JEE Main 2012 19 May Online
Exam JEE Main
Year 2012
Date 19 May
Mode Online
Subject Mathematics
Chapter Limits JEE Main 2012 · 19 May · Online · Mathematics · Limits
If f(x)=3 x^ 10 -7 x^8+5 x^6-21 x^3+3 x^2-7 , then _ α arrow 0...
If
f ( x ) = 3 x 10 − 7 x 8 + 5 x 6 − 21 x 3 + 3 x 2 − 7 f(x)=3 x^{10}-7 x^8+5 x^6-21 x^3+3 x^2-7 f ( x ) = 3 x 10 − 7 x 8 + 5 x 6 − 21 x 3 + 3 x 2 − 7 , then
lim α → 0 f ( 1 − α ) − f ( 1 ) α 3 + 3 α \lim _{\alpha \rightarrow 0} \frac{f(1-\alpha)-f(1)}{\alpha^3+3 \alpha} lim α → 0 α 3 + 3 α f ( 1 − α ) − f ( 1 ) is
Options A − 53 3 -\frac{53}{3} − 3 53 B 53 3 \frac{53}{3} 3 53 C − 55 3 -\frac{55}{3} − 3 55 D 55 3 \frac{55}{3} 3 55 Correct answer B. 53 3 \frac{53}{3} 3 53
Step-by-step solution
Let f ′ ( x ) = 3 x 10 − 7 x 8 + 5 x 6 − 21 x 3 + 3 x 2 − 7 f ′ ( x ) = 30 x 9 − 56 x 7 + 30 x 5 − 63 x 2 + 6 x f ′ ( 1 ) = 30 − 56 + 30 − 63 + 6 = 66 − 63 − 56 = − 53 \begin{aligned}\text{ Let }f^{\prime}(x) & =3 x^{10}-7 x⁸+5 x⁶-21 x³+3 x²-7 \\
f^{\prime}(x) & =30 x⁹-56 x⁷+30 x⁵-63 x²+6 x \\
f^{\prime}(1) & =30-56+30-63+6 \\
& =66-63-56=-53
\end{aligned} Let f ′ ( x ) f ′ ( x ) f ′ ( 1 ) = 3 x 10 − 7 x 8 + 5 x 6 − 21 x 3 + 3 x 2 − 7 = 30 x 9 − 56 x 7 + 30 x 5 − 63 x 2 + 6 x = 30 − 56 + 30 − 63 + 6 = 66 − 63 − 56 = − 53 Consider lim α → 0 f ( 1 − α ) − f ( 1 ) α 3 + 3 α = lim α → 0 f ′ ( 1 − α ) ( − 1 ) − 0 3 α 2 + 3 (By using \begin{aligned}& \text{ Consider } \lim _{\alpha \rightarrow 0} \frac{f(1-\alpha)-f(1)}{\alpha^3+3 \alpha}\\
& =\lim _{\alpha \rightarrow 0} \frac{f^{\prime}(1-\alpha)(-1)-0}{3 \alpha^2+3} \quad \text{ (By using }
\end{aligned} Consider α → 0 lim α 3 + 3 α f ( 1 − α ) − f ( 1 ) = α → 0 lim 3 α 2 + 3 f ′ ( 1 − α ) ( − 1 ) − 0 (By using L'hospital rule)
=
f ′ ( 1 − 0 ) ( − 1 ) 3 ( 0 ) 2 + 3 = − f ′ ( 1 ) 3 = 53 3 \frac{f^{\prime}(1-0)(-1)}{3(0)^2+3}=\frac{-f^{\prime}(1)}{3}=\frac{53}{3} 3 ( 0 ) 2 + 3 f ′ ( 1 − 0 ) ( − 1 ) = 3 − f ′ ( 1 ) = 3 53
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