y(x)=∫0x(2t2−15t+10)dt
y′(x)]x=a=[2x2−15x+10]a=2a2−15a+10
Slope of normal =−31
⇒2a2−15a+10=3⇒a=7
and a=21 (rejected)
b=y(7)=∫07(2t2−15t+10)dt
=[32t3−215t2+10t]07
⇒6b=4×73−45×49+60×7
∣a+6b∣=406
If the normal to the curve y(x)=∫0x(2t2−15t+10)dt at a point (a,b) is parallel to the line x+3y=−5,a>1, then the value of ∣a+6b∣ is equal to ________.
Held on 16 Mar 2021 · Verified 6 Jul 2026.
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