxed{4}$ real solutions: $x\in \{-2,-1,1,2\}$.

This gives $x=5$ or $x=-\frac{1}{5}$. Checking the options, the integer solution is $x=\pm 5$. Indeed $x=-5$: $\frac{-4}{-6}-\frac{-6}{-4}=\frac{2}{3}-\frac{3}{2}=-\frac{5}{6}$, and $x=5$ gives $\frac{5}{6}$. Thus $x=\boxed{\pm 5}$.

For $n=10$:

By definition, $a,b,c$ are in H.P. if and only if their reciprocals $\frac{1}{a},\frac{1}{b},\frac{1}{c}$ are in A.P. This means:

Option A is $\boxed{\text{correct by definition of H.P.}}$ Options C and D describe A.P. and G.P. respectively, not H.P.