Assertion - f x - -x2x3log t dt attains a minima at t- 1 Reason- f- x 0 for x - 1- -

The correct option is C Assertion is true but Reason is false f(x)=∫ _ x ^ 2 ^ x ^ 3 log t #160; dt f'(x)=3 x ^ 2 log x ^ 3 2xlog x ^ 2 f' ...

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Byju's Answer

Standard XII

Mathematics

Logarithmic Function

Assertion : ...

Question

Assertion : $f (x) = \int_{x^{2}}^{x^{3}} log t d t$ attains a minima at $t = 1$ Reason: $f^{'} (x) < 0, x ϵ (0, 1)$ and $f^{'} (x) > 0$ for $x ϵ (1, \infty)$ .

Both Assertion & Reason are individually true & Reason is correct explanation of Assertion

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Both Assertion & Reason are individually true but Reason is not the ,correct (proper) explanation of Assertion

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Assertion is true but Reason is false

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Assertion is false but Reason is true

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Solution

The correct option is C Assertion is true but Reason is false
$f (x) = \int_{x^{2}}^{x^{3}} log t d t f^{'} (x) = 3 x^{2} log x^{3} - 2 x log x^{2} f^{'} (x) = 9 x^{2} log x - 4 x log x 9 x^{2} log x - 4 x log x = 0 log x (9 x^{2} - 4 x) = 0 \Rightarrow log x = 0 \Rightarrow x = 1 f^{''} (x) = 10 x log x + 9 x - 4 log x - 4 f^{''} (1) = 5 f^{''} (1) > 0$
So it has minimum at $x = 1$
Assertion is correct . Reason is false.

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Assertion :f(x)=∫x3x2logtdt attains a minima at t=1 Reason: f′(x)<0,xϵ(0,1) and f′(x)>0 for xϵ(1,∞).

The correct option is C Assertion is true but Reason is falsef(x)=∫x3x2logtdtf′(x)=3x2logx3−2xlogx2f′(x)=9x2logx−4xlogx9x2logx−4xlogx=0logx(9x2−4x)=0⇒logx=0⇒x=1f′′(x)=10xlogx+9x−4logx−4f′′(1)=5f′′(1)>0So it has minimum at x=1Assertion is correct . Reason is false.

Assertion : $f (x) = \int_{x^{2}}^{x^{3}} log t d t$ attains a minima at $t = 1$ Reason: $f^{'} (x) < 0, x ϵ (0, 1)$ and $f^{'} (x) > 0$ for $x ϵ (1, \infty)$ .