 ### JEE Mains Previous Years Questions with Solutions

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1

### AIEEE 2003

If the function $$f\left( x \right) = 2{x^3} - 9a{x^2} + 12{a^2}x + 1,$$ where $$a>0,$$ attains its maximum and minimum at $$p$$ and $$q$$ respectively such that $${p^2} = q$$ , then $$a$$ equals
A
$${1 \over 2}$$
B
$$3$$
C
$$1$$
D
$$2$$

## Explanation

$$f\left( x \right) = 2{x^3} - 9a{x^2} + 12{a^2}x + 1$$

$$f'\left( x \right) = 6{x^2} - 18ax + 12{a^2};$$

$$\,\,\,\,\,\,\,\,\,\,f''\left( x \right) = 12x - 18a$$

For max. or min.

$$6{x^2} - 18ax + 12{a^2} = 0$$

$$\Rightarrow {x^2} - 3ax + 2{a^2} = 0$$

$$\Rightarrow x = a$$ or $$x=2a.$$

At $$x=a$$

$$f''(a) = 12a - 18a = -6a < 0$$

At $$x=a$$ maximum As f''($$a$$) < 0.

At $$x=2a$$

$$f''(a) = 24a - 18a = 6a > 0$$

At $$x=2a$$ minimum As f''(2$$a$$) > 0.

$$\therefore$$ $$p=a$$ and $$q=2a$$

As per question $${p^2} = q$$

$$\therefore$$ $${a^2} = 2a$$

$$\Rightarrow$$ $$a(a - 2) = 0$$

$$\therefore$$ $$a$$ = 0, 2

but $$a > 0,$$ therefore, $$a=2.$$
2

### AIEEE 2002

If $$2a+3b+6c=0,$$ $$\left( {a,b,c \in R} \right)$$ then the quadratic equation $$a{x^2} + bx + c = 0$$ has
A
at least one root in $$\left[ {0,1} \right]$$
B
at least one root in $$\left[ {2,3} \right]$$
C
at least one root in $$\left[ {4,5} \right]$$
D
none of these

## Explanation

Let $$f\left( x \right) = {{a{x^3}} \over 3} + {{b{x^2}} \over 2} + cx \Rightarrow f\left( 0 \right) = 0$$ and $$f(1)$$

$$= {a \over 3} + {b \over 2} + c = {{2a + 3b + 6c} \over 6} = 0$$

Also $$f(x)$$ is continuous and differentiable in $$\left[ {0,1} \right]$$ and

$$\left[ {0,1\left[ {.\,\,} \right.} \right.$$ So by Rolle's theorem $$f'\left( x \right) = 0.$$

i.e $$\,\,a{x^2} + bx + c = 0$$ has at least one root in $$\left[ {0,1} \right].$$
3

### AIEEE 2002

The maximum distance from origin of a point on the curve
$$x = a\sin t - b\sin \left( {{{at} \over b}} \right)$$
$$y = a\cos t - b\cos \left( {{{at} \over b}} \right),$$ both $$a,b > 0$$ is
A
$$a-b$$
B
$$a+b$$
C
$$\sqrt {{a^2} + {b^2}}$$
D
$$\sqrt {{a^2} - {b^2}}$$

## Explanation

Distance of origin from $$\left( {x,y} \right) = \sqrt {{x^2} + {y^2}}$$

$$= \sqrt {{a^2} + {b^2} - 2ab\cos \left( {t - {{at} \over b}} \right)} ;$$

$$\le \sqrt {{a^2} + {b^2} + 2ab}$$ $$\left[ {{{\left\{ {\cos \left( {t - {{at} \over b}} \right)} \right\}}_{\min }} = - 1} \right]$$

$$=a+b$$

$$\therefore$$ Maximum distance from origin $$=a+b$$

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