37. The interval in which y = x2 e–x is increasing is (A) (– ∞, ∞) (B) (– 2, 0) (C) (2, ∞) (D) (0, 2)

We have, f (x) = x2 e–xSo, f (x) = x2ddxe−x+e−xdx2dx=x2e−xddx (−x)+e−x2⋅xdxdx= -x2 e-x + e-x 2x.= x e-x ( x + 2).=x (2−x)eexIf f (x) = 0.⇒x (2−x)ex=0.⇒ x = 0, x = 2.Hence, we get there disjoint interval [−∞, 0), (0, 2) (2, ∞)When, x (−∞, 0), we have, f (x) = ( -ve) ( + ve) = ( -ve) < 0.So, f is strictly decreasing.When x ∈ (0,2), f (x) = ( + ve) ( + ve) = ( + ve) > 0.So, f is strictly increasing.And when x ∈ (2, ∞), f (x) = ( +ve) ( -ve) = ( -ve) < 0.So, f is strictly decreasing.Hence, option (D) is correct.

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Ncert Solutions Maths class 12th Maths Class 12th Application of Derivatives

37. The interval in which y = x² e^–x is increasing is (A) (– ∞, ∞) (B) (– 2, 0) (C) (2, ∞) (D) (0, 2)

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Vishal Baghel | Contributor-Level 10

4 months ago

We have, f (x) = x² e^–x

So, f (x) = $x^{2} \frac{d}{d x} e^{- x} + e^{- x} \frac{d x^{2}}{d x}$

$= x^{2} e^{- x} \frac{d}{d x} (- x) + e^{- x} 2 \cdot x \frac{d x}{d x}$

= -x² e^-x + e^-x 2x.

= x e^-x ( x + 2).

$= \frac{x (2 - x)}{e^{e x}}$

If f (x) = 0.

$\Rightarrow \frac{x (2 - x)}{e^{x}} = 0 .$

$\Rightarrow$ x = 0, x = 2.

Hence, we get there disjoint interval

$[- \infty, 0), (0, 2) (2, \infty)$

When, x $(- \infty, 0),$ we have, f (x) = ( -ve) ( + ve) = ( -ve) < 0.

So, f is strictly decreasing.

When x ∈ (0,2), f (x) = ( + ve) ( + ve) = ( + ve) > 0.

So, f is strictly increasing.

And when x ∈ $(2, \infty),$ f (x) = ( +ve) ( -ve) = ( -ve) < 0.

So, f is strictly decreasing.

Hence, option (D) is correct.

We have, f (x) = x2 e–xSo, f (x) = <math><mrow><msup><mrow><mi>x</mi></mrow><mrow><mn>2</mn></mrow></msup><mfrac><mrow><mi>d</mi></mrow><mrow><mi>d</mi><mi>x</mi></mrow></mfrac><msup><mrow><mi>e</mi></mrow><mrow><mo>−</mo><mi>x</mi></mrow></msup><mo>+</mo><msup><mrow><mi>e</mi></mrow><mrow><mo>−</mo><mi>x</mi></mrow></msup><mfrac><mrow><mi>d</mi><msup><mrow><mi>x</mi></mrow><mrow><mn>2</mn></mrow></msup></mrow><mrow><mi>d</mi><mi>x</mi></mrow></mfrac></mrow></math><math><mrow><mo>=</mo><msup><mrow><mi>x</mi></mrow><mrow><mn>2</mn></mrow></msup><msup><mrow><mi>e</mi></mrow><mrow><mo>−</mo><mi>x</mi></mrow></msup><mfrac><mrow><mi>d</mi></mrow><mrow><mi>d</mi><mi>x</mi></mrow></mfrac><mo stretchy="false"> (</mo><mo>−</mo><mi>x</mi><mo stretchy="false">)</mo><mo>+</mo><msup><mrow><mi>e</mi></mrow><mrow><mo>−</mo><mi>x</mi></mrow></msup><mn>2</mn><mo>⋅</mo><mi>x</mi><menclose notation="updiagonalstrike"><mrow><mfrac><mrow><mi>d</mi><mi>x</mi></mrow><mrow><mi>d</mi><mi>x</mi></mrow></mfrac></mrow></menclose></mrow></math>= -x2 e-x + e-x 2x.= x e-x ( x + 2).<math><mrow><mo>=</mo><mfrac><mrow><mi>x</mi><mo stretchy="false"> (</mo><mn>2</mn><mo>−</mo><mi>x</mi><mo stretchy="false">)</mo></mrow><mrow><msup><mrow><mi>e</mi></mrow><mrow><mi>e</mi><mi>x</mi></mrow></msup></mrow></mfrac></mrow></math>If f (x) = 0.<math><mrow><mo>⇒</mo><mfrac><mrow><mi>x</mi><mo stretchy="false"> (</mo><mn>2</mn><mo>−</mo><mi>x</mi><mo stretchy="false">)</mo></mrow><mrow><msup><mrow><mi>e</mi></mrow><mrow><mi>x</mi></mrow></msup></mrow></mfrac><mo>=</mo><mn>0</mn><mo>.</mo></mrow></math><math><mrow><mo>⇒</mo></mrow></math> x = 0, x = 2.Hence, we get there disjoint interval<math><mrow><mo stretchy="false"> [</mo><mo>−</mo><mo>∞</mo><mo>, </mo><mn>0</mn><mo stretchy="false">)</mo><mo>, </mo><mo stretchy="false"> (</mo><mn>0</mn><mo>, </mo><mn>2</mn><mo stretchy="false">)</mo><mo></mo><mo stretchy="false"> (</mo><mn>2</mn><mo>, </mo><mo>∞</mo><mo stretchy="false">)</mo></mrow></math>When, x <math><mrow><mo stretchy="false"> (</mo><mo>−</mo><mo>∞</mo><mo>, </mo><mn>0</mn><mo stretchy="false">)</mo><mo>, </mo></mrow></math> we have, f (x) = ( -ve) ( + ve) = ( -ve) < 0.So, f is strictly decreasing.When x ∈ (0,2), f (x) = ( + ve) ( + ve) = ( + ve) > 0.So, f is strictly increasing.And when x ∈<math><mrow><mo stretchy="false"> (</mo><mn>2</mn><mo>, </mo><mo>∞</mo><mo stretchy="false">)</mo><mo>, </mo></mrow></math> f (x) = ( +ve) ( -ve) = ( -ve) < 0.So, f is strictly decreasing.Hence, option (D) is correct.

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y (x) = ∫? (2t² - 15t + 10)dt
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37. The interval in which y = x² e^–x is increasing is (A) (– ∞, ∞) (B) (– 2, 0) (C) (2, ∞) (D) (0, 2)

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