1.Let,f(x)={ex−esinxax3;x<0b;x=0xln(1+4x);x>0\large{f(x) = \begin{cases} \LARGE{\frac{e^x - e^{\sin x}}{ax^3}} & ; x < 0 \\ b & ; x = 0 \\ \LARGE{\frac{x}{\ln(1 + 4x)}} & ; x > 0 \end{cases}}f(x)=⎩⎨⎧ax3ex−esinxbln(1+4x)x;x<0;x=0;x>0If fff is continuous at x=0x = 0x=0, then (3a+4b)(3a + 4b)(3a+4b) is equal to:a.−3-3−3b.000c.333d.444Login to continueOnly logged in users canattempt or see the solution.
