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1

MCQ (Single Correct Answer)

The intercepts on $$x$$-axis made by tangents to the curve,

$$y = \int\limits_0^x {\left| t \right|dt,x \in R,} $$ which are parallel to the line $$y=2x$$, are equal to :

$$y = \int\limits_0^x {\left| t \right|dt,x \in R,} $$ which are parallel to the line $$y=2x$$, are equal to :

A

$$ \pm 1$$

B

$$ \pm 2$$

C

$$ \pm 3$$

D

$$ \pm 4$$

Since, $$y = \int\limits_0^x {\left| t \right|} dt,x \in R$$

therefore $${{dy} \over {dx}} = \left| x \right|$$

But from $$y = 2x,{{dy} \over {dx}} = 2$$

$$ \Rightarrow \left| x \right| = 2 \Rightarrow x = \pm 2$$

Points $$y = \int\limits_0^{ \pm 2} {\left| t \right|dt} = \pm 2$$

$$\therefore$$ equation of tangent is

$$y - 2 = 2\left( {x - 2} \right)$$ or $$y + 2 = 2\left( {x + 2} \right)$$

$$ \Rightarrow $$ $$x$$-intercept $$ = \pm 1.$$

therefore $${{dy} \over {dx}} = \left| x \right|$$

But from $$y = 2x,{{dy} \over {dx}} = 2$$

$$ \Rightarrow \left| x \right| = 2 \Rightarrow x = \pm 2$$

Points $$y = \int\limits_0^{ \pm 2} {\left| t \right|dt} = \pm 2$$

$$\therefore$$ equation of tangent is

$$y - 2 = 2\left( {x - 2} \right)$$ or $$y + 2 = 2\left( {x + 2} \right)$$

$$ \Rightarrow $$ $$x$$-intercept $$ = \pm 1.$$

2

MCQ (Single Correct Answer)

A line is drawn through the point $$(1, 2)$$ to meet the coordinate axes at $$P$$ and $$Q$$ such that it forms a triangle $$OPQ,$$ where $$O$$ is the origin. If the area of the triangle $$OPQ$$ is least, then the slope of the line $$PQ$$ is :

A

$$-{1 \over 4}$$

B

$$-4$$

C

$$-2$$

D

$$-{1 \over 2}$$

Equation of a line passing through $$\left( {{x_1},{y_1}} \right)$$ having

slope $$m$$ is given by $$y - {y_1} = m\left( {x - {x_1}} \right)$$

Since the line $$PQ$$ is passing through $$(1,2)$$ therefore its

equation is

$$\left( {y - 2} \right) = m\left( {x - 1} \right)$$

where $$m$$ is the slope of the line $$PQ$$.

Now, point $$P\left( {x,0} \right)$$ will also satisfy the equation of $$PQ$$

$$\therefore$$ $$y - 2 = m\left( {x - 1} \right)$$

$$ \Rightarrow 0 - 2 = m\left( {x - 1} \right)$$

$$ \Rightarrow - 2 = m\left( {x - 1} \right)$$

$$ \Rightarrow x - 1 = {{ - 2} \over m}$$

$$ \Rightarrow x = {{ - 2} \over m} + 1$$

Also, $$OP = \sqrt {\left( {x - 0} \right){}^2 + {{\left( {0 - 0} \right)}^2}} = x = {{ - 2} \over m} + 1$$

Similarly, point $$Q\left( {0,y} \right)$$ will satisfy equation of $$PQ$$

$$\therefore$$ $$y - 2 = m\left( {x - 1} \right)$$

$$ \Rightarrow y - 2 = m\left( { - 1} \right) \Rightarrow y = 2 - m$$b

and $$OQ = y = 2 - m$$

Area of $$\Delta POQ = {1 \over 2}\left( {OP} \right)\left( {OQ} \right) = {1 \over 2}\left( {1 - {2 \over m}} \right)\left( {2 - m} \right)$$

( As Area of $$\Delta = {1 \over 2} \times $$ base $$\,\, \times \,\,$$ height )

$$ = {1 \over 2}\left[ {2 - m - {4 \over m} + 2} \right] = {1 \over 2}\left[ {4 - \left( {m + {4 \over m}} \right)} \right]$$

$$ = 2 - {m \over 2} - {2 \over m}$$

Let Area $$ = f\left( m \right) = 2 - {m \over 2} - {2 \over m}$$

Now, $$f'\left( m \right) = {{ - 1} \over 2} + {2 \over {{m^2}}}$$

Put $$f'\left( m \right) = 0$$

$$ \Rightarrow {m^2} = 4 \Rightarrow m = \pm 2$$

Now, $$f''\left( m \right) = {{ - 4} \over {{m^3}}}$$

$${\left. {f''\left( m \right)} \right|_{m = 2}} = - {1 \over 2} < 0$$

$${\left. {f''\left( m \right)} \right|_{m = - 2}} = {1 \over 2} > 0$$

Area will be least at $$m=-2$$

Hence, slope of $$PQ$$ is $$-2.$$

slope $$m$$ is given by $$y - {y_1} = m\left( {x - {x_1}} \right)$$

Since the line $$PQ$$ is passing through $$(1,2)$$ therefore its

equation is

$$\left( {y - 2} \right) = m\left( {x - 1} \right)$$

where $$m$$ is the slope of the line $$PQ$$.

Now, point $$P\left( {x,0} \right)$$ will also satisfy the equation of $$PQ$$

$$\therefore$$ $$y - 2 = m\left( {x - 1} \right)$$

$$ \Rightarrow 0 - 2 = m\left( {x - 1} \right)$$

$$ \Rightarrow - 2 = m\left( {x - 1} \right)$$

$$ \Rightarrow x - 1 = {{ - 2} \over m}$$

$$ \Rightarrow x = {{ - 2} \over m} + 1$$

Also, $$OP = \sqrt {\left( {x - 0} \right){}^2 + {{\left( {0 - 0} \right)}^2}} = x = {{ - 2} \over m} + 1$$

Similarly, point $$Q\left( {0,y} \right)$$ will satisfy equation of $$PQ$$

$$\therefore$$ $$y - 2 = m\left( {x - 1} \right)$$

$$ \Rightarrow y - 2 = m\left( { - 1} \right) \Rightarrow y = 2 - m$$b

and $$OQ = y = 2 - m$$

Area of $$\Delta POQ = {1 \over 2}\left( {OP} \right)\left( {OQ} \right) = {1 \over 2}\left( {1 - {2 \over m}} \right)\left( {2 - m} \right)$$

( As Area of $$\Delta = {1 \over 2} \times $$ base $$\,\, \times \,\,$$ height )

$$ = {1 \over 2}\left[ {2 - m - {4 \over m} + 2} \right] = {1 \over 2}\left[ {4 - \left( {m + {4 \over m}} \right)} \right]$$

$$ = 2 - {m \over 2} - {2 \over m}$$

Let Area $$ = f\left( m \right) = 2 - {m \over 2} - {2 \over m}$$

Now, $$f'\left( m \right) = {{ - 1} \over 2} + {2 \over {{m^2}}}$$

Put $$f'\left( m \right) = 0$$

$$ \Rightarrow {m^2} = 4 \Rightarrow m = \pm 2$$

Now, $$f''\left( m \right) = {{ - 4} \over {{m^3}}}$$

$${\left. {f''\left( m \right)} \right|_{m = 2}} = - {1 \over 2} < 0$$

$${\left. {f''\left( m \right)} \right|_{m = - 2}} = {1 \over 2} > 0$$

Area will be least at $$m=-2$$

Hence, slope of $$PQ$$ is $$-2.$$

3

MCQ (Single Correct Answer)

Let $$a,b \in R$$ be such that the function $$f$$ given by $$f\left( x \right) = In\left| x \right| + b{x^2} + ax,\,x \ne 0$$ has extreme values at $$x=-1$$ and $$x=2$$

**Statement-1 :** $$f$$ has local maximum at $$x=-1$$ and at $$x=2$$.

**Statement-2 :** $$a = {1 \over 2}$$ and $$b = {-1 \over 4}$$

A

Statement - 1 is false, Statement - 2 is true.

B

Statement - 1 is true , Statement - 2 is true; Statement - 2 is a correct explanation for Statement - 1.

C

Statement - 1 is true, Statement - 2 is true; Statement - 2 is **not** a correct explanation for Statement - 1.

D

Statement - 1 is true, Statement - 2 is false.

Given, $$f\left( x \right) = \ln \left| x \right| + b{x^2} + ax$$

$$\therefore$$ $$f'\left( x \right) = {1 \over x} + 2bx + a$$

At $$x=-1,$$ $$f'\left( x \right) = - 1 - 2b + a = 0$$

$$ \Rightarrow a - 2b = 1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( i \right)$$

At $$x=2,$$ $$\,\,f'\left( x \right) = {1 \over 2} + 4b + a = 0$$

$$ \Rightarrow a + 4b = - {1 \over 2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( {ii} \right)$$

On solving $$(i)$$ and $$(ii)$$ we get $$a = {1 \over 2},b = - {1 \over 4}$$

Thus, $$f'\left( x \right) = {1 \over x} - {x \over 2} + {1 \over 2} = {{2 - {x^2} + x} \over {2x}}$$

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

So maximum at $$x=-1,2$$

Hence both the statements are true and statement $$2$$ is a correct explanation for $$1.$$

$$\therefore$$ $$f'\left( x \right) = {1 \over x} + 2bx + a$$

At $$x=-1,$$ $$f'\left( x \right) = - 1 - 2b + a = 0$$

$$ \Rightarrow a - 2b = 1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( i \right)$$

At $$x=2,$$ $$\,\,f'\left( x \right) = {1 \over 2} + 4b + a = 0$$

$$ \Rightarrow a + 4b = - {1 \over 2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( {ii} \right)$$

On solving $$(i)$$ and $$(ii)$$ we get $$a = {1 \over 2},b = - {1 \over 4}$$

Thus, $$f'\left( x \right) = {1 \over x} - {x \over 2} + {1 \over 2} = {{2 - {x^2} + x} \over {2x}}$$

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

So maximum at $$x=-1,2$$

Hence both the statements are true and statement $$2$$ is a correct explanation for $$1.$$

4

MCQ (Single Correct Answer)

A spherical balloon is filled with $$4500\pi $$ cubic meters of helium gas. If a leak in the balloon causes the gas to escape at the rate of $$72\pi $$ cubic meters per minute, then the rate (in meters per minute) at which the radius of the balloon decreases $$49$$ minutes after the leakage began is :

A

$${{9 \over 7}}$$

B

$${{7 \over 9}}$$

C

$${{2 \over 9}}$$

D

$${{9 \over 2}}$$

Volume of spherical balloon $$ = V = {4 \over 3}\pi {r^3}$$

$$ \Rightarrow 4500\pi = {{4\pi {r^3}} \over 3}$$

( as Given, volume $$ = 4500\pi {m^3}$$ )

Differentiating both the sides, $$w.r.t't'$$ we get,

$${{dV} \over {dt}} = 4\pi {r^2}\left( {{{dr} \over {dt}}} \right)$$

Now, it is given that $${{dV} \over {dt}} = 72\pi $$

$$\therefore$$ After $$49$$ min, Volume -

$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \left( {4500 - 49 \times 72} \right)\pi $$

$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \left( {4500 - 3528} \right)\pi $$

$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 972\pi {m^3}$$

$$ \Rightarrow V = 972\,\,\pi {m^3}$$

$$\therefore$$ $$972\pi = {4 \over 3}\pi r{}^3$$

$$ \Rightarrow {r^3} = 3 \times 243 = 3 \times {3^5} = {3^6} = {\left( {{3^2}} \right)^3} \Rightarrow r = 9$$

Also, we have $${{dV} \over {dt}} = 72\pi $$

$$\therefore$$ $$72\pi = 4\pi \times 9 \times 9\left( {{{dr} \over {dt}}} \right) \Rightarrow {{dr} \over {dt}} = \left( {{2 \over 9}} \right)$$

$$ \Rightarrow 4500\pi = {{4\pi {r^3}} \over 3}$$

( as Given, volume $$ = 4500\pi {m^3}$$ )

Differentiating both the sides, $$w.r.t't'$$ we get,

$${{dV} \over {dt}} = 4\pi {r^2}\left( {{{dr} \over {dt}}} \right)$$

Now, it is given that $${{dV} \over {dt}} = 72\pi $$

$$\therefore$$ After $$49$$ min, Volume -

$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \left( {4500 - 49 \times 72} \right)\pi $$

$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \left( {4500 - 3528} \right)\pi $$

$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 972\pi {m^3}$$

$$ \Rightarrow V = 972\,\,\pi {m^3}$$

$$\therefore$$ $$972\pi = {4 \over 3}\pi r{}^3$$

$$ \Rightarrow {r^3} = 3 \times 243 = 3 \times {3^5} = {3^6} = {\left( {{3^2}} \right)^3} \Rightarrow r = 9$$

Also, we have $${{dV} \over {dt}} = 72\pi $$

$$\therefore$$ $$72\pi = 4\pi \times 9 \times 9\left( {{{dr} \over {dt}}} \right) \Rightarrow {{dr} \over {dt}} = \left( {{2 \over 9}} \right)$$

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