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MCQ (Previous Year Questions) - Tangent And Normal (Competition Level 1) - JEE MCQ


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15 Questions MCQ Test - MCQ (Previous Year Questions) - Tangent And Normal (Competition Level 1)

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MCQ (Previous Year Questions) - Tangent And Normal (Competition Level 1) - Question 1

A function y = f(x) has a second order derivative f‘‘ (x) = 6( x – 1). If its graph passes through the point (2,1) and at that point the tanget to the graph is y = 3x – 5, then the function, is- 

[AIEEE 2004]

Detailed Solution for MCQ (Previous Year Questions) - Tangent And Normal (Competition Level 1) - Question 1

Since, f ”(x) = 6(x – 1)
On integrating, we get

f ” (x) = 3 (x – 1)2 + c ...(i)
Also, at the point (2, 1) the tangent to graph is
y = 3x – 5
Slope of tangent = 3 ⇒ f ’ (2) = 3
∴f ’ (2) = 3(2 – 1)2 + c = 3 [from Eq (i)]
⇒3 + c = 3 ⇒ c = 0
Form Eq. (i),
f ’ (x) = 3(x – 1)2
On integrating, we get
f ’ (x) = 3(x – 1)3 + k ...(ii)
Since, graph passes through (2, 1).
∴ 1 = (2 – 1)2 + k
     k = 0
Hence, equation of function is
f(x) = (x – 1)3.

MCQ (Previous Year Questions) - Tangent And Normal (Competition Level 1) - Question 2

The normal to the curve x = a (1+ cos θ), y = a sin θ at ‘θ’ always passes through the fixed point-

[AIEEE 2004]

Detailed Solution for MCQ (Previous Year Questions) - Tangent And Normal (Competition Level 1) - Question 2

Given that,
x = a(1 + cosθ), y = a sin θ

On differentiating w.r.t. θ, we get

Equation of normal at the given points is

It is clear that in the given options normal passes through the point (a, 0).

MCQ (Previous Year Questions) - Tangent And Normal (Competition Level 1) - Question 3

If 2a + 3b + 6c = 0, then at least one root of the equation ax2 + bx + c = 0 lies in the interval-

[AIEEE 2004]

Detailed Solution for MCQ (Previous Year Questions) - Tangent And Normal (Competition Level 1) - Question 3

MCQ (Previous Year Questions) - Tangent And Normal (Competition Level 1) - Question 4

If the equation anxn + an–1 xn–1 +....+ a1x= 0 ; a1 ≠ 0, n ≥ 2, has a positive root x =α, then the equation nanxn–1 + (n – 1) an–1 xn–2 + .... + a1 = 0 has a positive root, which is -

[AIEEE-2005]

*Multiple options can be correct
MCQ (Previous Year Questions) - Tangent And Normal (Competition Level 1) - Question 5

The normal to the curve x = a (cos θ + θ sin θ), y = a (sin θ – θ cos θ) at any point 'θ' is such that -

Detailed Solution for MCQ (Previous Year Questions) - Tangent And Normal (Competition Level 1) - Question 5

Given that, x = a(cos θ + θsin θ)

and   y = a(sin θ– q cos θ)

On differentiating w.r.t. θ respectively, we get

On dividing Eqs. (ii) by (i), we get

Since, slope of normal

So, equation of normal is

MCQ (Previous Year Questions) - Tangent And Normal (Competition Level 1) - Question 6

A spherical iron ball 10 cm in radius is coated with a layer of ice of uniform thickness that melts at a rate of 50 cm3/min. When the thickness of ice is 5 cm, then the rate of which the thickness of ice decreases, is -

[AIEEE-2005]

Detailed Solution for MCQ (Previous Year Questions) - Tangent And Normal (Competition Level 1) - Question 6

Given that 

MCQ (Previous Year Questions) - Tangent And Normal (Competition Level 1) - Question 7

Angle between the tangents to the curve y=x2 – 5x+6 at the points (2, 0) and (3, 0) is–

[AIEEE 2006]

Detailed Solution for MCQ (Previous Year Questions) - Tangent And Normal (Competition Level 1) - Question 7

Hence, angle beteen the tangents is π/2

MCQ (Previous Year Questions) - Tangent And Normal (Competition Level 1) - Question 8

The equation of the tangent to the curve y = x + 4/x2 , that is parallel to the x–axis, is -  

[AIEEE 2010]

Detailed Solution for MCQ (Previous Year Questions) - Tangent And Normal (Competition Level 1) - Question 8

We have 

On differentiating with respect to x, wer get

Since, the tangent is parallel to x-axis, therefore

MCQ (Previous Year Questions) - Tangent And Normal (Competition Level 1) - Question 9

The shortest distance between line y – x = 1 and curve x = y2 is:

[AIEEE 2011]

Detailed Solution for MCQ (Previous Year Questions) - Tangent And Normal (Competition Level 1) - Question 9

MCQ (Previous Year Questions) - Tangent And Normal (Competition Level 1) - Question 10

The intercepts on x-axis made by tangents to the curve,   which are parallel to the line y =2x, are equal to :

 [JEE MAIN 2013]

Detailed Solution for MCQ (Previous Year Questions) - Tangent And Normal (Competition Level 1) - Question 10

y – 2 = 2(x – 2)

y = 2x – 2

x – intercept = ± 1

MCQ (Previous Year Questions) - Tangent And Normal (Competition Level 1) - Question 11

Find the equation of the straight line which is tangent at one point and normal at another point of the curve,
x = 3t2, y = 2t3.

[REE 2000 (Mains), 5]

Detailed Solution for MCQ (Previous Year Questions) - Tangent And Normal (Competition Level 1) - Question 11

MCQ (Previous Year Questions) - Tangent And Normal (Competition Level 1) - Question 12

If the normal to the curve, y = f(x) at the point (3, 4) makes an angle 3π/4 with the positive x-axis. The f ’(3)=

[JEE 2000 (Scr.), 1]

Detailed Solution for MCQ (Previous Year Questions) - Tangent And Normal (Competition Level 1) - Question 12

MCQ (Previous Year Questions) - Tangent And Normal (Competition Level 1) - Question 13

The point(s) on the curve y3 + 3x2 = 12y where the tangent is vertical, is (are)

 [JEE 2002 (Scr.), 3]

Detailed Solution for MCQ (Previous Year Questions) - Tangent And Normal (Competition Level 1) - Question 13

from (1)

MCQ (Previous Year Questions) - Tangent And Normal (Competition Level 1) - Question 14

The tangent to the curve y = ex drawn at the point (c, ec) intersects the line joining the points (c – 1, ec1) and (c + 1, ec + 1)

[JEE 2007, 3]

Detailed Solution for MCQ (Previous Year Questions) - Tangent And Normal (Competition Level 1) - Question 14

slope of the line joining the points (c – 1, ec – 1) and (c + 1, ec + 1) is

⇒ angent to the curve y = ex will intersect the given line to the left of the line x = c

 

 

MCQ (Previous Year Questions) - Tangent And Normal (Competition Level 1) - Question 15

Tangent to the curve y = x2 + 6 at a point P(1, 7) touches the circle x2 + y2 + 16x + 12y + c = 0 at a point Q. Then the coordinates of Q are    

[JEE 2005 (Scr.), 3]

Detailed Solution for MCQ (Previous Year Questions) - Tangent And Normal (Competition Level 1) - Question 15

(–6, –7)

y = x2 + 6

Tangent at (1, 7)

y = 2x + 5 ........(1)

which is also tangent to the circle

x2 + y2 + 16x + 12y + c = 0

i.e.  x2 + (2x + 5)2 + 16x + 12(2x + 5) + c = 0

must have equal roots i.e. a = b

⇒ 5x2 + 60x + 85 + c = 0

a = –6


⇒  x = –6 and y = 2x + 5 = – 7

Q (–6, –7)

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