Question
If the derivative of the function f(x)=bx2+ax+4;x≥−1ax2+b;x<−1′, is everywhere continuous, then
Answer: Option A
:
A
Wehave,f(x)={ax2+b,x<−1bx2+ax+4,x≥−1∴f′(x)={2ax,<−12bx+a,x≥−1Since,f(x)isdifferentiableatx=−1,thereforeitiscontinuousatx=−1andhence,limx→−1−f(x)=limx→−1+f(x)⇒a+b=b−a+4⇒a=2andalso,limx→−1−f(x)=limx→−1+f(x)⇒−2a=−2b+a⇒3a=2b⇒b=3(∵a=2)Hence,a=2,b=3
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:
A
Wehave,f(x)={ax2+b,x<−1bx2+ax+4,x≥−1∴f′(x)={2ax,<−12bx+a,x≥−1Since,f(x)isdifferentiableatx=−1,thereforeitiscontinuousatx=−1andhence,limx→−1−f(x)=limx→−1+f(x)⇒a+b=b−a+4⇒a=2andalso,limx→−1−f(x)=limx→−1+f(x)⇒−2a=−2b+a⇒3a=2b⇒b=3(∵a=2)Hence,a=2,b=3
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