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1

JEE Advanced 2019 Paper 1 Offline

MCQ (More than One Correct Answer)
Let $$\Gamma $$ denote a curve y = y(x) which is in the first quadrant and let the point (1, 0) lie on it. Let the tangent to I` at a point P intersect the y-axis at YP. If PYP has length 1 for each point P on I`, then which of the following options is/are correct?
A
$$xy' + \sqrt {1 - {x^2}} = 0$$
B
$$xy' - \sqrt {1 - {x^2}} = 0$$
C
$$y = {\log _e}\left( {{{1 + \sqrt {1 - {x^2}} } \over x}} \right) - \sqrt {1 - {x^2}} $$
D
$$y = - {\log _e}\left( {{{1 + \sqrt {1 - {x^2}} } \over x}} \right) + \sqrt {1 - {x^2}} $$

Explanation

Let a point P(h, k) on the curve y = y(x), so equation of tangent to the curve at point P is

$$y - k = {\left( {{{dy} \over {dx}}} \right)_{h,k}}(x - h)$$ ....(i)

Now, the tangent (i) intersect the Y-axis at Yp, so coordinates Yp is $$\left( {0,k - h{{dy} \over {dx}}} \right)$$,

where $${{{dy} \over {dx}}}$$ = $${\left( {{{dy} \over {dx}}} \right)}$$(h,k)

So, PYp = 1 (given)

$$ \Rightarrow \sqrt {{h^2} + {h^2}{{\left( {{{dy} \over {dx}}} \right)}^2}} = 1$$

$$ \Rightarrow {{dy} \over {dx}} = \pm {{\sqrt {1 - {x^2}} } \over x}$$

[on replacing h by x]

$$ \Rightarrow dy = \pm {{\sqrt {1 - {x^2}} } \over x}dx$$

On puting x = sin$$\theta $$, dx = cos$$\theta $$d$$\theta $$, we get

$$dy = \pm {{\sqrt {1 - {{\sin }^2}\theta } } \over {\sin \theta }}\cos \theta d\theta $$

$$ = \pm {{{{\cos }^2}\theta } \over {\sin \theta }}d\theta $$

$$ = \pm (\cos ec\theta - \sin \theta )d\theta $$

$$ \Rightarrow y = \pm [1n(\cos ec\theta - \cot \theta ) + \cos \theta ] + C$$

$$ \Rightarrow y = \pm \left[ {1n\left( {{{1 - \cos \theta } \over {\sin \theta }}} \right) + \cos \theta } \right] + C$$

$$ \Rightarrow y = \pm \left[ {1n\left( {{{1 - \sqrt {1 - {{\sin }^2}\theta } } \over {\sin \theta }}} \right) + \sqrt {1 - {{\sin }^2}\theta } } \right] + C$$

$$ \Rightarrow y = \pm \left[ {1n\left( {{{1 - \sqrt {1 - {x^2}} } \over x}} \right) + \sqrt {1 - {x^2}} } \right] + C$$

[$$ \because $$ x = sin$$\theta $$]

$$ = \pm \left[ { - 1n{{1 + \sqrt {1 - {x^2}} } \over x} + \sqrt {1 - {x^2}} } \right] + C$$

[on rationalization]

$$ \because $$ The curve is in the first quadrant so y must be positive, so

$$y = 1n\left( {{{1 + \sqrt {1 - {x^2}} } \over x}} \right) - \sqrt {1 - {x^2}} + C$$

As curve passes through (1, 0), so

$$0 = 0 - 0 + c \Rightarrow c = 0$$, so required curve is

$$y = 1n\left( {{{1 + \sqrt {1 - {x^2}} } \over x}} \right) - \sqrt {1 - {x^2}} $$

and required differential eaquation is

$${{dy} \over {dx}} = - {{\sqrt {1 - {x^2}} } \over x}$$

$$ \Rightarrow xy' + \sqrt {1 - {x^2}} = 0$$

Hence, options (a) and (c) are correct.
2

JEE Advanced 2017 Paper 2 Offline

MCQ (More than One Correct Answer)
If $$g(x) = \int_{\sin x}^{\sin (2x)} {{{\sin }^{ - 1}}} (t)\,dt$$, then
A
$$g'\left( { - {\pi \over 2}} \right) = 0$$
B
$$g'\left( { - {\pi \over 2}} \right) = - 2\pi $$
C
$$g'\left( {{\pi \over 2}} \right) = 2\pi $$
D
$$g'\left( {{\pi \over 2}} \right) = 0$$

Explanation

$$g(x) = \int_{\sin x}^{\sin (2x)} {{{\sin }^{ - 1}}} (t)\,dt$$

g'(x) = 2cos 2x sin$$-$$1(sin2 x) $$-$$ cos x sin$$-$$1(sin x)

$$g'\left( {{\pi \over 2}} \right) = - 2{\sin ^{ - 1}}(0) = 0$$

$$g'\left( { - {\pi \over 2}} \right) = - 2{\sin ^{ - 1}}(0) = 0$$

No option is matching.
3

JEE Advanced 2016 Paper 1 Offline

MCQ (More than One Correct Answer)

Let $$f:(0,\infty ) \to R$$ be a differentiable function such that $$f'(x) = 2 - {{f(x)} \over x}$$ for all $$x \in (0,\infty )$$ and $$f(1) \ne 1$$. Then

A
$$\mathop {\lim }\limits_{x \to {0^ + }} f'\left( {{1 \over x}} \right) = 1$$
B
$$\mathop {\lim }\limits_{x \to {0^ + }} xf\left( {{1 \over x}} \right) = 2$$
C
$$\mathop {\lim }\limits_{x \to {0^ + }} {x^2}f'(x) = 0$$
D
$$\left| {f(x)} \right| \le 2$$ for all $$x \in (0,2)$$

Explanation

$$f'(x) = 2 - {{f(x)} \over x}$$

$$ \Rightarrow f'(x) = {1 \over x}f(x) = 2$$ is linear differential equation.

Hence, $$I.F. = {e^{\int {{1 \over x}dx} }} = {e^{\ln x}} = x$$

Thus the solution is given by

$$x\,.\,f(x) = \int {2x\,dx + \lambda } $$

i.e., $$xf(x) = {x^2} + \lambda $$

As f(1) $$\ne$$ 1, we have $$\lambda$$ $$\ne$$ 0

$$\therefore$$ $$f(x) = x + {\lambda \over x}$$, $$\lambda$$ $$\ne$$ 0

Thus, $$f'(x) = 1 - {\lambda \over {{x^2}}}$$, $$\lambda$$ $$\ne$$ 0

Now, $$\mathop {\lim }\limits_{x \to {0^ + }} f'\left( {{1 \over x}} \right) = \mathop {\lim }\limits_{x \to {0^ + }} (1 - \lambda {x^2}) = 1$$

$$\mathop {\lim }\limits_{x \to {0^ + }} xf\left( {{1 \over x}} \right) = \mathop {\lim }\limits_{x \to {0^ + }} x\left( {{1 \over x} + \lambda x} \right) = \mathop {\lim }\limits_{x \to {0^ + }} (1 + \lambda {x^2}) = 1$$

$$\mathop {\lim }\limits_{x \to {0^ + }} {x^2}f'(x) = \mathop {\lim }\limits_{x \to {0^ + }} {x^2}\left( {1 - {\lambda \over {{x^2}}}} \right) = \mathop {\lim }\limits_{x \to {0^ + }} ({x^2} - \lambda ) = - \lambda $$

Again, $$\mathop {\lim }\limits_{x \to {0^ + }} f(x) \to \infty $$

Hence the function is not bounded.

Note that $$\lambda$$ can be $$-$$ve or +ve.

4

JEE Advanced 2016 Paper 1 Offline

MCQ (More than One Correct Answer)
A solution curve of the differential equation

$$\left( {{x^2} + xy + 4x + 2y + 4} \right){{dy} \over {dx}} - {y^2} = 0,$$ $$x>0,$$ passes through the

point $$(1,3)$$. Then the solution curve
A
intersects $$y=x+2$$ exactly at one point
B
intersects $$y=x+2$$ exactly at two points
C
intersects $$y = {\left( {x + 2} \right)^2}$$
D
does NOT intersect $$\,y = {\left( {x + 3} \right)^2}$$

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