Let the function f(x) = 2x2 – loge x, x > 0, be decreasing in (0, a) and increasing in (a, 4). A tangent to the parabola y2 = 4ax at a point P on it passes through the point (8a, 8a – 1) but does not pass through the point <math> <mrow> <mrow> <mo>(</mo> <mrow> <mo>−</mo> <mfrac> <mrow> <mn>1</mn> </mrow> <mrow> <mi>a</mi> </mrow> </mfrac> <mo>,</mo> <mtext> </mtext> <mtext> </mtext> <mn>0</mn> </mrow> <mo>)</mo> </mrow> <mo>.</mo> </mrow> </math> If the equation of the normal at P is <math> <mrow> <mfrac> <mrow> <mi>x</mi> </mrow> <mrow> <mi>α</mi> </mrow> </mfrac> <mo>+</mo> <mfrac> <mrow> <mi>γ</mi> </mrow> <mrow> <mi>β</mi> </mrow> </mfrac> <mo>=</mo> <mn>1</mn> <mo>,</mo> </mrow> </math> then a + b is equal to…………….

f ' ( x ) = 4 x 2 − 1 x so f (x) is decreasing in ( 0 , 1 2 ) a n d ( 1 2 , ∞ ) ⇒ a = 1 2 Tangent at y2 = 2x is y = mx + 1 2 m it is passing through (4, 3) therefore we get m = 1 2 o r 1 4 So tangent may be y = 1 2 x + 1 o r y = 1 4 x + 2 b u t y = 1 2 x + 1 passes through (-2, 0) so rejected.Equation of normal x 9 + y 3 6 = 1

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Maths NCERT Exemplar Solutions Class 11th Chapter Twelve Maths Class 12th Maths Applications of Derivatives

Let the function f(x) = 2x² – log_e x, x > 0, be decreasing in (0, a) and increasing in (a, 4). A tangent to the parabola y² = 4ax at a point P on it passes through the point (8a, 8a – 1) but does not pass through the point $(- \frac{1}{a}, 0) .$ If the equation of the normal at P is $\frac{x}{α} + \frac{γ}{β} = 1,$ then a + b is equal to…………….

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Raj Pandey | Contributor-Level 9

2 weeks ago

$f' (x) = \frac{4 x^{2} - 1}{x}$ so f (x) is decreasing in $(0, \frac{1}{2}) a n d (\frac{1}{2}, \infty)$ $\Rightarrow a = \frac{1}{2}$

Tangent at y² = 2x is y = mx + $\frac{1}{2 m}$ it is passing through (4, 3) therefore we get m = $\frac{1}{2} o r \frac{1}{4}$

So tangent may be $y = \frac{1}{2} x + 1 o r y = \frac{1}{4} x + 2 b u t y = \frac{1}{2} x + 1$ passes through (-2, 0) so rejected.

Equation of normal $\frac{x}{9} + \frac{y}{36} = 1$

<math> <mrow> <mi>f</mi> <mo>'</mo> <mrow> <mo> (</mo> <mrow> <mi>x</mi> </mrow> <mo>)</mo> </mrow> <mo>=</mo> <mfrac> <mrow> <mn>4</mn> <msup> <mrow> <mi>x</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msup> <mo>−</mo> <mn>1</mn> </mrow> <mrow> <mi>x</mi> </mrow> </mfrac> </mrow> </math> so f (x) is decreasing in <math> <mrow> <mrow> <mo> (</mo> <mrow> <mn>0</mn> <mo>, </mo> <mfrac> <mrow> <mn>1</mn> </mrow> <mrow> <mn>2</mn> </mrow> </mfrac> </mrow> <mo>)</mo> </mrow> <mi>a</mi> <mi>n</mi> <mi>d</mi> <mtext> </mtext> <mtext> </mtext> <mrow> <mo> (</mo> <mrow> <mfrac> <mrow> <mn>1</mn> </mrow> <mrow> <mn>2</mn> </mrow> </mfrac> <mo>, </mo> <mo>∞</mo> </mrow> <mo>)</mo> </mrow> </mrow> </math> <math> <mrow> <mo>⇒</mo> <mi>a</mi> <mo>=</mo> <mfrac> <mrow> <mn>1</mn> </mrow> <mrow> <mn>2</mn> </mrow> </mfrac> </mrow> </math>  Tangent at y2 = 2x is y = mx + <math> <mrow> <mfrac> <mrow> <mn>1</mn> </mrow> <mrow> <mn>2</mn> <mi>m</mi> </mrow> </mfrac> </mrow> </math> it is passing through (4, 3) therefore we get m = <math> <mrow> <mfrac> <mrow> <mn>1</mn> </mrow> <mrow> <mn>2</mn> </mrow> </mfrac> <mi>o</mi> <mi>r</mi> <mtext> </mtext> <mtext> </mtext> <mfrac> <mrow> <mn>1</mn> </mrow> <mrow> <mn>4</mn> </mrow> </mfrac> </mrow> </math>  So tangent may be  <math> <mrow> <mi>y</mi> <mo>=</mo> <mfrac> <mrow> <mn>1</mn> </mrow> <mrow> <mn>2</mn> </mrow> </mfrac> <mi>x</mi> <mo>+</mo> <mn>1</mn> <mtext> </mtext> <mtext> </mtext> <mi>o</mi> <mi>r</mi> <mtext> </mtext> <mtext> </mtext> <mi>y</mi> <mo>=</mo> <mfrac> <mrow> <mn>1</mn> </mrow> <mrow> <mn>4</mn> </mrow> </mfrac> <mi>x</mi> <mo>+</mo> <mn>2</mn> <mtext> </mtext> <mtext> </mtext> <mtext> </mtext> <mi>b</mi> <mi>u</mi> <mi>t</mi> <mtext> </mtext> <mtext> </mtext> <mi>y</mi> <mo>=</mo> <mfrac> <mrow> <mn>1</mn> </mrow> <mrow> <mn>2</mn> </mrow> </mfrac> <mi>x</mi> <mo>+</mo> <mn>1</mn> </mrow> </math>  passes through (-2, 0) so rejected.Equation of normal  <math> <mrow> <mfrac> <mrow> <mi>x</mi> </mrow> <mrow> <mn>9</mn> </mrow> </mfrac> <mo>+</mo> <mfrac> <mrow> <mi>y</mi> </mrow> <mrow> <mn>3</mn> <mn>6</mn> </mrow> </mfrac> <mo>=</mo> <mn>1</mn> </mrow> </math>

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