JEE Main 2022 · 30 Jul · Shift 2 · Mathematics Q69

MathematicsSingle correct+4 / -1Question 69 of 90

Let

$f(x)=\begin{cases}\dfrac{\sin(ax+3x)+\sin x}{x}, & x<0\\ 2, & x=0\\ \dfrac{(x+6bx^2)^{1/3}-x^{1/3}}{3x^{4/3}}, & x>0\end{cases}$

be continuous at x = 0. If $(1-k)\,f\left(\dfrac7{18}\right)=ab$ , then k is equal to

−6

−13

Answer and worked solution

Official answer: D

Continuity. Left: $a+3+1=2$ , so $a=-2$ . Right: $\dfrac{x^{1/3}\left[(1+6bx)^{1/3}-1\right]}{3x^{4/3}}\to\dfrac{2b}3=2$ , so $b=3$ and $ab=-6$ .

At $\frac7{18}$ : $x+18x^2=\frac{28}9=8x$ , so the numerator is $2x^{1/3}-x^{1/3}=x^{1/3}$ and $f=\dfrac1{3x}=\dfrac67$ .

So $(1-k)\frac67=-6$ , $k=8$ .

(D)

Topics:Continuity Limits

JEE Main 2022 · 30 Jul · Shift 2, Mathematics question 69