The equation of the tangent to the curvey=∫xx2 logtdt at x=2 is
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a
y–6 log 2=7log2(x–2)
b
y−log2e1/3=(log2)x
c
y−8log2e−1/3=5log2x
d
y+8 log 2+2=(7log 2)x
answer is D.
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Detailed Solution
y=∫xx2 logtdt=(tlogt−t)xx2=2x2−xlogx−x2+x⇒y(2)=6log2−2Also, y′(2)=7log2Thus, equation of tangent at x = 2 isy−(6log2−2)=(7log2)(x−2) or y+8log2+2=(7log2)(x)