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1

### JEE Main 2016 (Offline)

A wire of length $$2$$ units is cut into two parts which are bent respectively to form a square of side $$=x$$ units and a circle of radius $$=r$$ units. If the sum of the areas of the square and the circle so formed is minimum, then:
A
$$x=2r$$
B
$$2x=r$$
C
$$2x = \left( {\pi + 4} \right)r$$
D
$$\left( {4 - \pi } \right)x = \pi \,\, r$$

## Explanation

$$4x + 2\pi r = 2$$ $$\,\,\,$$ $$\Rightarrow 2x + \pi r = 1$$

$$S = {x^2} + \pi {r^2}$$

$$S = {\left( {{{1 - \pi r} \over 2}} \right)^2} + \pi {r^2}$$

$${{dS} \over {dr}} = 2\left( {{{1 - \pi r} \over 2}} \right)\left( {{{ - \pi } \over 2}} \right) + 2\pi r$$

$$\Rightarrow {{ - \pi } \over 2} + {{{\pi ^2}r} \over 2} + 2\pi r = 0$$

$$\Rightarrow r = {1 \over {\pi + 4}}$$

$$\Rightarrow x = {2 \over {\pi + 4}}\,$$

$$\Rightarrow x = 2r$$
2

### JEE Main 2015 (Offline)

Let $$f(x)$$ be a polynomial of degree four having extreme values
at $$x=1$$ and $$x=2$$. If $$\mathop {\lim }\limits_{x \to 0} \left[ {1 + {{f\left( x \right)} \over {{x^2}}}} \right] = 3$$, then f$$(2)$$ is equal to :
A
$$0$$
B
$$4$$
C
$$-8$$
D
$$-4$$

## Explanation

$$\mathop {\lim }\limits_{x \to 0} \left[ {1 + {{f\left( x \right)} \over {{x^2}}}} \right] = 3 \Rightarrow \mathop {Lim}\limits_{x \to 0} {{f\left( x \right)} \over {{x^2}}} = 2$$

So, $$f(x)$$ contains terms in $$x{}^2,{x^3}$$ and $${x^4}$$

Let $$f\left( x \right) = {a_1}{x^2} + {a_2}{x^3} + {a_3}{x^4}$$

Since $$\mathop {\lim }\limits_{x \to 0} {{f\left( x \right)} \over {{x^2}}} = 2 \Rightarrow {a_1} = 2$$

Hence, $$f\left( x \right) = 2{x^2} + {a_2}{x^3} + {a_3}{x^4}$$

$$f'\left( x \right) = 4x + 3{a_2}{x^2} + 4{a_3}{x^3}$$

As given: $$f'\left( 1 \right) = 0$$ and $$f'\left( 2 \right) = 0$$

Hence, $$4 + 3{a_2} + 4{a_3} = 0\,\,\,\,\,\,\,\,\,\,...\left( 1 \right)$$

and $$8 + 12{a_2} + 32{a_3} = 0\,\,\,\,\,...\left( 1 \right)$$

By $$4x\left( {eq1} \right) - eq\left( 2 \right),$$ we get

$$16 + 12{a_2} + 16{a_3} - \left( {8 + 12{a_2} + 32{a_3}} \right) = 0$$

$$\Rightarrow 8 - 16{a_3} = 0 \Rightarrow {a_3} = 1/2$$

and by eqn. $$\left( 1 \right),4 + 3{a_2} + 4/2 = 0 \Rightarrow {a_2} = - 2$$

$$\Rightarrow f\left( x \right) = 2{x^2} - 2{x^3} + {1 \over 2}{x^4}$$

$$f\left( 2 \right) = 2 \times 4 - 2 \times 8 + {1 \over 2} \times 16 = 0$$
3

### JEE Main 2014 (Offline)

If $$f$$ and $$g$$ are differentiable functions in $$\left[ {0,1} \right]$$ satisfying
$$f\left( 0 \right) = 2 = g\left( 1 \right),g\left( 0 \right) = 0$$ and $$f\left( 1 \right) = 6,$$ then for some $$c \in \left] {0,1} \right[$$
A
$$f'\left( c \right) = g'\left( c \right)$$
B
$$f'\left( c \right) = 2g'\left( c \right)$$
C
$$2f'\left( c \right) = g'\left( c \right)$$
D
$$2f'\left( c \right) = 3g'\left( c \right)$$

## Explanation

Since, $$f$$ and $$g$$ both are continuous function on $$\left[ {0,1} \right]$$

and differentiable on $$\left( {0,1} \right)$$ then $$\exists c \in \left( {0,1} \right)$$ such that

$$f'\left( c \right) = {{f\left( 1 \right) - f\left( 0 \right)} \over 1} = {{6 - 2} \over 1} = 4$$

and $$g'\left( c \right) = {{g\left( 1 \right) - g\left( 0 \right)} \over 1} = {{2 - 0} \over 1} = 2$$

Thus, we get $$f'\left( c \right) = 2g'\left( c \right)$$
4

### JEE Main 2013 (Offline)

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 :
A
$$\pm 1$$
B
$$\pm 2$$
C
$$\pm 3$$
D
$$\pm 4$$

## Explanation

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 $$= \pm 1.$$

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