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QUESTION: 1

Find slope of normal to the curve y=5x^{2}-10x + 7 at x=1

Solution:

y = 5x^{2} - 10x + 7

dy/dx = 10x - 10

(At x = 1) 10(1) - 10

m1 = 0

As we know that slope, m1m2 = -1

=> 0(m2) = -1

m2 = -1/0 (which is not defined)

QUESTION: 2

The equation of the tangent line to the curve y = which is parallel to the line 4x -2y + 3 = 0 is

Solution:

QUESTION: 3

Find the equation of tangent to which has slope 2.

Solution:

y = 1/(x-3)

dy/dx = d/dx(x-3)^{-1}

dy/dx = (-1) (x-3)^{(-1-1)} . d(x-3)/dx

dy/dx = -(x-3)^{(-2)}

dy/dx = - 1/(x-3)^{2}

Given, slope = 2, dy/dx = 2

- 1/(x-3)^{2} = 2

⇒ -1 = 2(x-3)^{2}

2(x-3)^{2} = -1

(x-3)^{2} = -½

We know that square of any number is always positive So, (x-3)^{2} > 0

(x-3)^{2} = -½ is not possible

No tangent to the curve has slope 2.

QUESTION: 4

If y = – sin2x. Find the values of x at which the tangents drawn to the graph of this function is parallel to the x- axis.

Solution:

QUESTION: 5

The normal at any point q to the curve x = a (cos q + q sin q), y = a (sin q – q cos q) is at distance from the origin that is equal to… .

Solution:

Clearly, dx/dy =tan q = slope of normal = −cotq

Equation of normal at θ' is y−a(sinq − qcosq) = −cotq(x−a(cosq + qsinq)

= ysinq − asin^{2}q + aqcosqsinq

= −xcosq + acos^{2}q + aqsinqcosq

= xcosq + ysinq = a

QUESTION: 6

The equation of the normal to the curve x^{2} = 4y which passes through the point (1, 2) is.

Solution:

h^{2 }= 4k

slope of normal=−1/(dy/dx) = −2h

equation of normal(y − k)= −2h(x−h)

k = 2 + 2/h(1 − h)

(h^{2}) / 4 = 2 + 2/h (1 − h)

h = 2, k = 1

equation of line (y - 1)= -1(x - 2)

x + y = 3

QUESTION: 7

The curve y = ax^{3} + bx^{2} + cx + 5 touches the x-axis at P(-2, 0) and cuts the y-axis at the point Q where its gradient is 3. the equation of the curve is… .

Solution:

y=ax^{3}+bx^{2}+cx+5

(−z,0) lies on curve y=ax^{3}+ bx^{2} + cx + 5d

⇒ 0= −8a + 4b − 2c + 5.....(1)

Also at (−2,0) curve touches x-axis ie dy/dx at (−2,0) is 0

∴ dy/dx=3ax^{2}+2bx+c

dy/dx (−2,0)=0=12a = −4b + c....(2)

Since curve crosses y-axis at Q, x-coordinate of point Q must be zero

⇒ y = 0 + 0 + 0 + 5

⇒y = 5

∴ coordinate of Q = (0,5)

According to Question dy/dx =3 at Q(0,5)

⇒ 3= 0 + 0 + c

⇒ c = 3

putting the value of c in eqn(1) & eq n(2)and gurthare solving we get

a=−1/2 & b=-3/4

∴Eqn of cover is y = −(x^{3}) / 2−3/4x^{2 }+ 3x + 5

QUESTION: 8

If a, b are real numbers such that x^{3}-ax^{2} + bx – 6 = 0 has its roots real and positive then minimum value of b is

Solution:

If p,q and r are the roots,

pq + qr + pr = b and pqr = 6

Since A.M.≥G.M . applying it to 3 numbers pq,qr and pr

(pq + qr + pr)/3 ≥ (p^{2}q^{2}r^{2})^{1/3} i.e.

(pq + qr + pr)/3 ≥ (36)^{1/3}

pq + qr + pr ≥ 3 (36)^{1/3}

So minimum value of b is (36)^{1/3}

QUESTION: 9

The equation of tangent to the curve y = x^{3} + 2x + 6 which is perpendicular to the line x + 14y + 4 = 0 is :

Solution:

y = x^{3} + 2x + 6

Slope of tangent = m1 = dy/dx = 3x^{3} + 2

Slope of perpendicular line = m2 = −1/14

m1 . m2 = -1

m1 = 14

3x^{2} + 2 = 14

x = ±2

Therefore the curve has tangents at x = 2 and x = -2

and these points also lie on the given curve

Equation of tangent - y = 14x + c

Coordinates of points of tangency

At x = 2 , y = 23 + 2(2) + 6 = 18

At x = -2 , y = (-2)3 + 2(-2) + 6 = -6

18 = 14(2) + c

c = -10

-6 = 14(-2) + c

c = 22

Equation of tangent - y = 14x - 10 and y = 14x + 22

Hence, 14x - y +22 = 0

QUESTION: 10

The equations of the tangents drawn to the curve y^{2} – 2x^{3} – 4y + 8 = 0 from the point (1, 2) is… .

Solution:

Let P(α, β) be any point on the curve

Now, the equation of the tangent at P is

Hence, the point of contacts are

(2, 2 + 2√3) and (2, 2 – 2√3)

Slope of the tangents are 2√3 , –2√3

Hence, the equations of tangents are

y – (2 + 2√3) = 2√3 (x – 2)

and

y –(2 – 2√3) = – 2√3 (x – 2)

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