Mathematics - MCQ Practice Questions

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.

Understanding:

We need to find ${\alpha }^3+{\beta }^3$ where $\alpha$ and $\beta$ are roots of $x^2-5x+6=0$.

Formula:

Using the identity:

Step 1: Identify sum and product of roots.

By Vieta's formulas:

Step 2: Substitute into the identity.

Answer:

The value of ${\alpha }^3+{\beta }^3$ is $35$.

Quick Tip:

A common error is computing $(\alpha +\beta {)}^3$ alone without subtracting $3\alpha \beta (\alpha +\beta )$. Always use the full identity.

Understanding:

Given $x+\frac{1}{x}=3$, we need to find $x^4+\frac{1}{x^4}$.

Formula:

Repeatedly use the squaring identity:

Step 1: Find $x^2+\frac{1}{x^2}$.

Step 2: Find $x^4+\frac{1}{x^4}$.

Answer:

The value of $x^4+\frac{1}{x^4}$ is $47$.

Quick Tip:

Square the given expression step by step rather than expanding directly. Each squaring increases the power by a factor of $2$.

Understanding:

We need to count real solutions of $x^2-5|x|+6=0$.

Since $x^2=|x{|}^2$, substituting $t=|x|\ge 0$ simplifies the equation.

Formula:

Let $t=|x|$. Then:

Step 1: Solve the quadratic in $t$.

Both values are non-negative, so both are valid.

Step 2: Recover $x$ from each value of $t$.

Step 3: Count total real solutions.

There are $4$ distinct real values: $\{-3,-2,2,3\}$.

Answer:

The equation has $4$ real solutions.

Quick Tip:

Whenever $|x|$ appears, substitute $t=|x|\ge 0$. Each positive value of $t$ gives two real values of $x$; $t=0$ gives only one.

Understanding:

For the quadratic $2x^2-3x+k=0$ to have equal roots, the discriminant must be zero.

Formula:

For equal roots, the discriminant condition is:

Step 1: Apply the condition.

Verification:

Substitute back: $\Delta =9-8\cdot \frac{9}{8}=9-9=0$. ✓

Answer:

The value of $k$ for equal roots is $\frac{9}{8}$.

Quick Tip:

Equal roots $\iff$ discriminant $=0$. Never confuse this with real roots ($\Delta \ge 0$) or distinct roots ($\Delta >0$).

Understanding:

We solve ${\log }_{2}x+{\log }_{4}x+{\log }_{8}x=11$.

Convert all logarithms to base $2$.

Formula:

Change-of-base rule:

Step 1: Express each term in base $2$.

Step 2: Substitute and simplify. Let $u={\log }_{2}x$.

Step 3: Recover $x$.

Answer:

The value of $x$ is $2^6=64$.

Quick Tip:

Convert all log terms to a single base before combining. The LCD of the denominators saves time when adding the fractions.

Understanding:

The roots of $x^2-3x+2=0$ are $\alpha$ and $\beta$. We need the equation with roots $2\alpha$ and $2\beta$.

Formula:

If the new roots are $2\alpha$ and $2\beta$, the new equation is:

Step 1: Find sum and product of original roots.

Step 2: Compute new sum and product.

Step 3: Write the new equation.

Verification:

Original roots: $x^2-3x+2=0\Rightarrow (x-1)(x-2)=0$, so $\alpha =1,\beta =2$.

New roots: $2,4$. Check: $(x-2)(x-4)=x^2-6x+8=0$. ✓

Answer:

The required quadratic equation is $x^2-6x+8=0$.

Quick Tip:

To obtain an equation with roots $k$ times those of $f(x)=0$, replace $x$ with $\frac{x}{k}$ and simplify. Here, replace $x$ with $\frac{x}{2}$: $\frac{x^2}{4}-\frac{3x}{2}+2=0\Rightarrow x^2-6x+8=0$.

Understanding:

Given $a+b+c=0$, evaluate $\frac{a^2}{bc}+\frac{b^2}{ca}+\frac{c^2}{ab}$.

Formula:

Combine fractions over a common denominator $abc$:

Step 1: Use the identity when $a+b+c=0$.

There is a known algebraic identity:

Step 2: Substitute.

Answer:

The value of the expression is $3$.

Quick Tip:

The identity $a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$ immediately gives $a^3+b^3+c^3=3abc$ when $a+b+c=0$. Memorise it — it appears very frequently.

Understanding:

The system is:

For infinitely many solutions, the two equations must be proportional.

Formula:

For $a_{1}x+b_{1}y=c_1$ and $a_{2}x+b_{2}y=c_2$, infinitely many solutions require:

Step 1: Set up the proportionality conditions.

Step 2: Solve $\frac{m}{12}=\frac{3}{m}$.

Step 3: Check with the third ratio using $m=6$.

All three ratios are equal. ✓

Step 4: Check $m=-6$.

The ratios are not equal, so $m=-6$ does not give infinitely many solutions. ✗

Answer:

The value of $m$ for infinitely many solutions is $m=6$.

Quick Tip:

Always check ALL three ratios. Getting two ratios equal is necessary but not sufficient — the constant ratio must also match.

Understanding:

We find the sum $S_n=\sum _{r=1}^n\frac{1}{1+2+3+\cdots +r}$.

Formula:

The sum of the first $r$ natural numbers is:

Step 1: Express the general term.

Step 2: Apply partial fractions.

Step 3: Sum using telescoping.

Verification with $n=1$:

$S_1=1$. Formula: $\frac{2(1)}{2}=1$. ✓

With $n=2$: $S_2=1+\frac{1}{3}=\frac{4}{3}$. Formula: $\frac{4}{3}$. ✓

Answer:

The sum of the series up to $n$ terms is $\frac{2n}{n+1}$.

Quick Tip:

Whenever the denominator of a general term is a product of two consecutive integers, partial fractions with a telescoping sum is the fastest approach.