Differential Calculus jee mains questions | JEE Main Maths Quadratic Equations Previous Year Question

Q 41 :

If $x = a (t - \frac{1}{t}), y = b (t + \frac{1}{t}),$ then $\frac{d y}{d x} =$

$\frac{x}{y}$
$\frac{b^{2} x}{a^{2} y}$
$\frac{b x}{a y}$
$\frac{a^{2} y}{b^{2} x}$

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(2)

Ans. $\frac{b^{2} x}{a^{2} y}$

Explanation:

We have

$x = a (t - \frac{1}{t})$

$y = b (t + \frac{1}{t})$

On differentiating both equations with respect to $t$ , we get

$\frac{d x}{d t} = a [\frac{d t}{d t} - \frac{d (t^{- 1})}{d t}]$

$\frac{d y}{d t} = b [\frac{d t}{d t} + \frac{d (t^{- 1})}{d t}]$

$\frac{d x}{d t} = a [1 + \frac{1}{t^{2}}]$ $\dots (i)$

$\frac{d y}{d t} = b [1 - \frac{1}{t^{2}}]$ $\dots (i i)$

On dividing equation (ii) by equation (i), we get

$\frac{d y}{d x} = \frac{\frac{d y}{d t}}{\frac{d x}{d t}} = \frac{b [\frac{t^{2} - 1}{t^{2}}]}{a [\frac{t^{2} + 1}{t^{2}}]}$

$\Rightarrow$ $\frac{d y}{d x} = \frac{b (t^{2} - 1)}{a (t^{2} + 1)} = \frac{b (t - \frac{1}{t})}{a (t + \frac{1}{t})}$

$\Rightarrow$ $\frac{d y}{d x} = \frac{b (\frac{x}{a})}{a (\frac{y}{b})} = \frac{b^{2} x}{a^{2} y}$

Q 42 :

Which of the following statements are correct?

(A) If $f : R \to R$ , then $f (x) =$ $| x |$ is continuous everywhere.

(B) If $f : R \to R$ , then $f (x) =$ $| x |$ is continuous everywhere but not differentiable at $x = 0$ .

(C) Let $f : R - {0} \to R,$ then $f (x) = \frac{1}{x}$ is continuous everywhere.

(D) Let $f : R \to R$ , then $f (x) =$ $| x - 1 |$ $+$ $| x - 2 |$ is continuous everywhere but not differentiable at exactly two points.

(E) If $f : R \to R$ , then $f (x) = \cot x$ is continuous everywhere.

Choose the correct answer from the options given below:

(A) only
(A), (C) only
(A), (B), (C), (D) only
(D), (E) only

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(3)

Ans. (A), (B), (C), (D) only

Explanation:

(A) - A modulus function is continuous everywhere.

(B) - For $f (x) =$ $| x |$ to be differentiable at $x = 0$ ,

$L.H.S. = \lim_{h \to 0} \frac{f (0 - h) - f (0)}{- h}$

$= \lim_{h \to 0} \frac{| 0 - h | - | 0 |}{- h}$

$= \lim_{h \to 0} \frac{h}{- h} = - 1$

$R.H.S. = \lim_{h \to 0} \frac{f (0 + h) - f (0)}{h}$

$= \lim_{h \to 0} \frac{| 0 + h | - | 0 |}{h}$

$= \lim_{h \to 0} \frac{h}{h} = 1$

$L.H.S. \neq R.H.S.$

So, $f (x) =$ $| x |$ is not differentiable at $x = 0$

(C) - $f (x) = \frac{1}{x}$ is continuous everywhere except at $x = 0$

(D) - Yes, and those points are $x = 1$ and $x = 2$

(E) - No, $\cot x$ is not continuous everywhere.

Q 43 :

If $f (x) =$ ${\begin{matrix} \frac{k \cos x}{π - 2 x}, & x \neq \frac{π}{2} \\ 3, & x = \frac{π}{2} \end{matrix}$ is continuous at $x = \frac{π}{2},$ then $k$ is:

6
4
3
2

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(1)

Ans. 6

Explanation:

If $f (x)$ is continuous at $x = \frac{π}{2}$

Then, $\frac{k \cos x}{π - 2 x} = 3$

$\Rightarrow$ $\frac{k \cos (\frac{π}{2})}{π - 2 (\frac{π}{2})} = 3$

Using L' Hospital rule,

$\frac{k (- \sin (\frac{π}{2}))}{- 2} = 3$

$\Rightarrow$ $\frac{k}{2} = 3$

$\Rightarrow$ $k = 6$

Q 44 :

The derivative of $\sec (\tan \sqrt{x})$ with respect to $x$ is:

$\frac{\sec (\tan \sqrt{x}) \tan (\tan \sqrt{x}) \sec^{2} \sqrt{x}}{2 \sqrt{x}}$
$\sec^{2} (\tan \sqrt{x})$
$\frac{\sec (\tan \sqrt{x}) \tan (\tan \sqrt{x}) \sec^{2} \sqrt{x}}{x}$
$\sec^{2} (\tan x^{1 / 3})$

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(1)

Ans. $\frac{s e c (t a n \sqrt{x}) t a n (t a n \sqrt{x}) s e c^{2} \sqrt{x}}{2 \sqrt{x}}$

Explanation:

Let $y = \sec (\tan \sqrt{x})$

So, $\frac{d y}{d x} = \frac{d [\sec (\tan \sqrt{x})]}{d (\tan \sqrt{x})} \times \frac{d (\tan \sqrt{x})}{d (\sqrt{x})} \times \frac{d (\sqrt{x})}{d x}$

$= \sec (\tan \sqrt{x}) \cdot \tan (\tan \sqrt{x}) \times \sec^{2} \sqrt{x} \times \frac{1}{2 \sqrt{x}}$

$= \frac{\sec (\tan \sqrt{x}) \tan (\tan \sqrt{x}) \sec^{2} \sqrt{x}}{2 \sqrt{x}}$

Topic Question Set

Q 41 :

If $x = a (t - \frac{1}{t}), y = b (t + \frac{1}{t}),$ then $\frac{d y}{d x} =$

Q 43 :

If $f (x) =$ ${\begin{matrix} \frac{k \cos x}{π - 2 x}, & x \neq \frac{π}{2} \\ 3, & x = \frac{π}{2} \end{matrix}$ is continuous at $x = \frac{π}{2},$ then $k$ is:

Q 44 :

The derivative of $\sec (\tan \sqrt{x})$ with respect to $x$ is:

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Topic Question Set

Q 41 : If x=a(t-1t), y=b(t+1t), then dydx=

Q 43 : If f(x)={kcosxπ-2x,x≠π23,x=π2 is continuous at x=π2, then k is:

Q 44 : The derivative of sec(tanx) with respect to x is:

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Q 41 :

If $x = a (t - \frac{1}{t}), y = b (t + \frac{1}{t}),$ then $\frac{d y}{d x} =$

Q 43 :

If $f (x) =$ ${\begin{matrix} \frac{k \cos x}{π - 2 x}, & x \neq \frac{π}{2} \\ 3, & x = \frac{π}{2} \end{matrix}$ is continuous at $x = \frac{π}{2},$ then $k$ is:

Q 44 :

The derivative of $\sec (\tan \sqrt{x})$ with respect to $x$ is: