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New answer posted

5 months ago

Maths Application of Derivatives

69. Find the approximate change in the surface area of a cube of side x metres caused by decreasing the side by 1%

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Vishal Baghel

Contributor-Level 10

We know that, the surface area 5 of a 'x' when length cube, is S = 6x².

So, $d S = \frac{d S}{d x} Δ x = 12 x \cdot Δ x .$

Given decrease in side, $Δ x = - 1 % x = - \frac{x}{100} .$

$∴ d S = 12 x (- \frac{x}{100}) = - 0.12 x^{2} m^{2} .$

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New answer posted

5 months ago

Maths Application of Derivatives

68. Find the approximate change in the volume V of a cube of side x metres caused by increasing the side by 1%.

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Vishal Baghel

Contributor-Level 10

We know that, the volume v of side 'a' mete of cube is v = x³.

So, $d = (\frac{d}{d x}) \cdot Δ x = 3 x^{2} \cdot Δ x .$

Given that, increase in side = 1% of x.

$\Rightarrow Δ x = \frac{x}{100}$

$∴ d v = 3 x^{2} (\frac{x}{100}) = 0.03 x^{3} m^{3} .$

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New answer posted

5 months ago

Maths Application of Derivatives

67. Find the approximate value of f (5.001), where f(x) = x³ – 7x2 + 15.

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Vishal Baghel

Contributor-Level 10

Given, y = f (x) = x³- 7x² + 15.

So, $\frac{d y}{d x} = f^{'} (x) = 3 x^{2} - 14 x .$

$\Rightarrow$ dy = (3x²- 14x) dx.

$\Rightarrow$ Δy = (3x²- 14x) Δx.

Let, x = 5 and Δx = 0.001. Then,

Δy = f (x + Δx) f (x).

$\Rightarrow$ f (x + Δx) = f (x) + Δy = f (x) + (3x²- 4x) Δx.

$\Rightarrow$ f (5 + 0.001) = 5³- 7 (5)² + 15 + [3 (5)² - 14 (5)]. (0.001).

$\Rightarrow$ f (5.001) = 125 - 175 + 15 + (75 - 70) (0.001)

= -35 + 0.005 = - 34.995.

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New answer posted

5 months ago

Maths Application of Derivatives

66. Find the approximate value of f(2.01), where f (x) = 4x² + 5x + 2.

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Vishal Baghel

Contributor-Level 10

Given, y = f (x) = 4x² + 5x + 2.

So, f (x) = 8x + 5. $\frac{d y}{d x}$ = 8x + 5 dy = (8x + 5) dx.

Let x = 2 and Δx = 0.01.Then,

f (x + Δx) = f (2 + 0.01) = f (2.01).

Δy = f. (x + Δx) f (Δx).

f (x +Δx) = f (x) +Δy.

= f (x) + dy = f (x) + (8x + 5) dx.

= f (2.01) = f (2) + (8 x 2 + 5). Δx {∴dx = Δx}

= 4 (2)² + 5 (2) + 2 + 21 (0.01)

= 16 + 10 + 2 + 0.21 = 28.21.

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New answer posted

5 months ago

Maths Application of Derivatives

65. Using differentials, find the approximate value of each of the following up to 3 places of decimal.

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Vishal Baghel

Contributor-Level 10

(i) Let y= ?x : Let x = 25 and x = 0. 3.

Then, ?y = ?x+?x $?x$

$=?25.3 ?25$

$=?25.3 ? 5$

$?25.3 = 5 + ? y ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? y ~ d y .$

= 5 + dy

= $5 + (\frac{d y}{d x}) A x .$

$= 5 + \frac{1}{2?x} ? x .$

$= 5 + \frac{1}{2?25} ? 0.3$

$= 5 + \frac{0.3}{2 * 5}$

= 5 + 0.03

$?25.3 = 5.030$

(ii) ?49.5

A.(ii)

Let y = ?x $.$ Let x = 49 and x = 0.5.

Then, $? y =?x+?x ?x .$

$=?49.5 ?49$

$?49.5 = 7 + ? y = 7 + {(\frac{d y}{d x})}_{}^{? x}$

$= 7 + \frac{1}{2?x} ? x$

$= 7 + \frac{1}{2?49} * (0.5)$

$= 7 + \frac{0.5}{14}$

= 7 + 0.0357.

$?49.5 = 7 ? 035$

(iii) ?0.6

A.(iii)

Let y = ?x $.$ Let x = 0.64 and ?x = 0.04.

Then, $? y =?x+Ax ?x$

$=?0.64-0.04 ?0.64 .$

$=?0.6 ? 0.8$

$?0.6= 0.8 + ? y = 0.8 + (\frac{d y}{d x}) ? x$

$= 0.8 + \frac{1}{2?x} ? x$

$= 0.8 + \frac{1}{2?0.64} * (? ? 0.04)$

$= 0.8 \frac{? 0.04}{2 * 0.8}$

= 0.8 - 0.025.

= 0.775.

(iv) ${(0.009)}^{\frac{1}{3}}$

A.(iv)

Let $y = x^{\frac{1}{3}}$ Let x = 0.008 and ?x = 0.00 1.

Then, ?y = ${[x + ? x]}^{\frac{1}{3}} ? {[x]}^{\frac{1}{3}}$

$= {(0.009)}^{\frac{1}{3}} ? {(0.008)}^{\frac{1}{3}}$

${(0.009)}^{\frac{1}{3}} = 0.2 + ? y = 0.2 + \frac{d y}{d x} ? ? x$

$= 0.2 + \frac{1}{3 {(x^{\frac{1}{3}})}^{2}} ? x$

$= 0.2 + \frac{1}{3 {[{(0.008)}^{\frac{1}{3}}]}^{2}} * (0.001)$

$= 0.2 + \frac{0.001}{3 * {(0.0)}^{2}} = 0.2 + \frac{0.001}{0.12}$

= 0.2 + 0.0083.

= 0.208.

(v) ${(0.999)}^{\frac{1}{10}}$

A.(v)

Let $y = x^{\frac{1}{10}} .$ Let x = 1 and ?x = -0.001

Then, $? y = {(x + ? x)}^{\frac{1}{10}} ? x^{\frac{1}{10}}$

$= {(0.999)}^{\frac{1}{10}} ? {(1)}^{\frac{1}{10}}$

$? 1 {(0.999)}^{\frac{1}{10}} = 1 + A y$

$= 1 + \frac{d y}{d x} ? x .$

$= 1 + \frac{1}{10 x^{\frac{9}{10}}} * ? x$

$= 1 ? \frac{0.001}{10 (. 1^{\frac{9}{10}})} = 1 ? \frac{0.001}{10} = 1 ? 0.0001$

= 0.999.

(vi) ${(15)}^{\frac{1}{4}}$

A.(vi)

Let $y = x^{\frac{1}{4}} .$ Then, x = 16 and ?x = 1.

Then, $? y = {(x + A x)}^{\frac{1}{4}} ? x^{\frac{1}{4}} .$

$= {(15)}^{\frac{1}{4}} ? {(16)}^{\frac{1}{4}}$

$? {(15)}^{\frac{1}{4}} = 2 + A y = 2 + \frac{d y}{d x} A x$

$= 2 + \frac{1}{4 x^{\frac{3}{4}}} ? (? 1)$

$= 4 ? \frac{1}{4 {(16)}^{\frac{3}{4}}}$

$= 2 ? \frac{1}{4 * 8}$

$= 4 ? \frac{1}{32}$

$= \frac{64 ? 1}{32} = \frac{63}{32} = 1.968$

(vii) ${(26)}^{\frac{1}{3}}$

A.(vii)

Let $y = x^{\frac{1}{3}} .$ Let x = 27 and ?x = 1.

Then, $? y = {(x + ? x)}^{\frac{1}{3}} ? x^{\frac{1}{3}}$

$= {(26)}^{\frac{1}{3}} ? {(27)}^{\frac{1}{3}}$

$? {(26)}^{\frac{1}{3}} = 3 + 4 y = 3 + \frac{d y}{d x} A x = 3 + \frac{1}{3 x^{\frac{2}{3}}} A x,$

$= 3 ? \frac{1}{3 {(2 F)}^{\frac{2}{3}}}$

$= 3 ? \frac{1}{27}$

$= \frac{81 ? 1}{27} = \frac{80}{27} = 2.962 .$

(viii) ${(255)}^{\frac{1}{4}}$

A.(viii)

Let $y = x^{\frac{1}{4}} .$ Let x = 256 and ?x = 1.

Then, $? y = {(x + ? x)}^{\frac{1}{4}} ? x^{\frac{1}{4}} .$

$= {(255)}^{\frac{1}{4}} ? {(256)}^{\frac{1}{4}}$

$? {(255)}^{\frac{1}{4}} = {(256)}^{\frac{1}{4}} + 4 y = 4 + \frac{d y}{d x} A x .$

$= 4 + \frac{1}{4 x^{\frac{3}{4}}} ? 4 x$

$= 4 \frac{? 1}{4 {(256)}^{\frac{3}{4}}}$

$= 4 ? \frac{1}{256}$

$= \frac{1024 ? 1}{256}$

$= \frac{1023}{256} = 3.996$

(ix) ${(82)}^{\frac{1}{4}}$

A.(ix)

...more

(i) Let y= ?x : Let x = 25 and x = 0. 3.

Then, ?y = ?x+?x $?x$

$=?25.3 ?25$

$=?25.3 ? 5$

$?25.3 = 5 + ? y ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? y ~ d y .$

= 5 + dy

= $5 + (\frac{d y}{d x}) A x .$

$= 5 + \frac{1}{2?x} ? x .$

$= 5 + \frac{1}{2?25} ? 0.3$

$= 5 + \frac{0.3}{2 * 5}$

= 5 + 0.03

$?25.3 = 5.030$

(ii) ?49.5

A.(ii)

Let y = ?x $.$ Let x = 49 and x = 0.5.

Then, $? y =?x+?x ?x .$

$=?49.5 ?49$

$?49.5 = 7 + ? y = 7 + {(\frac{d y}{d x})}_{}^{? x}$

$= 7 + \frac{1}{2?x} ? x$

$= 7 + \frac{1}{2?49} * (0.5)$

$= 7 + \frac{0.5}{14}$

= 7 + 0.0357.

$?49.5 = 7 ? 035$

(iii) ?0.6

A.(iii)

Let y = ?x $.$ Let x = 0.64 and ?x = 0.04.

Then, $? y =?x+Ax ?x$

$=?0.64-0.04 ?0.64 .$

$=?0.6 ? 0.8$

$?0.6= 0.8 + ? y = 0.8 + (\frac{d y}{d x}) ? x$

$= 0.8 + \frac{1}{2?x} ? x$

$= 0.8 + \frac{1}{2?0.64} * (? ? 0.04)$

$= 0.8 \frac{? 0.04}{2 * 0.8}$

= 0.8 - 0.025.

= 0.775.

(iv) ${(0.009)}^{\frac{1}{3}}$

A.(iv)

Let $y = x^{\frac{1}{3}}$ Let x = 0.008 and ?x = 0.00 1.

Then, ?y = ${[x + ? x]}^{\frac{1}{3}} ? {[x]}^{\frac{1}{3}}$

$= {(0.009)}^{\frac{1}{3}} ? {(0.008)}^{\frac{1}{3}}$

${(0.009)}^{\frac{1}{3}} = 0.2 + ? y = 0.2 + \frac{d y}{d x} ? ? x$

$= 0.2 + \frac{1}{3 {(x^{\frac{1}{3}})}^{2}} ? x$

$= 0.2 + \frac{1}{3 {[{(0.008)}^{\frac{1}{3}}]}^{2}} * (0.001)$

$= 0.2 + \frac{0.001}{3 * {(0.0)}^{2}} = 0.2 + \frac{0.001}{0.12}$

= 0.2 + 0.0083.

= 0.208.

(v) ${(0.999)}^{\frac{1}{10}}$

A.(v)

Let $y = x^{\frac{1}{10}} .$ Let x = 1 and ?x = -0.001

Then, $? y = {(x + ? x)}^{\frac{1}{10}} ? x^{\frac{1}{10}}$

$= {(0.999)}^{\frac{1}{10}} ? {(1)}^{\frac{1}{10}}$

$? 1 {(0.999)}^{\frac{1}{10}} = 1 + A y$

$= 1 + \frac{d y}{d x} ? x .$

$= 1 + \frac{1}{10 x^{\frac{9}{10}}} * ? x$

$= 1 ? \frac{0.001}{10 (. 1^{\frac{9}{10}})} = 1 ? \frac{0.001}{10} = 1 ? 0.0001$

= 0.999.

(vi) ${(15)}^{\frac{1}{4}}$

A.(vi)

Let $y = x^{\frac{1}{4}} .$ Then, x = 16 and ?x = 1.

Then, $? y = {(x + A x)}^{\frac{1}{4}} ? x^{\frac{1}{4}} .$

$= {(15)}^{\frac{1}{4}} ? {(16)}^{\frac{1}{4}}$

$? {(15)}^{\frac{1}{4}} = 2 + A y = 2 + \frac{d y}{d x} A x$

$= 2 + \frac{1}{4 x^{\frac{3}{4}}} ? (? 1)$

$= 4 ? \frac{1}{4 {(16)}^{\frac{3}{4}}}$

$= 2 ? \frac{1}{4 * 8}$

$= 4 ? \frac{1}{32}$

$= \frac{64 ? 1}{32} = \frac{63}{32} = 1.968$

(vii) ${(26)}^{\frac{1}{3}}$

A.(vii)

Let $y = x^{\frac{1}{3}} .$ Let x = 27 and ?x = 1.

Then, $? y = {(x + ? x)}^{\frac{1}{3}} ? x^{\frac{1}{3}}$

$= {(26)}^{\frac{1}{3}} ? {(27)}^{\frac{1}{3}}$

$? {(26)}^{\frac{1}{3}} = 3 + 4 y = 3 + \frac{d y}{d x} A x = 3 + \frac{1}{3 x^{\frac{2}{3}}} A x,$

$= 3 ? \frac{1}{3 {(2 F)}^{\frac{2}{3}}}$

$= 3 ? \frac{1}{27}$

$= \frac{81 ? 1}{27} = \frac{80}{27} = 2.962 .$

(viii) ${(255)}^{\frac{1}{4}}$

A.(viii)

Let $y = x^{\frac{1}{4}} .$ Let x = 256 and ?x = 1.

Then, $? y = {(x + ? x)}^{\frac{1}{4}} ? x^{\frac{1}{4}} .$

$= {(255)}^{\frac{1}{4}} ? {(256)}^{\frac{1}{4}}$

$? {(255)}^{\frac{1}{4}} = {(256)}^{\frac{1}{4}} + 4 y = 4 + \frac{d y}{d x} A x .$

$= 4 + \frac{1}{4 x^{\frac{3}{4}}} ? 4 x$

$= 4 \frac{? 1}{4 {(256)}^{\frac{3}{4}}}$

$= 4 ? \frac{1}{256}$

$= \frac{1024 ? 1}{256}$

$= \frac{1023}{256} = 3.996$

(ix) ${(82)}^{\frac{1}{4}}$

A.(ix)

Let $y = x^{\frac{1}{4}} .$ let x = 81 and ?x = 1.

Then, $? y = {(x + 4 x)}^{\frac{1}{4}} ? x^{\frac{1}{4}}$

$= {(82)}^{\frac{1}{4}} ? {(81)}^{\frac{1}{4}}$

$7 {(82)}^{\frac{1}{4}} = 3 + 4 y = 3 + \frac{d y}{d x} ? x$

$= 3 + \frac{1}{4 x^{\frac{3}{4}}} ? ? x$

$= 3 + \frac{1}{4 {(81)}^{\frac{3}{4}}}$

$= 3 + \frac{1}{4 * 3^{3}}$

$= 3 + \frac{1}{108} = \frac{324 + 1}{108} = \frac{325}{108} = 3.001 .$

(x) ${(401)}^{\frac{1}{2}}$

A.(x)

Let y = $x^{\frac{1}{2}}$ = ?x $.$ Let x = 400 and ?x = 1.

Then, $? y = {(x + ? x)}^{\frac{1}{2}} ? x^{\frac{1}{2}} .$

$? ? ? y = {(401)}^{\frac{1}{2}} ? {(400)}^{\frac{1}{2}}$

$? ? {(401)}^{\frac{1}{2}} = 20 ? + ? y = 20 + \frac{d y}{d x} ? x$

$= 20 + \frac{1}{2?x} ? x$

$= 20 + \frac{1}{2?400}$

$= 20 + \frac{1}{2 * 20}$

$= 20 + \frac{1}{40}$

$= \frac{800 + 1}{40} = \frac{801}{40} = 20.025 .$

(xi) ${(0.0037)}^{\frac{1}{2}}$

A.(xi)

Let $y = x^{\frac{1}{2}}$ Let x = 0.0036 and ?x = 0.0001

Then, $? y =?x+?x ?x$

$=?0.0037 ?0.0036$

$?0.0037 = 0.06 + D y = 0.06 + \frac{d y}{d x} ? x = 0.06 + \frac{x}{2?x}$

$= 0.06 + \frac{0.001}{2?0.0036}$

$= 0.06 + \frac{0.001}{2 * 0.06}$

$= 0.06 + \frac{0.001}{0.12 .}$

= 0.06 + 0.0008.

= 0.0608.

(xii) ${(26.57)}^{\frac{1}{3}}$

A.(xii)

Let $y = x^{\frac{1}{3}} .$ Let x = 27 and ?x = 0.43.

Then, $? y = {(x + A x)}^{\frac{1}{3}} ? {(x)}^{\frac{1}{3}}$

$= {(26.57)}^{\frac{1}{3}} ? {(27)}^{\frac{1}{3}}$

$? {(26.57)}^{\frac{1}{3}} = 3 + 4 y = 3 + \frac{d y}{d x} ? x = 3 + \frac{1}{3 x^{\frac{2}{3}}} ? ? x .$

$= 3 + \frac{1}{3 {(27)}^{\frac{2}{3}}} ? less00New answer posted 5 months ago Maths Application of Derivatives 64. The line y = x + 1 is a tangent to the curve y 2 = 4x at the point (A) (1, 2) (B) (2, 1) (C) (1, - 2) (D) (- 1, 2) 0 Follower 2 Views Follow Characters 0 /2500 Disclaimer: Please note that external links or URLs are not allowed. Non adherence will lead to removal of the comment Cancel Post V Vishal Baghel Contributor-Level 10 The given eqn of curve is y^{2} = 4 x . Then, differentiating wrt x we get, 2 y \frac{d y}{d x} = 4 \Rightarrow \frac{d y}{d x} = \frac{2}{y} which is the slope of the tangent to the curve. The line y = x + 1 compared to y = m x + c gives slope of line = 1. Since, tangent is the line we have, \frac{2}{y} = 1 . \Rightarrow y = 2 Putting y = 2 in y^{2} = 4 x we get, 4 x = 2^{2} \Rightarrow x = 1 . Hence, the required point is (1, 2) Option (A) is correct. 00New answer posted 5 months ago Maths Application of Derivatives 63. The slope of the normal to the curve y = 2x2 + 3 sin x at x = 0 is (A) 3 (B) \frac{1}{3} (C) -3 (D) - \frac{1}{3} 0 Follower 2 Views Follow Characters 0 /2500 Disclaimer: Please note that external links or URLs are not allowed. Non adherence will lead to removal of the comment Cancel Post V Vishal Baghel Contributor-Level 10 Given, y = 2 x^{2} + 3 s i n x Slope of tangent, \frac{d y}{d x} = 4 x + 3 c o s x {\frac{d y}{d x} |}_{x = 0} = 4 * 0 + 3 * c o s 0 = 3 So, slope of normal = \frac{- 1}{3} Option (D) is correct. 00New answer posted 5 months ago Maths Application of Derivatives 62. Find the equation of the tangent to the curve y = \sqrt3x - 2 which is parallel to the line 4x - 2y + 5 = 0. 0 Follower 1 View Follow Characters 0 /2500 Disclaimer: Please note that external links or URLs are not allowed. Non adherence will lead to removal of the comment Cancel Post V Vishal Baghel Contributor-Level 10 Kindly go through the solution 00New answer posted 5 months ago Maths Application of Derivatives 61. Find the equations of the tangent and normal to the hyperbola \frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1 at the point (x_{0}, y_{0}) 0 Follower 2 Views Follow Characters 0 /2500 Disclaimer: Please note that external links or URLs are not allowed. Non adherence will lead to removal of the comment Cancel Post V Vishal Baghel Contributor-Level 10 The given eq n of the hyperbola is \frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1______(1) Differentiating eqn (1) wrt 'x' we get, \frac{2 x}{a^{2}} - \frac{2 y}{b^{2}} \frac{d y}{d x} = 0 \Rightarrow \frac{2 y}{b^{2}} \frac{d y}{d x} = \frac{2 x}{a^{2}} \Rightarrow \frac{d y}{d x} = \frac{b^{2} x}{a^{2} y} {\frac{d y}{d x} |}_{(x, y)} = (x_{0}, y_{0}) \frac{b^{2} x_{0}}{a^{2} y_{0}} is the reqd slope of tangent to the curve So, eq n of tangent at point (x_{0}, y_{0}) is y - y_{0} = \frac{b^{2} x_{0}}{a^{2} y_{0}} (x - x_{0}) . \Rightarrow \frac{y y_{0}}{b^{2}} - \frac{{y_{0}}^{2}}{b^{2}} = \frac{x x_{0}}{a^{2}} - \frac{{x_{0}}^{2}}{a^{2}} \Rightarrow \frac{x x_{0}}{a^{2}} - \frac{y y_{0}}{b^{2}} = \frac{{x_{0}}^{2}}{a^{2}} - \frac{{y_{0}}^{2}}{b^{2}} As (x_{0}, y_{0}) lies on the parabola given by eqn (1) we write, \frac{{x_{0}}^{2}}{a^{2}} - \frac{{y_{0}}^{2}}{b^{2}} = 1 Hence, \frac{x x_{0}}{a^{2}} - \frac{y y_{0}}{b^{2}} = 1 00New answer posted 5 months ago Maths Application of Derivatives 60. Prove that the curves x = y 2 and xy = k cut at right angles* if 8k 2 = 1. 0 Follower 2 Views Follow Characters 0 /2500 Disclaimer: Please note that external links or URLs are not allowed. Non adherence will lead to removal of the comment Cancel Post V Vishal Baghel Contributor-Level 10 The given eq n of the curves are x = y^{2}________(1) and x y = x___________(2) Differentiating eqn (1) and (2) wrt 'x' we get, \frac{d y^{2}}{d x} = \frac{d x}{d x} \to 2 y \frac{d y}{d x} = 1 \to \frac{d y}{d x} = \frac{1}{2 y}_________(3) and \frac{d}{d x} (x y) = \frac{d}{d x} (x) \to x \frac{d y}{d x} + y = 0 \to \frac{d y}{d x} = - \frac{y}{x}_________(4) Since the two curves cut each other at right angles we get, (\frac{1}{2 y}) * (\frac{- y}{x}) = - 1 . \to 2 x = 1________(5) The point of intersection can be solve from eqn (1) and (2), y^{2} . y = k \to y^{3} = k \to y = k^{1 / 3} ∴ x = y^{2} = r^{2 / 3} . Hence, using eq n (5) we get, 2 {(r)}^{2 / 3} = 1 . \Rightarrow {[2 r^{2 / 3}]}^{3} = 1^{3} \Rightarrow 8 r^{2} = 1 Hence proved 00❮ 488489490491492493494495496 ❯Taking an Exam? Selecting a College? Get authentic answers from experts, students and alumni that you won't find anywhere else Sign Up on Shiksha On Shiksha, get access to 65k Colleges 1.2k Exams 679k Reviews 1800k Answers .write-a-review-widget h2.tbSec2 {
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loadNewHomePageWidget('/shiksha/loadAjaxForm/abroadtab','hmpgrdsn_tababroad');
}\times if(typeof(googletag) != 'undefined'){ googletag.cmd.push(function() { googletag.display("div-gpt-ad-1547104965023-0"); }); }This website uses Cookies and related technologies for the site to function correctly and securely, improve & personalise your browsing experience, analyse traffic, and support our marketing efforts and serve the Core Purpose. By continuing to browse the site, you agree to Privacy Policy and Cookie Policy . OKGet our experts to answer your questions within 24 Hrs Ask Question Shiksha About Us Management Team Careers Shiksha Authors Colleges Client login Advertise with us Add Colleges Contact Us Others FAQs Grievances Notices / Summons Privacy Sitemap Terms & Conditions Our Group Infoedge.in Naukri.com Naukri Campus Naukrigulf 99acres.com Jeevansathi.com AmbitionBox.com Naukri Learning Job Hai MBA : MBA Home Top MBA Colleges MBA Colleges Executive MBA Colleges MBA Exams Engineering : Engineering Top Engineering Colleges Engineering Colleges Engineering Exams JEE Main JEE Advanced Medicine : NEET NEET PG AIIMS PG JIPMER PG PGIMER Exam Top Medical Colleges Medical Colleges Medical Exams Other Courses : Animation B Com B Sc BBA CA Fashion & Textile Design Hotel Management Law Mass Communication MBBS MCA M.Tech BCA Pharmacy M Sc Study Abroad : Study Abroad Home BTech abroad MBA abroad MS abroad GRE GMAT SAT IELTS TOEFL Sarkari Exams : SSC CGL IBPS PO SBI PO NDA UPSC Civil Services Exam IES SSC CHSL AFCAT IBPS CWE RRB SSC JE CDS SBI Clerk IBPS Clerk Resources : Careers after 12th Courses After 12th Education Boards Ask a Question Discussions Write a college review Articles Shiksha Ask & Answer App Education Trends Important Updates : BITSAT JEE Main 2026 CAT 2025 Admit Card XAT Registration 2026 Last Date XAT 2026 Exam CAT Syllabus 2025 CAT Score vs Percentile CAT 2025 NEET 2026 JEE Main PYQ Trade Marks belong to the respective owners. Copyright © 2025 Info edge India Ltd. All rights reserved.try{ overlayViewsArray.push(new Array('common/calendardiv','calendardivId')); }catch(e){ }try{ overlayViewsArray.push(new Array('network/commonOverlay','addRequestOverlay')); }catch(e){ }window.dataLayer = window.dataLayer || [];
function gtag(){dataLayer.push(arguments);}
gtag('js', new Date());

gtag('config', 'G-F0XEJFWK9T');// added for switch tracking of search results on and off
var track_search_results_flag = "1";
var facebook_channel_path = "https://www.shiksha.com/public/channel.html";

function proceedToPostQuestionFromHome(objForm,id){
objForm.setAttribute("method",'get');
objForm.setAttribute("action",'https://ask.shiksha.com/messageBoard/MsgBoard/postQuestionFromCafeForm');
objForm.submit();
}try{ jsb9eraseCookie("jsb9Track");
var prevWindowOnunload = window.onbeforeunload; window.onbeforeunload = function(){ try{ if(typeof(prevWindowOnunload) == "function") { prevWindowOnunload.call(); } jsb9onUnloadTracking(); }catch(e){ } }
function jsb9onUnloadTracking() { jsb9eraseCookie("jsb9Track"); var date = new Date(); var presentTime = date.getTime(); var presentUrl = window.location.href; jsb9createCookie("jsb9Track",presentTime+"|"+presentUrl,5); } var date = new Date(); jsb9TrackEndTime = date.getTime(); function jsb9TrackTime() { var jsb9date = new Date(); jsb9TrackFinalLoad=jsb9date.getTime(); if(typeof(jsb9TrackVal) == "string") { var cookieArr = jsb9TrackVal.split('|'); var prevTime = cookieArr[0]; var refererUrl = cookieArr[1]; var jsb9Iframe = document.createElement('div');jsb9Iframe.style.cssText='height:0px;overflow:hidden;'; jsb9Iframe.id = 'jsb9Div'; var style = 'border:0;width:0;height:0;display:none'; var jsb9ServerTime = 386.78693771362; var presentUrl = window.location.href; var customTrack = ""; for(var i in jsb9recordTimes) { customTrack += "|"+i+":"+jsb9recordTimes[i]; } var data = presentUrl+"|"+refererUrl+"|"+prevTime+"|"+jsb9TrackStartTime+"|"+jsb9TrackEndTime+"|"+jsb9TrackFinalLoad+"|"+jsb9ServerTime+customTrack; jsb9Iframe.innerHTML = '<iframe border="0" height=0 widht=0 style="visibility: hidden" src="https://track.99acres.com/images/zero.gif?data='+data+'"></iframe>';
document.body.appendChild(jsb9Iframe); } } }catch(e){ }var shikshaProduct = 'anaDesktopV2';/**
* Method loads all required scripts for Unified Registration
* callback function is loadRequiredDataForUnifiedRegistrationProcess
* callback API will be alwys called either file is already loaded or not
*/
function initForUnifiedRegistration()
{
LazyLoad.loadOnce([
'//js.shiksha.ws/public/js/build/ajax-api.1e2339.js'
],loadRequiredDataForUnifiedRegistrationProcess,null,null,true);
}

/*
Function which put values into global vars. like is_ldb user or cross btn click
*/

function loadRequiredDataForUnifiedRegistrationProcess()
{
/* ajax to set if user register or not START */
//checkLdbUser();
/* ajax to set if user register or not END */
/* set variable to check whether user has clicked unified overlay or not*/
unified_form_overlay1_cancel_clicked = getCookie('is_unified_overlay1_clicked');
unified_form_overlay2_cancel_clicked = getCookie('is_unified_overlay2_clicked');
unified_form_overlay3_cancel_clicked = getCookie('is_unified_overlay3_clicked');
/* set Form submit url for diff types of overlays */
if(typeof(arr_unified) !== 'undefined') {
ShikshaUnifiedRegistarion.url_unified = ShikshaUnifiedRegistarion.ajaxUrlHelper(arr_unified);
}
if($('homepagePromotionCarousel')) {
//if($('questionText')) { $('questionText').value = 'Make an informed career choice, ask the expert now!';}
}
// showResetPasswordLayer();
}
/* UNIFIED REGISTRATION APIs END */$j = $.noConflict();var lazyLoadJsFiles = []
function lazyLoadJsFilesInParallel() {
if(lazyLoadJsFiles.length > 0) {
LazyLoad.loadOnce(lazyLoadJsFiles,function(){},null,null,true);
}
}

function LazyLoadAnAContactUsCallback(){
$LAB.script('//js.shiksha.ws/public/js/build/quesDiscPosting.8361d6.js').wait(function(){
LazyLoadAnAContactUs();
});
}

function loadJsFilesInParallel(){
$j(document).trigger('labLoadedJs');
$LAB
.script("//js.shiksha.ws/public/js/build/AutoSuggestor.512dd5.js")
.wait(function(){
if(SHOW_AUTOSUGGESTOR_JS){
autosuggestorInstanceCheck = setInterval(function(){
var fileLoaded = false;
try{
var aso = new AutoSuggestor();
fileLoaded = true;
} catch(e) {
fileLoaded = false;
}
if(fileLoaded){
clearInterval(autosuggestorInstanceCheck);
if(typeof(initializeAutoSuggestorInstance) == 'function') {
initializeAutoSuggestorInstance();
}
if(typeof(initializeAutoSuggestorInstanceSearchV2) == 'function') {
try {
initializeAutoSuggestorInstanceSearchV2('{"suggestionContainerClass":"search-college-layer","bucketCount":4,"searchVersion":"2","tabPressedHandler":"handleAutoSuggestorTabPressed","enterPressedHandler":"handleAutoSuggestorEnterPressedSearch","mouseClickedHandler":"handleAutoSuggestorMouseClickedSearch","navigationKeysHandler":"handleNavigationKeysHandler","objectName":"autoSuggestorInstanceSearch","searchBoxId":"searchby-college","suggestionContainerId":"search-college-layer","typeAhead":0,"suggestionType":"course_and_institute","suggestionCount":8,"autoSuggestorUrl":"\/search\/AutoSuggestorV2\/getSuggestionsFromSolr\/","isMobileSearch":false,"showGrouping":1,"getTrendingSuggestions":1}');
initializeAutoSuggestorInstanceSearchV2('{"suggestionContainerClass":"search-college-layer","bucketCount":4,"searchVersion":"2","tabPressedHandler":"handleAutoSuggestorTabPressedCareers","enterPressedHandler":"handleAutoSuggestorEnterPressedCareers","mouseClickedHandler":"handleAutoSuggestorEnterPressedCareers","objectName":"autoSuggestorInstanceCareer","searchBoxId":"searchby-career","suggestionContainerId":"search-career-layer","getTrendingSuggestions":1,"typeAhead":0,"disableCache":1,"suggestionType":"career","suggestionCount":8}');
initializeAutoSuggestorInstanceSearchV2('{"suggestionContainerClass":"search-college-layer","bucketCount":4,"searchVersion":"2","tabPressedHandler":"handleAutoSuggestorTabPressedQuestion","enterPressedHandler":"handleAutoSuggestorEnterPressedQuestion","mouseClickedHandler":"handleAutoSuggestorEnterPressedQuestion","mouseClickedHandlerLastElement":"handleAutoSuggestorEnterPressedAskQuestion","askNewQuestionURL":"https:\/\/ask.shiksha.com","mouseClickedHandlerCTA":"handleAutoSuggestorEnterPressedCTA","backKeyHandler":"handleAutoSuggestorBackKeyPressedQuestion","objectName":"autoSuggestorInstanceQuestion","searchBoxId":"searchby-question","suggestionContainerId":"search-question-layer","removeExactMatch":1,"suggestionType":"question","suggestionCount":8,"isMobileSearch":false,"autoSuggestorUrl":"\/search\/AutoSuggestorV2\/getSuggestionsFromSolr\/","showGrouping":1,"getTrendingSuggestions":1}');
initializeAutoSuggestorInstanceSearchV2('{"suggestionContainerClass":"search-college-layer","bucketCount":4,"searchVersion":"2","tabPressedHandler":"handleAutoSuggestorTabPressedExams","enterPressedHandler":"handleAutoSuggestorEnterPressedExams","mouseClickedHandler":"handleAutoSuggestorEnterPressedExams","backKeyHandler":"handleAutoSuggestorBackKeyPressedExams","objectName":"autoSuggestorInstanceExam","searchBoxId":"searchby-exam","suggestionContainerId":"search-exam-layer","suggestionType":"exam","suggestionCount":8,"autoSuggestorUrl":"\/search\/AutoSuggestorV2\/getSuggestionsFromSolr\/","showGrouping":1,"getTrendingSuggestions":1}');
tagsASConfig = '{"suggestionContainerClass":"search-college-layer","bucketCount":4,"searchVersion":"2","enterPressedHandler":"handleAutoSuggestorEnterPressedTags","mouseClickedHandler":"handleAutoSuggestorEnterPressedTags","objectName":"autoSuggestorInstanceForTags","searchBoxId":"tagSearch","suggestionContainerId":"tagSearch_container","typeAhead":0,"suggestionType":"tag","suggestionCount":10,"autoSuggestorUrl":"\/Tagging\/TaggingDesktop\/getAutoSuggestor","showGrouping":1,"disableCache":1,"removePartialMatch":0}';
initializeAutoSuggestorInstanceSearchV2(tagsASConfig);
initializeAutoSuggestorInstanceSearchV2('{"suggestionContainerClass":"search-college-layer","searchVersion":"2","enterPressedHandler":"handleAutoSuggestorEnterPressedTags","mouseClickedHandler":"handleAutoSuggestorEnterPressedTags","objectName":"autoSuggestorInstanceForTagsQDP","searchBoxId":"tagSearchQDP","suggestionContainerId":"tagSearchQDP_container","typeAhead":0,"suggestionType":"tag","suggestionCount":6,"autoSuggestorUrl":"\/Tagging\/TaggingDesktop\/getAutoSuggestor","showGrouping":1,"disableCache":1,"callBackFunctionAfterBuildingSuggestionContainer":"showSuggestedTags","callBackFunctionOnInputKeysPressed":"showSuggestedTags"}');
}
catch(e) {
console.log(e);
}
}
if(typeof(initializeAutoSuggestorInstancesForCollgeReviews) == 'function') {
initializeAutoSuggestorInstancesForCollgeReviews();
}
if(typeof(initializeAutoSuggestorInstancesForCampusConnect) == 'function') {
initializeAutoSuggestorInstancesForCampusConnect();
}

if(typeof(initializeAutoSuggestorInstanceAlt) == 'function') {
initializeAutoSuggestorInstanceAlt();
}
}
},1000);
}
});
LazyLoadAnADesktopCallback();
}
(function(g,b,d){var c=b.head||b.getElementsByTagName("head"),D="readyState",E="onreadystatechange",F="DOMContentLoaded",G="addEventListener",H=setTimeout;
function f(){ loadJsFilesInParallel(); /*if(typeof CTACallBackSearch == 'function') CTACallBackSearch();*/} H(function(){if("item"in c){if(!c[0]){H(arguments.callee,25);return}c=c[0]}var a=b.createElement("script"),e=false;a.onload=a[E]=function(){if((a[D]&&a[D]!=="complete"&&a[D]!=="loaded")||e){return false}a.onload=a[E]=null;e=true;f()};
a.src="//js.shiksha.ws/public/js/LAB.min.js";c.insertBefore(a,c.firstChild)},0);if(b[D]==null&&b[G]){b[D]="loading";b[G](F,d=function(){b.removeEventListener(F,d,false);b[D]="complete"},false)}})(this,document);lazyLoadCss('//css.shiksha.ws/public/css/build/socialSharingInner.min.5de709.css');var reset_password_token = '';
var reset_usremail = '';
var reset_password = '';
var usergroup = '';

//var registration_context = '';
var user_logged_in_pref_data = "";
var userInfo = null;
var firstTimePageLoad = true;var serviceWorker = "service-worker.js";
if ('serviceWorker' in navigator && typeof serviceWorker == 'string' && serviceWorker.length > 0 && typeof registerServiceWorker == 'function') {
registerServiceWorker(serviceWorker);
}
/* function to bind autosuggestor */
if(typeof(bindAutoSuggestorEvents) == 'function'){
bindAutoSuggestorEvents();
}
/* responsible for showing GNB sub-menu on mouse hover */
if(typeof(activateGnbMouseOver) == 'function'){
activateGnbMouseOver();
}
var homepageMiddlePanel, instituteDetailPageClass, cookieDomain = '.shiksha.com';
$j(document).ready(function(){
shikshaOnreadyEventPC();
getSocialSharingLayer();
});
$j(window).on('load',function(){
shikshaOnloadEventPC();
if(typeof isUserLoggedIn !='undefined' && isUserLoggedIn && typeof getPredictorList != 'undefined' && typeof getPredictorList == 'function'){
getPredictorList();
}
});window.pushNotiConf = {
deviceType : 'desktop'
};
if(document.getElementById("pushNotificationContainer") != null){
var d = new Date();
var dStr = d.getDate()+''+d.getMonth()+''+d.getFullYear()+'v10';
var randomId = Math.floor(Math.random() * 100);

var loadNotificationSDK = function() {
var obj = document.createElement('script');
/* obj.src = 'https://localshiksha.com/public/js/shikshaPushNotification.js?'+dStr+randomId;*/
obj.src = '//js.shiksha.ws/public/js/build/shikshaPushNotification.b0fa7c.js?'+dStr+randomId;
obj.async = true;
document.head.appendChild(obj);
};

var helperObj = document.createElement('script');
/* helperObj.src = 'https://localshiksha.com/pwa/public/js/push-notification-helpers.js?'+dStr;*/
helperObj.src = '//js.shiksha.ws/pwa/public/push-notification-helpers.js?'+dStr+randomId;
helperObj.async = true;
document.head.appendChild(helperObj);

if (typeof helperObj.onload == 'object') {
helperObj.onload = function(e) {
console.log('notification helper sdk loaded');
loadNotificationSDK();
};
} else {
loadNotificationSDK();
}
if(typeof gaTrackEventCustom == 'function'){
window.phpGATrack = gaTrackEventCustom;
}
}setTimeout(function(){if(typeof loadViewsOnPageLoad == 'function'){loadViewsOnPageLoad();} },2000);var multiLocationCourses = [0];if(shikshaProduct=='CMSExam'){
var elem = document.getElementById("cookieAdSlot-d");
elem.parentNode.removeChild(elem);
}facebook Twitter Google +if (document.addEventListener){
document.addEventListener('mouseup', handleShareLayerHide, false);
} else if (document.attachEvent){
document.attachEvent('onmouseup', handleShareLayerHide);
}

function handleShareLayerHide(e)
{
var target = (e && e.target) || (event && event.srcElement);
var obj = $('socialLayer');
if(target!=obj){
obj.style.display='none';
}
}\times//Load website tour
$j(document).ready(function(){
window.contentMapping = {"Filters":"Filter rankings by publisher, exams, location, specialization etc.","DEB":"Download details of eligibility, admissions, fees, infra etc.","Shortlist":"Save your shortlist, get updates, college recommendations etc.","Compare":"Compare colleges on ranking, placements, reviews, fees etc.","Reviews":"Get honest reviews from students who studied in this college","Questions":"Get answers from the students currently studying in this college","Institute":"View college details, courses offered & recommendations","Course":"View course details, admission process & insights","Exam":"View Exam Detail","QuestionSearch":"Now you can also search from 5 lakh+ answered questions","DownloadGuide":"Get all details offline, receive important updates & more.","QuestionPapers":"Get prep tips & previous years question papers."};
initializeWebsiteTour('cta','anaDesktopV2',contentMapping);
});
lazyLoadCss('//css.shiksha.ws/public/css/build/quesDiscPosting.min.99ce37.css');$

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