If $$ f(x) = \begin{cases} \dfrac{1}{|x|}, & |x| > 2 \\ A + Bx^2, & |x| \leq 2 \end{cases} $$ Then $f(x)$ is differentiable at $x = -2$ for
(a) A = $\tfrac{3}{4}$ ,B = $-\tfrac{1}{16}$
(b) A = $-\tfrac{3}{4}$ , B = $\tfrac{1}{16}$
(c) A = $-\tfrac{3}{4}$ ,B = $-\tfrac{1}{16}$
(d) A = $\tfrac{3}{4}$, B = $\tfrac{1}{16}$
$G = U(98) = \{k \in \mathbb{N} : k \leq 98,\ \gcd(k,98)=1 \}$ be a group, where $\mathbb{N}$ is the set of all natural numbers.
The number of generators of the largest cyclic subgroup of $G$ is
(a) 12
(b) 32
(c) 42
(d) 49
Given series: $6,\,20,\,42,\,72,\,110,\,\_\_$
First differences: $14,\,22,\,30,\,38$
Second differences: $8,\,8,\,8$ (constant)
Next first difference $= 38 + 8 = 46$
Next term $= 110 + 46 = 156$
Answer: $\boxed{156}$ ✅
Let father's age be $F$ and child's age be $C$.
Given $C = \dfrac{F}{2}$
Twenty years ago: $F - 20 = 10(C - 20)$
Substitute $C = \dfrac{F}{2}$:
$F - 20 = 10\left(\dfrac{F}{2} - 20\right)$
$F - 20 = 5F - 200$
$180 = 4F \Rightarrow F = 45$
Answer: $\boxed{45\text{ years}}$ ✅ (Option d)
$\dfrac{v_A}{v_B}=\dfrac{1000}{950}=\dfrac{20}{19}$, $\dfrac{v_B}{v_C}=\dfrac{500}{450}=\dfrac{10}{9}$
$\Rightarrow \dfrac{v_A}{v_C}=\dfrac{200}{171}$
When A runs $400$ m, C runs $400\times\dfrac{171}{200}=342$ m.
Difference $=400-342=58$ m.
Answer: 58 meters ✅
Rates: $A=\frac{1}{15},\; B=\frac{1}{10},\; C=\frac{1}{12}$ (work/day)
Work done by $B+C$ in 4 days: $4\!\left(\frac{1}{10}+\frac{1}{12}\right)=4\cdot\frac{11}{60}=\frac{11}{15}$
Remaining work: $1-\frac{11}{15}=\frac{4}{15}$
Time for $A$: $\dfrac{\frac{4}{15}}{\frac{1}{15}}=4$ days
Answer: $\boxed{4\text{ days}}$ ✅
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