NEW
New Website Launch
Experience the best way to solve previous year questions with mock tests (very detailed analysis), bookmark your favourite questions, practice etc...
1

WB JEE 2020

MCQ (More than One Correct Answer)
English
Bengali
Let $$y = {{{x^2}} \over {{{(x + 1)}^2}(x + 2)}}$$. Then $${{{d^2}y} \over {d{x^2}}}$$ is
A
$$2\left[ {{3 \over {{{(x + 1)}^4}}} - {3 \over {{{(x + 1)}^3}}} + {4 \over {{{(x + 2)}^3}}}} \right]$$
B
$$3\left[ {{2 \over {{{(x + 1)}^3}}} + {4 \over {{{(x + 1)}^2}}} - {5 \over {{{(x + 2)}^3}}}} \right]$$
C
$${6 \over {{{(x + 1)}^3}}} - {4 \over {{{(x + 1)}^2}}} + {3 \over {{{(x + 1)}^3}}}$$
D
$${7 \over {{{(x + 1)}^3}}} - {3 \over {{{(x + 1)}^2}}} + {2 \over {{{(x + 1)}^3}}}$$

Explanation

Given, $$y = {{{x^2}} \over {{{(x + 1)}^2}(x + 2)}}$$

By using partial fraction,

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

$$ \therefore $$ $${4 \over {x + 2}} - {3 \over {x + 1}} + {1 \over {{{(x + 1)}^2}}}$$

On differentiating both sides w.r.t. x, we get

$${{dy} \over {dx}} = {{ - 4} \over {{{(x + 2)}^2}}} + {3 \over {{{(x + 1)}^2}}} - {2 \over {{{(x + 1)}^3}}}$$

Again differentiating w.r.t. x

$${{{d^2}y} \over {d{x^2}}} = {8 \over {{{(x + 2)}^3}}} - {6 \over {{{(x + 1)}^3}}} + {6 \over {{{(x + 1)}^4}}}$$

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

মনে করো $$y = {{{x^2}} \over {{{(x + 1)}^2}(x + 2)}}$$। সেক্ষেত্রে $${{{d^2}y} \over {d{x^2}}}$$ হবে

A
$$2\left[ {{3 \over {{{(x + 1)}^4}}} - {3 \over {{{(x + 1)}^2}}} + {4 \over {{{(x + 2)}^3}}}} \right]$$
B
$$3\left[ {{2 \over {{{(x + 1)}^4}}} + {4 \over {{{(x + 1)}^2}}} - {5 \over {{{(x + 2)}^3}}}} \right]$$
C
$${6 \over {{{(x + 1)}^3}}} - {4 \over {{{(x + 1)}^2}}} + {3 \over {{{(x + 1)}^3}}}$$
D
$${7 \over {{{(x + 1)}^3}}} - {3 \over {{{(x + 1)}^2}}} + {2 \over {{{(x + 1)}^3}}}$$

Explanation

$${{{x^2}} \over {{{(x + 1)}^2}(x + 2)}} = {A \over {x + 1}} + {B \over {{{(x + 1)}^2}}} + {C \over {x + 2}}$$

$$\therefore$$ $${x^2} = A(x + 1)(x + 2) + B(x + 2) + C{(x + 1)^2}$$

$$x = - 1$$ বসিয়ে

$$\therefore$$ $$1 = B$$

$$x = - 2$$ বসিয়ে, $$4 = C$$

x2-এর সহগ তুলনা করে পাই

$$1 = A + C \Rightarrow A = 1 - C = - 3$$

$$\therefore$$ $$y = {{ - 3} \over {x + 1}} + {1 \over {{{(x + 1)}^2}}} + {4 \over {x + 2}}$$

$$\therefore$$ $${{dy} \over {dx}} = {3 \over {{{(x + 1)}^2}}} - {2 \over {{{(x + 1)}^3}}} - {4 \over {{{(x + 2)}^2}}}$$

$${{{d^2}y} \over {d{x^2}}} = {{ - 6} \over {{{(x + 1)}^3}}} + {6 \over {{{(x + 1)}^4}}} + {8 \over {{{(x + 2)}^3}}}$$

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

2

WB JEE 2019

MCQ (More than One Correct Answer)
English
Bengali
Let f and g be differentiable on the interval I and let a, b $$ \in $$ I, a < b. Then,
A
If f(a) = 0 = f(b), the equation f'(x) + f(x)g'(x) = 0 is soluble in (a, b)
B
If f(a) = 0 = f(b), the equation f'(x) + f(x)g'(x) = 0 may not be soluble in (a, b)
C
If g(a) = 0 = g(b), the equation g'(x) + kg(x) = 0 is soluble in (a, b), k $$ \in $$ R
D
If g(a) = 0 = g(b), the equation g'(x) + kg(x) = 0 may not be soluble in (a, b), k $$ \in $$ R

Explanation

From option (a)

We have,

$$f(a) = f(b) = 0$$

$$ \Rightarrow f'(a)\,.\,f'(b) < 0$$

Again, let $$h(x) = f'(x) + f(x)g'(x)$$

$$ \Rightarrow h(a) = f'(a) + f(a)g'(a) = f'(a)$$

and $$h(b) = f'(b) + f(b)g'(b) = f'(b)$$

$$ \therefore $$ $$h(a)\,.\,h(b) = f'(a)\,.\,f'(b) < 0$$

$$ \therefore $$ h(x) = 0 has root between (a, b)

From option (c)

We have, $$g(a) = g(b) = 0$$

$$ \Rightarrow g'(a)\,.\,g'(b) < 0$$

Again, let $$m(x) = g'(x) + kg(x)$$

$$ \Rightarrow m(a) = g'(a) + kg(a) = g'(a)$$

$$ \Rightarrow m(b) = g'(b) + kg(b) = g'(b)$$

$$ \therefore $$ $$m(a)\,.\,m(b) = g'(a)\,.\,g'(b) < 0$$

$$ \therefore $$ m(x) = 0 has root between (a, b).

So, option (a) and (c) are correct.

মনে করো f ও g অন্তরাল I তে অন্তরকলনযোগ্য এবং a, b $$\in$$ I, a < b । সেক্ষেত্রে -

A
যদি f (a) = 0 = f (b) হয়, তবে f' (x) + f (x) g' (x) = 0 সমীকরণটি (a, b) তে সমাধানযোগ্য
B
যদি f (a) = 0 = f (b) হয়, তবে f' (x) + f (x) g' (x) = 0 সমীকরণটি (a, b) তে সমাধানযোগ্য নাও হতে পারে
C
যদি g(a) = 0 = g(b) হয়, তবে g'(x) + kg(x) = 0 সমীকরণটি (a, b) তে সমাধানযোগ্য, k $$\in$$ R
D
যদি g(a) = 0 = g(b) হয়, তবে g'(x) + kg(x) = 0 সমীকরণটি (a, b) তে সমাধানযোগ্য নাও হতে পারে k $$\in$$ R
3

WB JEE 2017

MCQ (More than One Correct Answer)
English
Bengali
If f(x) = xn, being a non-negative integer, then the values of n for which f'($$\alpha$$ + $$\beta$$) = f'($$\alpha$$) + f'($$\beta$$) for all $$\alpha$$, $$\beta$$ > 0 is
A
1
B
2
C
0
D
5

Explanation

We have,

f(x) = xn

$$ \Rightarrow $$ f'(x) = nxn$$-$$1

Now, f'($$\alpha$$ + $$\beta$$) = f'($$\alpha$$) + f'($$\beta$$)

$$ \Rightarrow $$ n($$\alpha$$ + $$\beta$$)n$$-$$1 = n$$\alpha$$n$$-$$1 + n$$\beta$$n$$-$$1

$$ \Rightarrow $$ ($$\alpha$$ + $$\beta$$)n$$-$$1 = $$\alpha$$n$$-$$1 + $$\beta$$n$$-$$1

From options we see that n = 2, satisfy the above equation.

$$ \therefore $$ n = 2

যদি $$f(x) = {x^n}$$ হয় যেখানে $$n$$-অঋণাত্মক পূর্ণসংখ্যা, তবে সব $$\alpha ,\beta > 0$$-এর জন্য $$f'(\alpha + \beta ) = f'(\alpha ) + f'(\beta )$$ হলে

A
$$1$$
B
$$2$$
C
$$0$$
D
$$5$$

Explanation

$$n = 1$$ হলে $$f(x) = x \Rightarrow f'(x) = 1$$

$$f'(\alpha + \beta ) = 1$$ এবং $$f'(\alpha ) + f'(\beta ) = 1 + 1 = 2$$

$$\therefore$$ $$n \ne 1$$

$$n = 2$$ হলে $$f(x) = {x^2}$$

$$\therefore$$ $$f'(x) = 2x$$

$$f'(\alpha + \beta ) =2(\alpha+\beta)= f'(\alpha ) + f'(\beta )$$

$$\therefore$$ $$n = 2$$ সম্ভব।

$$n = 0$$ হলে $$f(1) = 1 \Rightarrow f'(x) = 0$$

$$f'(\alpha + \beta ) = f'(\alpha ) + f'(\beta ) = 0$$

$$\therefore$$ $$n = 0$$ হতে পারে

$$n = 5$$ হলে $$f(x) = {x^5} \Rightarrow f'(x) = 5{x^4}$$

$$f'(\alpha + \beta ) = 5{(\alpha + \beta )^4}$$

$$f'(\alpha ) + f'(\beta ) = 5({\alpha ^4} + {\beta ^4})$$

$$\therefore$$ $$n \ne 5$$

Joint Entrance Examination

JEE Main JEE Advanced WB JEE

Graduate Aptitude Test in Engineering

GATE CSE GATE ECE GATE EE GATE ME GATE CE GATE PI GATE IN

Medical

NEET

CBSE

Class 12